• ATOMIC STRUCTURE • STOICHIOMETRY-I • PERIODICITY IN PROPERTIES
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THEORY AND EXERCISE BOOKLET CONTENTS S.NO.
TOPIC
PAGE NO.
ATOMIC STRUCTURE
THEORY WITH SOLVED EXAMPLES ............................................................. 5 – 40
CLASS ROOM PROBLEMS ........................................................................... 41 – 43
EXERCISE - I (JEE Main) ................................................................................ 44 – 50
EXERCISE - II (JEE Advanced – Objective) .................................................... 51 – 60
EXERCISE - III (JEE Advanced) ...................................................................... 61 – 65
EXERCISE - IV (JEE Advanced – Previous Years)................ ......................... 66 – 75
ANSWER KEY................................................................................................. 76 – 77
STOICHIOMETRY-I
THEORY WITH SOLVED EXAMPLES ........................................................... 78 – 100
CLASS ROOM PROBLEMS ......................................................................... 101 – 105
EXERCISE - I (JEE Main) .............................................................................. 106 – 112
EXERCISE - II (JEE Advanced – Objective) .................................................. 113 – 117
EXERCISE - III (JEE Advanced) .................................................................... 118 – 123
EXERCISE - IV (JEE Main & JEE Advanced – Previous Years) .................... 124 – 125
ANSWER KEY............................................................................................... 126 – 127
PERIODICITY IN PROPERTIES
THEORY WITH SOLVED EXAMPLES .......................................................... 128 – 155
EXERCISE - I (JEE Main) .............................................................................. 156 – 161
EXERCISE - II (JEE Advanced – Objective) .................................................. 162 – 164
EXERCISE - III (JEE Advanced) .................................................................... 165 – 169
EXERCISE - IV (JEE Main & JEE Advanced – Previous Years) .................... 170 – 174
ANSWER KEY............................................................................................... 175 – 176
Chemistry ( Booklet-1 )
Page # 4
JEE SYLLABUS
• ATOMIC STRUCTURE . JEE - ADVANCED Atomic structure and chemical bonding: Bohr model, spectrum of hydrogen atom, quantum numbers; Wave-particle duality, de Broglie hypothesis; Uncertainty principle; Qualitative quantum mechanical picture of hydrogen atom, shapes of s, p and d orbitals; Electronic configurations of elements (up to atomic number 36); Aufbau principle; Pauli’s exclusion principle and Hund’s rule; • STOICHIOMETRY-I JEE - ADVANCED Mole concept; Chemical formulae; Balanced chemical equations; Calculations (based on mole concept) involving common oxidation-reduction, neutralisation, and displacement reactions; Concentration in terms of mole fraction, molarity, molality and normality. • PERIODICITY IN PROPERTIES JEE - ADVANCED Modem periodic law and present form of the periodic table, s, p, d and f block elements, periodic trends in properties of elements atomic and ionic radii, ionization enthalpy, electron gain enthalpy, valence, oxidation states and chemical reactivity.
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ATOMIC STRUCTURE
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ATOMIC STRUCTURE Different models of atom (1)
Dalton's atomic model. Dalton’s atomic model is one of the fundamentals of physics and chemistry. This theory of atomic composition was hypothesized and partially confirmed by the English chemist and Physicist John Dalton. Dalton came with his Atomic theory as a result of his research into gases. He discovered that certain gases only could be combined in certain proportions even if two different compounds shared the same common element or group of elements. Through deductive reasoning and experimentation, he made an interesting discovery. His findings led him to hypothesize that elements combine at the atomic level in fixed ratios. This ratio would naturally differ in compounds due to the unique atomic weights of the elements being combined. This was a revolutionary idea but further experimentation by himself and others confirmed his theory. The findings became the basis of Dalton’s Atomic Laws or Model. These laws focus on five basic theorems.
(1)
Pure Elements consist of particles called atoms.
(2)
Atoms of an element are all the same for that element. That means gold is gold and oxygen is oxygen down to the last atom.
(3)
Atoms of different elements can be told apart by their atomic weights.
(4)
Atoms of elements unite to form chemical compounds.
(5)
Atoms can neither be created or destroyed in chemical reaction. The grouping only changes.
(2)
Thomson’s model or Plum Pudding model or Raisin Pudding model or water melon model. Thomson stated that atom is uniform sphere of positive charge with negatively charged electron embedded in it in such a way that atom becomes electrically neutral.
•
Important features : It shows electrical neutrality of atom. The mass of the atom is assumed to be uniformly distributed over the atom. The model was rejected as it was not consistent with the results of alpha scattering experiment.
(3)
Rutherford’s Nuclear Model of Atom : -particle scattering experiment :-
A beam of -particle was projected towards this gold foil. The foil was surrounded by circular fluorescent ZnS screen which produces flash when -particle collide with it. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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ATOMIC STRUCTURE
Observations of experiment : Most of -particles passed through the foil un deflected or deflected by very small angles [less than 1º]. Few particles were deflected through angles as large as 90º. Very few particles [ 1 in 20 thousand] bounce back or rejected through an angle almost equal to 180º. Angle of deflection increases with increase in atomic number.
•
Conclusions Since, most of the -particles passed through the foil un deflected, therefore there must be large empty space present in the atom. -particles are positively charged and having considerable mass could be deflected by only some heavy positively charged centre by repulsion.[Rutherford named this positively charged heavy centre as nucleus] Since, very few -particles were thrown back therefore the size of nucleus should be very small and is should be rigid because -particle can recoil back only if it undergo direct collision with the heavy positively charged centre.
•
Rutherford’s model On the basis of scattering experiment, Rutherford proposed an atomic model or nuclear atomic model or
nuclear model of atoms.
The electrons are revolving around the nucleus in circular path and electrostatics force of attraction between the electrons and nucleus is balanced by the centrifugal force acting on revolving electron. The -ve charge on electrons are equal to the total positive charge on the nucleus and therefore as a whole atom is electrically neutral. Size of the nucleus is extremely small as compared to size of the atom. Radius of nucleus Radius of nucleus is of the order of 10–15 m and radius of atom is order of 10–10 m. r = r0A1/3 A = Mass number = no. of protons + no. of neutrons. r0 = constant = 1.2 × 10–13 cm.
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ATOMIC STRUCTURE •
Page # 7
Distance of closest approach : (3)
rmin
(2)
(1)
r
If the -particle is projected with some velocity -v from very large distance then at point (1), energy of the -particle E
1 mv 2 O 2 (K.E.) (P.E.)
As the particle is moving closer to the nucleus of other particle, repulsive coulombic force increase (P.E. increases) and velocity of the particle decrease kinetic energy is converted into P.E. At point (2) E = K.E. + P.E. =
kq q 1 mv '2 1 2 2 r
[As no ext. force is acting, Total energy of system remains constant] velocity becomes zero at rmin distance where columbic potential energy of repulsion becomes equal to the initial kinetic energy. At point (3), E P.E. 0
k q1 q2 rmin
1 k q1 q2 mv 2 2 rmin
rmin
2 k q1 q2 2
mv
2k z1 z2e2 mv2
q1 = z1 e ; q2 = z2 e e = electronic charge = 1.6 × 10–19 C z1 = z2 = Atomic number •
Failures of Rutherford’s Model :
(1)
According to maxwell’s wave theory, an accelerated charged particles (like electron) revolving in the field of another charged particle like nucleus looses energy in the form of electromagnetic radiation and if this happen then orbit of the revolving electron should keep on decreasing and ultimately electron should fall into the nucleus and atom should collapse but this dosen’t happen actually.
(2)
Rutherford’s model didn’t give any idea about electronic arrangement of an atom and since he didn’t mention specific energy levels he couldn’t explain the discontinuous atomic hydrogen spectrum. WAVES AND ITS CHARACTERISTICS It is a periodical disturbance causing transfer of energy without transfer of matter.
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ATOMIC STRUCTURE
Characteristics of waves : Amplitude : Maximum displacement from mean position it remains constant with distance except stationary or standing waves. Wavelength : It is the distance between two adjacent crest or troughs. Frequency (or ) : The no of waves passing through a point in 1second, unit-sec–1 or Hz Wave number ( or ) : No. of waves present in unit distance. 1 metre 1,cm 1 etc. Velocity : linear distance travelled by wave in one second.
v NATURE OF LIGHT •
Maxwell electromagnetic wave theory (wave nature of light) : An accelerated electrically charged particle produces and transmits electrical and magnetic field. These are transmitted in the form of waves known as electromagnetic waves or electromagnetic radiations. He stated that light also possess electrical and magnetic field and it is also known as electromagnetic radiations or e.m.w.
•
Characteristics of Electromagnetic Radiations : 1. In these electromagnetic radiation electrical and magnetic field oscillates er to each other and it also propagates er to both field. 2. All these electromagnetic radiation do not require any medium and can travel in vacuum. 3. Velocity of all electromagnetic radiation is 3 × 108 m/s in vacuum. 4. Energy of an electromagnetic wave is directly proportional to intensity and it is independent of frequency. 5. There are also showing diffraction and interference And therefore. Maxwell concluded light to be wave nature. But Maxwell theory couldn’t explain the results of photoelectric effect and black body radiations.
Photon Energies for EM Spectrum
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ATOMIC STRUCTURE
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A Blackbody A blackbody is an object that emits a well defined spectrum of radiation solely based on its temperature. We see from figure at right that the hotter the blackbody, the more intense it is, and the shorter the peak wavelength. The picture does not say anything about what the object is made of, or how heavy it is, etc. It doesn't matter! The only property that determines the spectrum of a blackbody is its temperature. Brick, iron or a dense gas will emit the same spectrum as long as they are at the same temperature. That spectrum will have a peak that lies at a particular wavelength.
•
Planck’s Quantum Theory : [Particle nature of light] : He stated that a body radiates energy in the form of discontinuous energy packets or bundles. Each bundle of energy is known as quantum and quantum of light is known as photons. Energy of each quantum is directly proportions to frequency of radiation.
E E = h
h = 6.62 × 10–34 Js. Plank’s constant
Total energy absorbed or emitted by a body will be whole no. integral multiple of energy of quantum Eabs or Eemitted = nh Ex.
Calculate the no. of photons emitted by 60 watt bulb in 10 hrs. When light of wavelength 6000 Å is emitted by it.
Sol.
E
nhc = 6.5 × 1024 : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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ATOMIC STRUCTURE
Energies in Electron Volts Room temperature thermal energy of a molecule..................................0.04 eV Visible light photons......................................................................1.5-3.5 eV Energy for the dissociation of an NaCl molecule into Na + and Cl – ions:.......4.2 eV Ionization energy of atomic hydrogen ..............................................13.6 eV Approximate energy of an electron striking a color television screen (CRT display) ....................20,000 eV High energy diagnostic medical x-ray photons..................200,000 eV (=0.2 MeV) Typical energies from nuclear decay: (1) gamma..................................................................................0-3 MeV (2) beta.....................................................................................0-3 MeV (3) alpha....................................................................................2-10 MeV Cosmic ray energies ...........................................................1 MeV - 1000 TeV 1 MeV = 10 6 eV, 1 GeV = 10 9 eV, 1 TeV = 10 12 eV •
Explanation of black body radiations using Planck’s quantum theory : When a solid substance like iron piece is heated it emits radiations. As heating is continued, more and more energy is being absorbed by the atom and hence, more energy will be emitted and therefore energy of e.m.w. increases and frequency of e.m.w. increases and therefore body first becomes red then yellow and finally white. Therefore, it can be concluded that light posses particle nature and energy of electromagnetic radiation depends upon frequency.
•
Explanation of Photo electric Effect using Planck’s Quantum Theory : When a metal sheet is subjected to electromagnetic radiation of suitable frequency then some electrons are ejected from metal surface and these electrons are known as photoelectron and the effect is known as photoelectric effect. If electromagnetic radiation of low frequency are used then there is no ejection of electron inspite of continuous increasing intensity. This observation was contradicting to maxwell theory according to which energy electromagnetic radiation I but can be explained using Planck’s quantum theory i.e. E .
DUAL NATURE OF LIGHT Since, wave nature of light explains diffraction interference phenomenon while particle nature explains black body radiation and photo electric effect light was considered to have dual nature particle nature as well as wave nature.
•
Electromagnetic spectrum : The arrangement of all the electromagnetic radiation in a definite order (decreasing or increasing of wavelength or frequency is known as electron magnetic spectrum SPECTRUM When light coming from a source is passed through a prism, the radiation of different wavelength deviated through different angles and get separated. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
ATOMIC STRUCTURE 1
VIB .
.... .
Angle of deviation
Page # 11
this process is called dispersion.
And such a collection or dispersed light giving its wavelength composition is known as spectrum. •
Spectrum are of two types.
Continuous
(1)
(2)
Emission
Absorption :
Discontinuous
EMISSION SPECTRUM This is the spectrum of radiations emitted by any source or atom or molecule of any substance (Which is excited by heating or electric discharge).
(i)
Continuous Spectrum : When white light from a source is dispersed, a bright spectrum continuously distributed on the dark background is obtained. The colours are continuously changing from violet to red and there is no Sharp boundaries between various colour. These colours appear to be merge into each other and therefore spectrum is known as continuous spectrum.
(ii)
Discontinuous Spectrum : When a atom is subjected to electromagnetic radiation, it cause excitation of electron to higher : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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ATOMIC STRUCTURE
energy level. When electrons return back to lower energy level, it emits certain radiations corresponding to difference in energy level and therefore spectrum obtained in case of atom is discontinuous spectrum having specific wavelength is known as Atomic spectrum or line spectrum.
•
Absorption Spectrum : When an atom is subjected to white light it absorbs some specific radiation corresponding to the difference in energy level the remaining radiations (transmitted radiations) are devoid of certain specific frequencies which is observed in the form of missing line and this spectrum is known as absorption spectrum. It is photographic negative of emission spectrum i.e. those bright lines which are present in emission spectrum of an atom are missing. In the absorption spectrum and observed in the form of dark lines.
(4)
Bohr's Model It was the first model based on Planck's quantum theory and the model explained stability of atom and line spectrum of hydrogen.
•
Postulates of Bohr's model
1.
An atom consist of centrally located small, dense, positively charged nucleus and electrons are revolving around the nucleus in circular paths known as circular orbits and coulombic force of attraction between nucleus and electron is balanced by centrifugal force of the revolving electron.
2.
Out of the infinite circular orbits only those circular orbits are possible in which angular momentum of electron is integral multiple of h/2 i.e. angular momentum of an electron can have fixed values like h 2h 3h , , etc. i.e., angular momentum of electron is quantised. 2 2 2 mvr
nh 2
m and v are mass and velocities of electron respectively and r is radius of orbit and n is integer which is later related with orbit number and shell number and h is Plank's constant. 3.
The energy of these circular orbits have fixed values and hence electron in an atom can have only certain values of energy. It is characteristic of an orbit and it cannot have any orbit value of its own. And therefore energy of an electron is also quantised.
4.
As long as electrons remains in these fixed orbits, it dosen't lose energy i.e. energy of an electron is stationary (not changing with time) and therefore these orbits are known as allowed energy levels or stationary states and this explains the stability of atom.
5.
The energy levels are designated as K, L, M, N and numbered as 1, 2, 3, 4 etc from nucleus outwards and as the distance of the shell's or energy level from the nucleus increases the energy of the energy level also increases i.e., EN > EM > EL > Ek
6.
The emission or absorption of energy in the form of radiations can only occur when an electron jump from one stationary states to other. E = Ehigher – Elower Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
ATOMIC STRUCTURE
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E = h where h is the energy of absorbed photon or emitted photon which corresponds to the difference in energy levels. Energy is absorbed when electron jumps from lower energy level (normal state) to higher energy level (exited, unstable state) and energy is emitted when electrons jumps from higher energy level to lower energy level. •
Bohr's model is applicable for a one electron species only, like H, He+, Li+2, Be+3 etc.
•
Derivation of Radius of different orbits in one electron species (using Bohr's model) : nh 2
mvr
q1 = e,
....(1) q2 = Ze
mv 2 kq1q2 kze 2 = r r2 r2
mn 2h2 4 2m 2r 2 .r
r
r
....(2)
kZe 2 r2
mn 2h 2 4 2m 2kZe 2 n 2h 2 4 2mkZe 2
(6.625 10–34 )2
2
4 9.1 10
r = 0.529 ×
r
–31
9
–19 2
9 10 (1.6 10
)
n2 Z
n2 A z
n2 z
for a particular atom
r n2
Radius of 1st orbit of H atom r = 0.529 Å Ex.
Calculate ratio of radius of 1st orbit of H atom to Li+2 ion :
Sol.
=
n 2 z1 4 = z 2 = n1 3 atom
Radius of 2nd orbit of H atom Radius of 3rd orbit of Li2
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ATOMIC STRUCTURE
Derivation of Velocity of electron in Bohr's orbit : v
nh , putting value of r.. 2mr
v=
nh 42mkZe2 2mn2h2
v=
2kZe2 2ke 2 Z = h.n. h n
v = 2.18 × 106
v
•
Z m/s n
Z n
Derivation of total energy of electron / system : T.E. of system = K.E. of e– + P.E. of system (nucleus and e–) kinetic energy of electron =
PE =
T.E
1 1 kze2 mv 2 = 2 2 r
kze2 r
– KZe 2 2r
T.E. = – 13.6 ×
z2 n2
eV/atom
T.E. = – 2.18 × 10–18
z2
J/atom n2 As shell no. or distance increases, the value of T.E. and P.E. increases (however magnitude decreases) and becomes maximum at infinity i.e., zero. negative sign indicates that electron is under the influence of attractive forces of nucleus. K.E. = –
P.E. 2
P.E 2 T.E = – K.E. T.E.
Calculation of energy of energy level in H atom (i)
When n = 1 (ground level) K.E. = 13.6 eV (atom) P.E. = –27.2 eV / atom T.E. = –13.6 eV/atom Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
ATOMIC STRUCTURE (ii)
Page # 15
When n = 2/2nd energy level / 1st excited state K.E. =
13.6 eV/atom = 3.4 eV / atom 4
P.E. = – 6.8 eV/atom T.E. = – 3.4 eV/atom E2 – E1 = –3.4 + 13.6 eV/atom = 10.2 eV/atom
(iii)
When n = 3 / 3rd energy level/2nd excited state.
K.E. =
13.6 eV/atom = 1.51 eV/atom 9
P.E. = –3.02 eV/atom T.E. = –1.51 eV/atom –E3 – E2 = –1.51 eV/atom + 3.4 eV/atom = 1.89 eV atom.
(iv)
when n = 4 / 3rd excited state K.E. =
13.6 4
2
13.6 16 10
P.E. = –1.70 eV/atom
= 0.85 eV/atom
T.E. = –0.85 eV/atom
E4 – E3 = 0.66 As distance increases (n increases) energy of the energy level increases but energy difference between consecutive energy level keeps on increasing i.e., maximum energy difference between 2 to 1 (consecutive) If reference value (P.E at ) is assigned value other than zero. – (i) All K.E. data remains same. (ii) P.E./T.E. of each shell will be changed however difference in P.E./T.E. between 2 shell will remain unchanged. •
Hydrogen spectrum (Before Bohr's Model) The spectrum of H atom is observed as discontinue line spectra. The line spectra of different in 3 region UV, visible and I R. The different lines observed H spectrum were classified in to different series and named after their discoverers. Rydberg gave an empirical formula to calculate wavelength, which is applicable to all series. 1 1 1 RH 2 – 2 n1 n 2
RH = 109677 cm–1 Rydberg's constant.
n1 and n2 are integers.
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Page # 16 Explanation of Hydrogen spectrum using Bohr's model
When electron in an excited atom comes back from higher energy level (n2) to lower ever level (n1) then it emits a photon, having energy equal to difference in energy levels. h = E = En 2 – En1
n2 h =
– 22mk 2 e 4 z 2 n22h 2
22mk 2 e 4 z 2 – – n12h2
n1
2mk 2e4 z2 1 1 2 2 h = 2 h n1 n2
1 2mk 2 e 4 z 2 h3 C
1 1 2 – 2 n1 n 2
1 1 1 RH 2 – 2 n1 n 2 The theoretical value is very closed to observed value Bohr's model provides theoretical explanation of H spectrum. Wavelength or wave no. of any line of any one electron species can be calculated as 1 = RHZ2
1 1 2 – 2, n1 n 2
124 0 E (ev )
hc E
nm DIFFERENT SERIES IN HYDROGEN SPECTRUM. MER BAL
PASCHEN BR AC KE TT PF UN D HR MP HU
LY MA N
•
ATOMIC STRUCTURE
EY
0
–3.4 –13.6
1
2 3
4
5
6
7
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LYMAN SERIES :
nf = 1,
ni = 2, 3, 4, 5, - - - - - - - ULTRAVIOLET
BALMER SERIES :
nf = 2,
ni = 3, 4, 5, 6, - - - - - - - VISIBLE
PASCHEN SERIES :
nf = 3,
ni = 4, 5, 6, 7, - - - - - - - INFRARED
BRACKETT SERIES :
nf = 4,
ni = 5, 6, 7, - - - - - - - INFRARED
PFUND SERIES :
nf = 5,
ni = 6, 7, - - - - - - - FARINFRARED
HUMPHREY SERIES :
nf = 6,
ni = 7, 8, - - - - - - - FARINFRARED
Ex.
Calculation of different types of maximum possible radiations which can be observed in (from higher energy level to lower energy level)
Sol.
n2 = n,
n1 = 1
Total no. of radiation obtained having difference wave lengths = Transition ending at 1st level + Transite ending at 2nd level .................................................. Transition ending at (n – 1) energy levels. = Total no. of radiation in lyman series + total no. of radiation in Balmer series ...................+ Total no. of radiation in (n – 1) series. = (n – 1) + (n – 2) + (n – 3) ........................ + 2 + 1
(n – 1)(n – 1 1) 2
n(n – 1) 2
(n1 – n1 ) n2 – (n1 1) ............................ +2 + 1 Total no. of different radiation
(n2 – n1 )(n2 – n1 1) N(N 1) = 2 2
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ATOMIC STRUCTURE
Ex.
The energy level of an atom for 1st, 2nd and third levels are E, 4E, 2E respectively. If photon of w avelengt h is emitted for a transition 3 to 1. Calculate the wavelength for transition 2 to 1 in terms of .
Sol.
2E – E = E =
Ex.
Let 1 be the frequency of series limit of lyman series, 2 be the frequency of 1st line of lyman series, 3 be the frequency of series limit of Balmer series. Then find relation between 1, 2 and 3
hc 1
4E E hc –E 3 3 2
3 = 2
E1
E2
Sol.
2
E3
E1 = E2 + E3 h1 h 2 h 3 1
Ex.
Hydrogen like species is observed to emit 6 wavelengths originating from all possible transitions between a group of levels these energy levels have energy between – 0.85 eV and –0.544 eV (including both these energy levels) calculate
(i)
The quantum number of levels between which transition is taking place.
(ii)
Find the atomic number of species.
Sol.
(n2 – n1) (n2 – n1 + 1) = 12 n2 – n1 = x x(x + 1) = 12 n2 = n1 + 3 – 13.6
– 13.6
z2
– 13.6
n12 z2 (n1 3)2
z2 n12
–0.85
n1 = 12 ; n2 = 15
•
Failures of Bohr’s Model :
(i)
It is applicable only for one electron species.
(ii)
He couldn’t explain the fine spectrum of hydrogen when a powerful spectroscope is used then several lines which are very closely placed are observed in addition to the expected lines later this problem was solved by Sommerfeld who introduced the concept of sub shell.
(iii)
He couldn’t explain splitting of spectral lines into a group of finer lines under the influence of external magnetic field (Zeeman effect) or under the influence of external electric field (stark effect).
(iv)
He couldn’t give theoretical reason for most of his assumption, for example he couldn’t justify why angular momentum of electron should be quantised.
(v)
Spectrum of isotopes of hydrogen were expected to be same according to the Bohr’s model but found different experimentally. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
ATOMIC STRUCTURE
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(5)
Sommerfeld model of Atom
(a) (b) (c)
Three refinements of the Bohr model were worked out by Arnold. Sommerfeld and are jointly responsible for giving rise to small deviations in the allowed energies from the values En These refinements brought the calculated frequencies into even better agreement with observation (from four to five significant digits). The refinements were: the introduction of elliptical orbits; allowance for an orbiting motion of the nucleus; and the consideration of relativistic mass effects.
PHOTO ELECTRIC EFFECT Emission of electron from the metal surface when the light of suitable frequency is subjected to the metal surface. The effect is known as photoelectric effect and the ejected electrons are known as photoelectrons.
•
Terms used in photo electric effect
(1)
Work function (w) : It is the minimum amount of energy required to cause a photo emission from the metal surface. It is also known as threshold energy or Binding energy. [Work function depends upon ionisation energy and thereforew is minimum for alkali metals).
(2)
Threshold frequency ( 0) : The minimum value of frequency that can cause photo emissions. If < 0, then there is no photo emission. hc w = h 0 w 0
(3)
Threshold wavelength : (0) The maximum value of wavelength that can cause photo emission. If > 0, then photo emission is not possible.
(4)
Intensity (I) : Energy falling on metal surface of unit area of unit time I
E nh At At
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ATOMIC STRUCTURE
Photo intensity (IP) : It is number of photons falling per unit area per unit time. Ip
n At
Relation between I and Ip : I = Ip h Photo intensity is independent of frequency while intensity depends on frequency.
Power : Total energy radiated per unit time. P
E nh t t
EFFECT OF VARIATION OF FREQUENCY (a)
Effect of Photo Emission : I = Iph If frequency of subjected photon increases (intensity increases keeping photo intensity constant) then there is no change in no of ejected photo electrons as well as no change in photo current.
Photo current
No Photo current
0
(b) Effect on kinetic Energy : Average k. E. as well as k. E. max increases with increases in frequency. h – W = K Emax. K Emax = h – h 0
y = mx + c
hc K Emax = h – 0
Kmax = h – w
tan slope h K. Emax
W or
0
h0
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EFFECT OF VARIATION OF PHOTO INTENSITY On increasing intensity, keeping frequency constant (i.e. increasing photo intensity) no of ejected photo electrons increases as well as photo current increases. IP3 IP2 IP1
Photo current
IP3 > IP2 > IP1
0
•
Effect on Kinetic Energy Average K.E. and K.Emax remains constant with change in photo intensity > 0
K.Emax
IP or I (keeping constant)
•
Stopping Potential or Retarding Potential (V0) It is the minimum potential require to stop the fastest moving electrons completely or it is the minimum potential at which photo current becomes zero. eV0 = h – w tan slope h/e V0
eV0 = h – h0
V0
h 0 w or e e
0
h h 0 e e
It can be commented that stopping potential increase with increase in frequency how ever if photo intensity is changed there is no effect on stopping potential.
saturation current
>0
reference beam
V0
IP or I (keeping constant)
Ex.
In ultraviolet light of wavelength 200 nm is used in an experiment of photoelectric effect with lithium cathode (w = 2.5 eV). Then calculate (i) K.Emax
(ii) Stopping potential
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ATOMIC STRUCTURE
Sol.
K.Emax = h – w
hc 2 .5
1240 2 .5 2400
6.2 – 2.5
3.7 eV
Stopping Potential = 3.7 V DeBroglie’s hypothesis (dual nature of matter) : deBroglie proposed that like light, electron possess dual nature i.e. behave like both as material particle and as wave. The concept of dual character of metal evolved a wave mechanical theory of atom according to which electrons. protons and even atoms possess wave properties when moving.
Derivation of deBroglie Relation ship : For a photon, E
hc
...(i)
E mc 2 mc 2
...(ii)
hc
h h mc p
He concluded that as electromagnetic radiation have some associated mass or associated momentum, in the same way every moving particle of mass ‘m’ and velocity ‘v’ is associated with waves and these associated waves are known as matter waves or deBroglie’s waves. h / mv h / p Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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Experimentation verification of deBroglie’s hypothesis : Given by Davission and Germer. They observed that a beam of electrons are diffracted by nickel crystal just like the x-rays. More over it wavelength of the electrons determined by the diffraction experiments are equal to the values calculated from deBroglie’s relations ship. DERIVATION OF BOHR’S POSTULATE OF QUANTISATION OF ANGULAR MOMENTUM FROM DEBROGLIE’S EQUATIONSAccording to deBroglie moving electron is associated with waves must be completely in phase and there fore only those orbits are possible where circumference of the orbit is integral multiple of i.e. 2 r n , where n is the no. of waves
2r
nh mv
mvr
nh 2
From the above expression i t can be commented that no of waves in a shell waves = shell no.
•
Calculation of deBroglie wavelength if K.E. of the particle is E : E
1 mv 2 2
2E = mv2
2mE = m2 v2 = p2
p 2mE h dB
dB
2mE
h 2mE
If a charge particle at rest (having charge ‘q’) is accelerated by potential difference ‘V’ volt then h
dB
2mqv ,
dB
150 Å (Only for electron) V
Ex.
Calculate dB of electron have K.E. 3eV
Sol.
dB
150 50 Å 3
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ATOMIC STRUCTURE
The Uncertainty Principle
The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. There is a minimum for the product of the uncertainties of these two measurements. There is likewise a minimum for the product of the uncertainties of the energy and time. xp
h 4
Et
h 4
The uncertainty principle contains implications about the energy that would be required to contain a particle within a given volume. The energy required to contain particles comes from the fundamental forces, and in particular the electromagnetic force provides the attraction necessary to contain electrons within the atom, and the strong nuclear force provides the attraction necessary to contain particles within the nucleus. But Planck's constant, appearing in the uncertainty principle, determines the size of the confinement that can be produced by these forces. Another way of saying it is that the strengths of the nuclear and electromagnetic forces along with the constraint embodied in the value of Planck's constant determine the scales of the atom and the nucleus.
Q.
Does uncertainty principle violates energy conservation principle?
(6)
Wave mechanical model of Atom Schordinger equation : The Schrodinger equation plays the role of Newton's laws and conservation of energy in classical mechanics - i.e., it predicts the future behavior of a dynamic system. It is a wave equation in terms of the wave function which predicts analytically and precisely the probability of events or outcome. The detailed outcome is not strictly determined, but given a large number of events, the Schrodinger equation will predict the distribution of results. H = E The kinetic and potential energies are transformed into the Hamiltonian which acts upon the wave function to generate the evolution of the wave function in time and space. The Schrodinger equation gives the quantized energies of the system and gives the form of the wave function so that other properties may be calculated. 2
82m (E U) 0 h2
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E = total energy of electron U = potential energy of electron 2 2 2 = Laplacian operator = x2 y2 z2 2
and
= wave function of electron = amplitude of matter wave.
Schordinger equation can be solved completely for hydrogen atom and hydrogen-like species like He+, Li2+. The Schordinger equation can also be written in terms of spherical polar coordinates (r, , ) in addition to Cartesian coordinate x, y, z.
r cos
Since, H-atom posses spherical symmetry, it is easier to solve schordinger equation if it is represented in polar coordinate. z z r cos x r sin . cos y r sin sin r y r sin
x x 2 y 2 z 2 R2 When schordinger equation in polar coordinate is solved for H-atom then the solution obtained can be expressed as
r . . r represents radial wave function which depends on n, .
. Angular wave function which depends on m, . (Psi) : [Wave function or Amplitude of Electron Wave] It is also known as atomic orbitals. It is a mathematical function whose value depends on coordinates of electron in an atom. It may be +ve or –ve depending upon value of coordinates and it has no physical significance.
[Probability Density] : According to electromagnetic wave theory, intensity of light is proportional to square of amplitude. In the same way 2 gives an idea of intensity of electron wave i.e. probability of finding electron at that point. Q.
What is the significance of
(x).dx ?
Q.
What is the significance of
.dV ?
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ATOMIC STRUCTURE
ORBITALS Orbitals are defined as that region or zone in the space where probability of finding electrons is possible (more than 90%). In an atom large no. of permissible orbitals are present. These orbitals are designated by a set of 3 quantum number (n, , m) which arise as a natural consequence in the solution of Schordinger equation i.e. the values of 3 quantum numbers are restricted by the solution of Schordinger equation.
The Postulates of Quantum Mechanics 1. Associated with any particle moving in a conservative field of force is a wave function which determines everything that can be known about the system. 2. With every physical observable q there is associated an operator Q, which when operating upon the wave function associated with a definite value of that observable will yield that value times the wave function. 3. Any operator Q associated with a physically measurable property q will be Hermitian. 4. The set of eigen functions of operator Q will form a complete set of linearly independent functions. 5. For a system described by a given wave function, the expectation value of any property q can be found by performing the expectation value integral with respect to that wave function. 6. The time evolution of the wave function is given by the time dependent Schrodinger equation.
The Wave function Postulate
It is one of the postulates of quantum mechanics that for a physical system consisting of a particle there is an associated wave function. This wave function determines everything that can be known about the system. The wave function is assumed here to be a single-valued function of position and time, since that is sufficient to guarantee an unambiguous value of probability of finding the particle at a particular position and time. The wave function may be a complex function, since it is its product with its complex conjugate which specifies the real physical probability of finding the particle in a particular state. (x, t) single-valued probability amplitude at (x, t) * (x, t)(x, t) probability of finding particle at x at time t provided the wave function is normalized. Q.
How can a continuous description such as wave lead to discrete energy levels ?
Constraints on Wave function
In order to represent a physically observable system, the wave function must satisfy certain constraints:
1. Must be a solution of the Schrodinger equation.
2. Must be normalizable. This implies that the wave function approaches zero as x approaches infinity.
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4. The slope of the function in x must be continuous. Specifically
(x) must be continuous. x
These constraints are applied to the boundary conditions on the solutions, and in the process help determine the energy eigenvalues. Few Radial Functions Rn, 1 n = 1, = 0 R10(r) = 2 a0
3 /2
1 1 n = 2, = 0 R20(r) = 8 a0
n = 2, = 1 R21(r) =
e r / a0
1 1 24 a0
3/2
3/2
1 n = 3, = 0 R30(r) = 81 3 a0 2
n = 3, = 1 R31(r) =
2 1 9 6 a0
r r / 2a0 2 e a0
3 /2
3 /2
1 n = 3, = 2 R32(r) = 81 30 a0 4
r r / 2a0 e a0
2 27 18 r 2 r e r / 3a0 a0 a20
r r2 6 2 2 a a0 0
3/2
r2 a20
r / 3a 0 e
e r / 3a0
Hydrogenic orbitals (a) Radial wave functions Rn l (r) = f (r) (Z/a0)3/2 e–/2 Where a0 is the bohr radius (0.53Å) and = 2Zr/na0. n l f (r) 1 0 2 2 2
0 1
(1/ 2 2 ) (2 – ) (1/ 2 6 )
3
0
(1/ 9 3 ) (6 – 6+ 2)
3
1
(1/ 9 6 ) (4 – )
3
2
(1/ 9 30 )2
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ATOMIC STRUCTURE
(b) Angular wave functions Yl,ml (, ) = (1/4)1/2 y(,) l ml y(, ) 0 0 1 1 0 31/2 cos 1 ±1 (3/2)1/2 sin e± i 2 0 (5/4)1/2 (3 cos2 – 1) 2 ±1 (15/4)1/2 cos sin e± i 2 ±2 (15/8)1/2 sin2 e± 2 i Q.
If a particle of mass m located in one -Dimensional box as shown.
m 0 2 2 2 wave function n (x) | K | sin
(A) Ans.
2 a
(B)
a 2
x
a
nx for 0 x a then correct value of |K|2 is : a
(C)
4 a
(D)
a 4
A
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Q.
The angular wave function for = 0 & m = 0 is Y(, ) =
Ans.
(A) Wave fuction is independent of & (B) Two demensional plot will be circle in x–z plane (C) Three demensional plot will be spherical (D) Y().Y() plot in two dimensional will be elliptical D
1 4
. Select the incorrect statement
QUANTUM NUMBERS In the solution to the Schrodinger equation for the hydrogen atom, three quantum numbers arise from the space geometry of the solution and a fourth arises from electron spin. No two electrons can have an identical set of quantum numbers according to the Pauli exclusion principle , so the quantum numbers set limits on the number of electrons which can occupy a given state and therefore give insight into the building up of the periodic table of the elements. (1)
Principal Quantum No. (n) : The principal quantum number or total quantum number n arises from the solution of the radial part of the Schrodinger equation for the hydrogen atom. The bound state energies of the electron in the hydrogen atom are given by En
13.6eV n2
(i) Permissible values of n : all integers from 1 to (infinity) (ii) This no. identifies shell in an atom. (iii) It gives an idea of average distance ‘R’ [size of any orbital from the nucleus higher the value of, n, higher will be the average distance and hence greater will be the size]. (iv) It gives idea of energy of electron upto some extent. Higher the value of n, higher will be the energy (if ‘’ is constant). for example order of energy 4s > 3s > 2s > 1s (v) It gives value of total no. of orbitals present in any shell and that is equal to n2. (vi) It gives value of total no. of electrons which may he present in the given shell = 2n2. (vii) It gives value of angular momentum of electron i.e.
nh 2
(viii) It give variation of radial probability distribution. (2)
Azimuthal Quantum No. or Angular momentum Quantum no. () : Permissible values of l is 0 to (n -1). i.e. value of is restricted by n. n = 1,
=0
n = 2,
= 0, 1. ; i.e. 2 values of .
n = 3, = 0, 1, 2. i.e. 3 values of . No. of values of ‘’ is equal to ‘n’ principal quantum number It identify sub shell in an atom The value of ‘’ gives name of sub-shell and shape of orbital. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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(3)
ATOMIC STRUCTURE Notation s p d f
Name Sharp Principal diffuse fundamental
Shape Spherical dumbell shaped double dumbell complex
magnetic quantum number (m) The direct implication of this quantum number is that the z-component of angular momentum is quantized according to L z mlh / 2 It is called the magnetic quantum number because the application of an external magnetic field causes a splitting of spectral lines called the Zeeman effect. The different orientations of orbital angular momentum represented by the magnetic quantum number can be visualized in terms of a vector model.
(4)
Spin quantum number (s) An electron, besides charge and mass, has also spin angular momentum commonly called spin. The spin angular momentum of the electron is constant and cannot be changed. An electron spin s = 1/2 is an intrinsic property of electrons. Electrons have intrinsic angular momentum characterized by quantum number 1/2. In the pattern of other quantized angular momenta, this gives total angular momentum The resulting fine structure which is observed corresponds to two possibilities for the z-component of the angular momentum.
This causes an energy splitting because of the magnetic moment of the electron
Orbitals diagram
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AUFBAU's PRINCIPLE The electrons are added progressively to Various orbitals in the order of increasing energy, starting with the orbital of lowest energy, generally. ENERGY OF ORBITALS (i) Hydrogen atom : In case of hydrogen atom, energy of orbital is mainly determined by principle quantum number. n, E For H atom 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p
(ii)
(iii) (iv)
Multielectronic atoms : In case of multielectronic atoms, energy of orbitals depends upon both n and and hence energy of orbitals is compared on the basis of (n + ) rules generally n + rule : As the value (n + ) increases the total energy of orbitals also increases. If value of (n + ) is same or different orbital then orbital with lower value of n, have lower energy.
1s 2s 3s 4s 5s 6s 7s
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
2p 3p 4p 5p 6p 7p
3d 4d 5d 6d
4f 5f 6f
< 4f < 5d < 6p < 7s
PAULI'S EXCLUSIONS PRINCIPLE The principle states that no two electrons in an atom can have same set of all 4 quantum no. Therefore spin quantum No. have only two values therefore it can also be commented that orbital can have maximum two electrons with opposite spin. HUND'S RULE OF MAXIMUM MULTIPLICITY While filing the orbitals of same energy (i.e. orbitals of same subshell) pairing of electron start only when each orbital have got 1 electron that too with same spin or parallel spin. Multiplicity is expressed as 2|s| + 1 where s represent total spin. s
1 1 1 1 – 2 2 2 2
s
1 1 1 1 – , 2|s| + 1 = 2 2 2 2 2
s
3 1 1 1 3 ,2 +1=4 2 2 2 2 2
2|s| + 1 = 2
(It has maximum multiplicity)
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ATOMIC STRUCTURE
Total spin resulting from d7 configuration ?
ELECTRONIC CONFIGURATION OF ATOM The distribution of electrons of an atom in its various shell, sub shells and orbitals is known as electron configuration. Valence shell or outer most shell nth shell Penultimate shell-Inner to outermost shell (n –1) shell Anti Penultimate shell (n – 2) shell Electronic configuration can be expressed in the following ways : (i) Orbital notation method : nx Li 3 1s2 2s1
(ii) Orbital diagram method : Li 3 1s
2s
(iii) Condensed form : [He] 2s1
H 1 1s1 1s
He 2 1s2 1s
Li 3 1s 2 2s1 1s
2s
Be 4 1s 2 2s 2 1s
2s
B 5 1s2 2s2 2 p1 1s
2s
1s
2s
2p
1s
2s
2p
2p
C 6 1s 2 2s 2 2 p 2
N 7 1s 2 2s 2 2 p3
•
Electronic Configuration of ions : Al [Ne]3s2 3p1, A1 [Ne ]3s2 ,
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Similarly in case of transition elements electrons are removed from nth shell for e.g. 4th shell in case of 3d series. Sc [Ar] 4s2 3d1 , Sc2 [Ar]3d1 4s0 Cl : [Ne]3s2.3p5 Cl– : [Ne]3s2.3p6 Exceptional configuration Expected confi.
Cr 24 [Ar]4s2 3d4
Observed confi.
Cr 24 [Ar]4s1 3d5
Expected confi.
Cu 29 [Ar] 4s2 3d9
Observed confi.
Cu 29 [Ar] 4s1 3d10
Above exceptional configuration can be explained on the basis of following factor. (i)
Symmetrical Electronic Configuration : It is well known fact that symmetry leads to stability, orbitals of a sub shell having half-filled or fullfilled configuration (i.e. symmetrical distribution of electrons) are relatively more stable. This effect is dominating in d and f subshell. d5, d10, f7, f14 configurations are relatively more stable.
(ii)
Exchange Energy : It is assumed that electrons in de-generate orbitals, do not remain confined to a particular orbital rather it keep on exchanging its position with electron having same spin and same energy (electrons present in orbitals having same energy). Energy is released in this process known as exchange energy which imparts stability to the atom. More the number of exchanges, more will be the energy released, more the energy released stability will be more. Cr 24
[ Ar ] 4S 2 3d4
[ Ar ] 4S1 3d5
3+2+1=6
n(n 1) 2
when n represents no. of electron having same spin.
The relative value of exchange energy + depairing energy and transger energy ( energy relased during depairing)
[ 4S 3d]
decides the final configuration. For example, if magnitude of Dep. energy + Ex. energy > transfer. energy as in cases of Cu and Cr then configuration is
Cr 24 [ Ar ] 4 s1 3d5
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ATOMIC STRUCTURE
Cu 29 [ Ar ] 4 s1 3d10
respectively
In case of carbon magnitude of transfer energy > dep. energy + exch. energy and configuration of carbon is
1s 2 2s 2 2p 2
Magnetic moment () : It is the measure of the magnetic nature of substance n (n 2) B.M.
where n is no. of unpaired electrons.
When = 0, then there is no unpaired electron in the species and it is known as diamagnetic specie. These species are repelled by magnetic field. When 0 , for paramagnetic species. These species have unpaired electrons and are weakly attracted by the magnetic field. Calculate magnetic moment of these species (i) Cr
(ii) N
(iii) Cl
(iv) Ar
(i) Cr : No. of unpaired electrons = 6. =
6(6 2) 48 BM
Generally species having unpaired electrons are coloured or impart colour to the flame. Species having unpaired electrons can be easily excited by the wavelengths corresponding to visible lights and hence they emit radiations having characteristic colour.
(A)
Radial wave function (R) In all cases R approaches zero as r approaches infinity. One finds that there is a node in the 2s radial function. In general, it has been found that ns-orbitals have (n–1) nodes, np-orbitals have (n–2) nodes etc. The importance of these plots lies in the fact that they give information about how the radial wave function changes with distance r and about the presence of nodes where the change of sign of R occurs.
R
R
r
(B)
2p R
2s
1s
r
r
Radial Probability density (R2) The square of the radial wave function R2 for an orbital gives the radial density. The radial density gives the probability density of finding the electron at a point along a particular radius line. To get such a variation, the simplest procedure is to plot R2 against r. These plots give useful information about probability density or relative electron density at a point as a function of radius. It may be noted that while for s-orbitals the maximum electron density is at the nucleus. all other orbitals have zero electron Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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density at the nucleus.
1s
2s
r
(C)
2p
r
r
Radial probability functions 4r2R2 The radial density R2 for an orbital, as discussed above, gives the probability density of finding the electron at a point at a distance r from the nucleus. Since the atoms have spherical symmetry, it is more useful to discuss the probability of finding the electron in a spherical shell between the spheres of radius (r + dr) and r. The volume of the shell is equal to 4/3(r + dr)3–4/3r3 = 4r2dr. This probability which is independent of direction is called radial probability and is equal to 4r2drR2. Radial probability function (4r2drR2) gives the probability of finding the electron at a distance r from the nucleus regardless of direction.
1s
r
2s
2p
r
r
The radial probability distribution curves obtained by plotting radial probability functions versus distance r from the nucleus for 1s, 2s and 2p orbitals are shown in fig.
Hydrogen 1s Radial Probability
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ATOMIC STRUCTURE
Hydrogen 3s Radial Probability
Hydrogen 2p Radial Probability
Hydrogen 3p Radial Probability
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Hydrogen 3d Radial Probability
Angular wave Function . The angular wave function ' ' depends only on the quantum number and ml and is independent of the principal quantum number n for a given type of orbital. It therefore means that all s orbitals will have same angular wave function.
(A)
Angular Wave Function, '' For an s-orbital, the angular part is independent or angle and is, therefore, of constant value. Hence this graph is circular or, more properly, in three dimensions i.e., spherical. For the pz orbital, we get two tangent spheres. The px and py orbitals are identical in shape but are oriented along the x and y axes respectively. The angular, wave function plots for d and f-orbitals are four lobed and six-lobed respectively. It is necessary to keep in mind that in the angular wave function plots, the distance from the center is proportional to the numerical values of '' in that direction and is not the distance from the center of the nucleus.
(B)
Angular Wave Function, ||2 The angular probability density plots can be obtained by squaring the angular function plots shown in fig. On squaring, different orbitals change in different ways. For an s orbital, the squaring causes no change in shape since the function everywhere is the same: thus another sphere is obtained.
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ATOMIC STRUCTURE
Plots of Total probability density: Shapes of Atomic Orbitals. The problems associated with the representations of the variations of ||2 in space have been circumvented by the following two approaches. (a) Charge cloud diagrams (b) Boundary surface diagrams
(A)
Charge cloud diagrams In this approach, the probability density ||2 is shown as a collection of dots such that the density of dots in any region represents the electron probability density in that region. Fig. shows such plots for some orbitals. These give some ideal about the shapes of the orbitals.
1s
(B)
3s
Boundary surface digrams In these diagrams, the shape of an orbital is defined as a surface of constant probability density that encloses some large fraction (say 90 %) of the probability of finding the electron. The probability density is ||2 . When ||2 is constant, so is ||. Hence || is constant on the surface for an s orbital (l = 0) has the shape of a spherical shell centered on the nucleus, fig. For each value of n, there is one s orbital. As n increases, there are (n-1) concentric spherical shells like the successive layers in an onion.
Fig. Boundary surface digrams for the 1 s orbital.
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SHAPES OF ATOMIC ORBITALS
The spherical Polar Coordinates
pX
d z2
py
d
x 2 y2
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ATOMIC STRUCTURE
dxz
fxyz
dyz
f
f
y( z2 x 2 )
f
f z3
f z( x 2 y2 )
x3
f
x ( y2 z2 )
y3
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Class Room Problems 1.
Sol.
To what minimum distance will an alpha particle with kinetic energy T=0.40 MeV approach in the case of a head-on collision to a stationary Pb nucleus: [0.0216 pm]
4.
Calculate for a hydrogen atom and a He+ ion: (a) the radius of the first Bohr orbit and the velocity of an electron moving along it: (b) the kinetic energy and the binding energy of an electron in the ground state : (c) the ionization potential, the first excitation potential and the wavelength of the resonance line (n' = 2 n =1).
Sol.
2.
Sol.
A H-like neutral species is in some excited state (A) and on absorbing a photon of energy 3.066 eV gets promoted to a new state B. When the electron from state B return back, photons of a maximum ten different wavelengths can be observed in which some photons have energy smaller than 3.066 eV. Some of the equal energy and only four photon having energy greater than 3.066 eV. Determine the orbit number of states A and B and ionization energy. [nB = 5, nA=2, I.E.=14.6 eV]
5. Sol.
6.
3.
Sol.
At what temperature the transitional kinetic energy of atomic hydrogen equal that for n2 =1 to n2 =2 transition. [T = 78.7 ×103k ]
Sol.
Calculate the angular frequency of an electron occupying the second Bohr orbit of He+ ion. [w = 2.07×1016s–1]
A hypothetical element "positronium" consists of an electron moving in space around a nucleus consisting a positron. Using the Bohr's atomic model, determine the first Bohr radius, Positron is a subatomic particle similar to electron in all respect except possessing a positive charge. [1.06 Å]
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Sol.
To what series does the spectral line of atomic hydrogen belong if its wave number is equal to the difference between the wave numbers of the following two lines of the balmer series: 486.1 and 410.2 nm? What is the wavelength of that line? [2.63 ×10–4 cm]
ATOMIC STRUCTURE 12.
Sol.
13. 8.
Sol.
The threshold wavelength for photoelectric emi ss i on i n tungsten i s 2300 Å. What wavelength of light must be used to eject electrons with a maximum energy of 1.5 eV? [ = 179.9 nm]
Sol.
14.
9.
Sol.
De te rm i ne the de -Brogl i e w av el engt h associated with an electron in the 3rd Bohr's orbit of He+ ion. [5 Å]
Sol.
15.
10.
Sol.
Find the velocity of photoelectrons liberated by electromagnetic radiation of wavelength = 18.0 nm from stationary He+ ions in the ground state. [V = 2.3×106ms–1]
Sol.
16. 11.
Sol.
At what minimum kinetic energy must a hydrogen atom move for its inelastic head-on collision with another, stationary, hydrogen atom to make one of them capable of emitting a photon? Both atoms are supposed to be in the ground state prior to the collision. [20.5 eV ]
Sol.
17.
Calculate the de Broglie wavelengths of an electron, proton, and uranium atom, all having the same kinetic energy 100 eV. [1.23 Å, 2.86×10–2 Å, 1.86 × 10–3 Å]
What amount of energy should be added to an electron to reduce its de Brogie wavelengths from 100 to 50 pm? [450 eV]
Find the de Broglie wavelength of hydrogen molecules, which corresponds to their most probable velocity at room temperature. [127 pm ]
An automobile of mass 500 kg is moving with speed of 50±0.001 km hr –1 . Determine uncertainty in position of moving automobile and interpret the result. [V = 2.778 ×10–4 m/s, X = 3.8 × 10–38 m]
Find the quantum number n corresponding to the excited state of He+ ion if on transition to the ground state that ion emits two photons in succession with wavelengths 108.5 and 30.4 nm. [n = 5]
Estimate the minimum errors in determining the velocity of an electron, a proton, and a ball of mass of 1 mg if the coordinates of the particles and of the centre of the ball are known with uncertainly 1 m.
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ATOMIC STRUCTURE Sol.
18.
Sol.
Page # 43 22.
If a photon of wavelength 200 pm strikes an atom and one of the inner bound electrons is injected out with a velocity of 2×107 ms–1 . Calculate the energy with which it is bound to nucleus? [5075 eV]
Sol.
Calculate the electrostatic potential energy of two electrons separated by 3 Å in vacuum? [4.8 eV]
23.
The wave function for electron in ground state 1
of hydrogen atom is (a3 ) 2 e 1s 0
r a0
, where
"a0" is radius of Bohr's orbit. Calculate the probability of finding the electrons somewhere between 0 and 2a0. Sol.
19.
Sol.
When radiation of wavelength 253.7nm falls on a copper surface, electrons are ejected. Calculate work function if the stopping potential is 0.5 V. [w0 4.39 eV]
24.
The normalized wave function of the hydrogen 1
atom for the 1s orbital is (a3 ) 2 e 1s 0
r a0
.
Show that in such a state the most probable distance from nucleus to electron is a0. Sol.
20.
Sol.
The size of an atomic nucleus is 10–14 m. Calculate uncertainty in momentum of an electron if it were to exist inside the nucleus. [5.27 ×10–21 kgm/s–1]
25.
Locate the nodal surfaces in 1/2
310 (r, , )
1 2 81
1 a 0
3 /2
r r 2 r / 3a0 cos 6 2 e a a0 0
Sol. 21.
Sol.
Calculate de-Brogie wavelength of a hydrogen atom with translational energy corresponding to a temperature of 27ºC. [ = 1.46 Å]
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Page # 44
EXERCISE – I
ATOMIC STRUCTURE
OBJECTIVE PROBLEMS (JEE MAIN)
Single Correct 1. Radiation of = 155 nm was irradiated on Li (work function = 5eV) plate. The stopping potential (in eV) is. (A) 3eV (B) 4eV (C) 0.3eV (D) 0.5 eV Sol.
5.
Sol.
6. 2.
Increasing order of magnetic moment among the following species is __________ . Na+, Fe+3, Co2+, Cr+2 (A) Na+ E3 > E4 (D) The order of velocity of electron in H, He+, Li+, Be3+ species in second Bohr orbit is Be3+ > Li+2 > He+ > H
Sol.
3.
Correct statement(s) regarding 3py orbital is/are (A) Angul ar part of wave function i s independent of angles ( and ). (B) No. of maxima when a curve is plotted between 4r2R2(r) vs r are ‘2’. (C) ‘xz’ plane acts as nodal plane. (D) Magnetic quantum number must be ‘–1’. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 52 6.
ATOMIC STRUCTURE
Select the correct curve(s) : If v = velocity of electron in Bohr’s orbit. r = Radius of electron in Bohr’s orbit. P.E. = Potential energy of electron in Bohr’s orbit. K.E. = Kinetic energy of electron in Bohr’s orbit.
v
r
(A)
orbital is R(r) =
(B)
n
n
P.E.
(C)
Assertion and Reason : 8. It is a data sufficiency problem in which it is to be decided on the basis of given statements whether the given question can be answered or not. No matter whether the answer is yes or no. Question : Is the orbital of hydrogen atom 3px? Statement–1 : The radial function of the
(D) n
(4 – )e–/2, =
r 2
Sol.
Sol.
7.
9
6a30 / 2
Statement–2 : The orbital has 1 radial node & 0 angular node. (A) Statement (1) alone is sufficient. (B) Statement (2) alone is sufficient. (C) Both together is sufficient. (D) Neither is sufficient.
K.E. 1/n
1
Which is/are correct statement. (A) The difference in angular momentum associated with the electron present in
9.
h 2 (B) Energy difference between energy levels will be changed if, P. E. at infinity assigned value other than zero. (C) Frequency of spectral line in a H-atom is in the order of (2 1) < (3 1) < (4 1) (D) On moving away from the nucleus, kinetic energy of electron decreases.
consecutive orbits of H- atom is (n–1)
Sol.
Statement–1 : Energy emitted when an electron jump from 5 2 (energy level) is less than when an electron jump from 2 1 in all ‘H’ like atom. Statement–2 : The |total energy difference| between 1st & 2nd energy level is greater than that of any two energy level provided level ‘1’ is not part of those two energy levels. (A) Statement–1 is true, statement–2 is true and statement–2 is correct explanation for statement–1. (B) Statement–1 is true, statement–2 is true and statement–2 is not the correct explanation for statement–1. (C) Statement–1 is true, statement–2 is false. (D) Statement–1 is false, statement–2 is true.
Sol.
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ATOMIC STRUCTURE 10.
Statement–1 : Emitted radiations will fall in visible range when an electron jump from higher level to n = 2 in Li+2 ion. Statement–2 : Balmer series radiations belong to visible range in all H–like atoms. (A) Statement–1 is true, statement–2 is true and statement–2 is correct explanation for statement–1. (B) Statement–1 is true, statement–2 is true and statement–2 is not the correct explanation for statement–1. (C) Statement–1 is true, statement–2 is false. (D) Statement–1 is false, statement–2 is false.
Page # 53 Sol.
13.
Sol.
Assertion : Cathode rays do not travel in straight lines. Reason : Cathode rays can penetrate through a thick metal sheet. (A) Statement–1 is true, statement–2 is true and statement–2 is correct explanation for statement–1. (B) Statement–1 is true, statement–2 is true and statement–2 is not the correct explanation for statement–1. (C) Statement–1 is true, statement–2 is false. (D) Statement–1 is false, statement–2 is false.
Sol.
11.
Assertion : The fraction of total number of electrons in p-sub levels of Mg is 50% Reason : The fraction of total number of electron in p-sub levels of Be is 40% (A) Statement–1 is true, statement–2 is true and statement–2 is correct explanation for statement–1. (B) Statement–1 is true, statement–2 is true and statement–2 is not the correct explanation for statement–1. (C) Statement–1 is true, statement–2 is false. (D) Statement–1 is false, statement–2 is true.
14.
quency, v =
Assertion : Energy of an electron is determined by its principal quantum number. Reason : Principal quantum number is a measure of the most probable distance of finding the electrons around the nucleus. (A) Statement–1 is true, statement–2 is true and statement–2 is correct explanation for statement–1. (B) Statement–1 is true, statement–2 is true and statement–2 is not the correct explanation for statement–1. (C) Statement–1 is true, statement–2 is false. (D) Statement–1 is false, statement–2 is true.
c )
(A) Statement–1 is true, statement–2 is true and statement–2 is correct explanation for statement–1. (B) Statement–1 is true, statement–2 is true and statement–2 is not the correct explanation for statement–1. (C) Statement–1 is true, statement–2 is false. (D) Statement–1 is false, statement–2 is true.
Sol.
12.
Assertion : Energy of radiation is large if its wavelength is large. Reason : Energy = hv(v = fre-
Sol.
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Page # 54
ATOMIC STRUCTURE
Match the column : 15. Match the column Column–I and Column–II contain data on Schrondinger Wave–Mechanical model, where symbols have their usual meanings. Match the columns. Column–I
Comprehension–1(3 questions) The French physicist Louis de-Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength of a material particle, its linear momentum p and planck constant h. =
r
(A) r
(B)
r2 4 r2
(C)
(, ) = K (independent of & )
(D)
atleast one angular node is present
(P) (Q) (R) (S)
4s 5px 3s 6dxy
Column–II
Sol.
16.
(P) (Q)
Match the column Column–I Electron moving in 2nd orbit in He+ electron is Electron moving in 3rd orbit in H–atom Electron moving in 1st orbit in Li+2 ion Electron moving in 2nd orbit in Be+3 ion Column–II Radius of orbit in which moving is 0.529 Å Total energy of electron is (–)13.6 × 9 eV
(R)
Velocity of electron is
(S)
De–broglie wavelength of electron is
(A) (B) (C) (D)
The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves. These waves di ffer from the electromagnetic waves as they, (i) have lower velocities (ii) have no electrical and magnetic fields and (ii i ) are not emi tted by the parti cl e under consideration. The experimental confirmation of the de–Broglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by de–Broglie. Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron. He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also ture. This is summed up in what we now call the “Heisenberg uncertainty principle : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of undertainty in the position, x and the uncertainity in the momentum (mv) must be greater than or equal to h/4. i.e. h 4 The correct order of wavelength of Hydrogen (1H1), Deuterium (1H2) and Tritium (1H3) moving with same kinetic energy is :
x (mv)
17.
2.188 10 6 m/s 3 150 Å 13.6
h h = p mv
(A) H > D > T (C) H < D < T
(B) H = D = T (D) H < D > T
Sol.
Sol.
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ATOMIC STRUCTURE 18.
Page # 55
The transiti on, so that the de–Broglie wavelength of electron becomes 3 times of its initial value in He+ ion will be : (A) 2 5 (B) 3 2 (C) 2 6 (D) 1 2
If the electron comes back from energy level having energy E2 to energy level having energy E1, then the difference may be expressed in terms of energy of photon as : hc E
Sol.
E2 – E1 = E, =
19.
Since h and c are constants, E corresponds to definite energy; thus each transition from one energy level to another will produce a light of definite wavelength. This is actually observed as a line in the spectrum of hydrogen atom. Wave number of line is given by the formula
If the uncertainty in velocity & position is same, then the uncertainty in momentum will be : (A)
hm 4
(B) m
(C)
h 4m
(D)
1 m
1 1 . R n2 n 2 2 1
h 4 h 4
20.
Sol.
where R is a Rydberg’s constant (R = 1.1 × 107 m–1) The energy photon emitted corresponding to transition n = 3 to n = 1 is [h = 6 × 10–34 J-sec] (A) 1.76 × 10–18 J (B) 1.98 × 10–18 J (C) 1.76 × 10–17 J (D) None of these
Sol. Comprehension–2 (4 questions) The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively. Maximum number of lines produced when an electron jumps from nth level to ground level n(n 1) . For examle, in the case 2 of n = 4, number of lines produced is 6. (4 3, 4 2, 4 1, 3 2, 3 1, 2 1). When an electron returns from n2 to n1 state, the number of lines in the spectrum will be equal to
is equal to
(n 2 n1 )(n2 n1 1) 2
21.
In a collection of H–atom, electrons make transition from 5th excited state to 2nd excited state then maximum number of different types of photons observed are : (A) 3 (B) 4 (C) 6 (D) 15
Sol.
22.
The difference in the wavelength of the 1st line of Lyman series and 2nd line of Balmer series in a hydrogen atom is : (A)
9 2R
(B)
(C)
88 15R
(D) None
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Page # 56 Sol.
ATOMIC STRUCTURE 26.
The wavelength required to remove the electron from first Bohr’s orbit to infinity? (A) 36.5 Å (B) 146 Å (C) 73 Å (D) 18.5 Å
Sol.
23.
The wave number of electromagnetic radiation emitted during the transition of electron in between two levels of Li2+ ion whose principal quantum numbers sum is 4 and difference is 2 is : (A) 3.5 R (B) 4 R (C) 8 R (D) 8/9 R
Sol.
The Kinetic energy of electron in first Bohr’s orbit is? (A) 2.25 10–12 erg (B) 11 10–12 erg (C) 5.5 10–10 erg (D) 27.2 10–11 erg
Sol.
Comprehension–3 (4 questions) A single electron atom has a nuclear charge +Ze where Z is the atomic number and ‘e’ is the electronic charge. It requires 47.2ev to excite the electron from 2nd Bohr’s orbit to 3rd Bohr’s orbit from above given data solve the following question: 24.
27.
Integer Type 28. If electron shows transition from n2 = 4 to n1 = 2 Then, number of fine lines in the spectral line according to Sommerfeld model is ? Sol.
The atomic no. of the element is: (A) 4 (B) 3 (C) (D) 5
Sol.
29.
25.
The energy required for transition of electron from third to fourth Bohr’s orbit is: (A) 83 10–12 erg (B) 13.25 10–12 erg (C) 26.5 10–12 erg (D) 10.6 10–11 erg
If the spin quantum number can have the 1 3 3 1 following values , ,0, , then, 2 2 2 2 th number of elements in 10 period of periodic table.
Sol.
Sol. 30.
What is the atomic number of first atom in which last electron has entered 5g1 electronic configuration according to n + rule
Sol.
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ATOMIC STRUCTURE 31.
Find the maximum number of electrons in ground state configuration of Zn which have l = any positive integral value and m = any non zero value.
Page # 57 36.
Sol.
A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is: (take the proton mass, mp = (5/3) x -27
10
-15
kg; h/e = 4.2 x 10
9
-15
10 m/F; 1 fm = 10
J.s/C;
1 =9X 4 o
m)
Sol. 32.
Find the difference in the value of (n + ) for 19th electron of Cr and 21st electron Sc.
Sol. 37.
33.
Fe+x ion have spin magnetic moment m =
35 B.M. Find the maximum multiplicity of the electrons in d-subshell in Fe+x.
In photoelectric effect, stopping potential depends on (A) frequency of the incident light (B) intensity of the incident light by varies source distance (C) emitter’s properties (D) frequency and intensity of the incident light
Sol.
Sol.
34.
Total number of subshell vacant upto outer most of Ca.
Sol. 38.
35.
Calculate Zeff(effective nuclear charge) using slater rules on last electron of V+4. (Periodic)
In the experiment on photoelectric effect using light having frequency greater than the threshold frequency, the photocurrent will certainly increase when (A) Anode voltage is increased (B) Area of cathode surface is increased (C) Intensity of incident light is increased (D) Distance between anode and cathode is increased.
Sol.
Sol.
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Page # 58 39.
An electron in hydrogen atom first jumps from second excited state to first excited state and then, from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons in the two cases by x, y and z, then select the wrong answers : (A) z = 1/x (B) x = 9/4 (C) y = 5/27 (D) z = 5/27
ATOMIC STRUCTURE Sol.
43.
Sol.
40.
An electron is in an excited state in hydrogenlike atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is , then (A) E = 6.8 eV, = 6.6 × 10–10 m (B) E = 3.4 eV, = 6.6 × 10–10 m (C) E = 3.4 eV, = 6.6 × 10–11 m (D) E = 6.8 eV, = 6.6 × 10–11 m
A beam of ultraviolet light of all wavelengths pas ses t hroug h hyd roge n gas at room temperature, in the x-direction. Assume that all photons emitted due to electron transition inside the gas emerge in the y-direction. Let A and B denote the lights emerging from the gas in the x and y directions respectively. (A) Some of the incident wavelengths will be absent in A (B) Only those wavelengths will be present in B which are absent in A (C) B will contain some visible light. (D) B will contain some infrared light.
Sol.
Sol.
44. 41.
A particular hydrogen like atom has its ground state binding “energy 122.4 eV. Its is in ground state. Then : (A) Its atomic number is 3 (B) An electron of 90eV can excite it. (C) An electron of kinetic energy nearly 91.8 eV can be brought to almost rest by this atom. (D) An electron of kinetic energy 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atom.
If radiation of allow wavelengths from ultraviolet to infrared is passed through hydrogen gas at room temperature, absorption lines will be observed in the : (A) Lyman series (B) Balmer series (C) Both (A) and (B) (D) neither (A) nor (B)
Sol.
Sol.
45.
42.
The electron in a hydrogen atom makes a transition n1 n2, where n1 & n2 are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final state. The possible values of n1 & n2 are : (A) n1 = 4, n2 = 2 (B) n1 = 8, n2 = 2 (C) n1 = 8, n2 = 1 (D) n1 = 6, n2 = 3
In the hydrogen atom, if the reference level of potential energy is assumed to be zero at the ground state level. Choose the incorrect statement. (A) The total energy of the shell increases with increase in the value of n (B) The total energy of the shell decrease with increase in the value of n. (C) The difference in total energy of any two shells remains the same. (D) The total energy at the ground state becomes 13.6 eV.
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ATOMIC STRUCTURE Sol.
Page # 59 48.
A free hydrogen atom in ground state is at rest. A neutron of kinetic energy ‘K’ collides with the hydrogen atom. After collision hydrogen atom emits two photons in succession one of which has energy 2.55 eV. (Assume that the hydrogen atom and neutron has same mass) (A) minimum value of ‘K’ is 25.5 eV.
46.
(B) minimum value of ‘K’ is 12.75 eV.
Choose the correct statement(s) for hydrogen and deuterium atoms (considering motion of nucleus) (A) The radius of first Bohr orbit of deuterium is less than that of hydrogen
(C) the other photon has energy 10.2 eV (D) the upper energy level is of excitation energy 12.75 eV. Sol.
(B) The speed of electron in the first Bohr orbit of deuterium is more than that of hydrogen. (C) The wavelength of first Balmer line of deuterium is more than that of hydrogen (D) The angular momentum of electron in the first Bohr orbit of deuterium is more than that of hydrogen. Sol.
49.
47.
Let An be the area enclosed be the nth orbit in a hydrogen atom. The graph of ln (An/A1) agains ln (n).
A neutron collides head-on with a stationary hydrogen atom in ground state. Which of the following statements are correct (Assume that the hydrogen atom and neutron has same mass) (A) If kinetic energy of the neutron is less than 20.4 eV collision must be elastic (B) If kinetic energy of the neutron is less than 20.4 eV collision may be inelastic
(A) will pass through origin (B) will be a stright line will slope 4
(C) Inelastic collision may be take place only when initial kinetic energy of neutron is greater than 20.4 eV.
(C) will be a monotonically increasing nonlinear curve (D) will be a circle.
(D) Perfectly inelastic collision can not take place.
Sol. Sol.
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Page # 60 50.
ATOMIC STRUCTURE
A hydrogen atom in a state having a binding energy 0.85 eV
52.
makes a transition to a state of excitation energy 10.2 eV. The wave length of emitted photon is ................ nm. Sol.
If the mass of the particle is m = 1.0 × 10–30 kg and a = 6.6 nm, the energy of the particle in its ground state is closest to (A) 0.8 meV
(B) 8 meV
(C) 80 meV
(D) 800 meV
Sol.
Paragraph for Questions 51 to 53 When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of p2 mass m is related to its linear momentum as E = . 2m Thus, the energy of the particle can be denoted by a quantum number ‘n’ taking values 1, 2, 3 … (n = 1, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a.Take h = 6.6 × 10–34 Jsand e= 1.6 × 10–19 C. [JEE 2009] 51.
53.
The speed of the particle, that can take discrete values, is proportional to (A) n –3/2 (C) n1/2
(B) n (D) n
–1
The allowed energy for the particle for a particular value of n is proportional to (A) a–2 (C) a–1
(B) a–3/2 (D) a2
Sol.
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ATOMIC STRUCTURE
EXERCISE – III 1.
Page # 61
OBJECTIVE PROBLEMS (JEE ADVANCED)
Bohr’s Model Calculate the wavelength of radiation emitted, producing a line in Lyman series, when an electron falls from fourth stationary state in hydrogen atom. (A) 9.6 × 10–8 m (B) 9.7 × 10–8 m –8 (C) 9.7 × 10 m (D) 9.6 × 10–8 cm
Sol.
2.
Sol.
5.
Calculate energy of electron which is moving in the orbit that has its radius sixteen times the radius of first Bohr orbit for H–atom. (A) 4.36 × 1018 J (B) –1.36 × 10–19 J (C) –1.36 × 10–29 KJ (D) –1.36 × 10–4 J
Sol.
Sol.
6.
Sol. 3.
Calculate the Rydberg constant R if He+ ions are known to have the wavelength difference between the first (of the longest wavelength) lines of Balmer and Lyman series equal to 133.7 nm. (A) 1.96 × 10–8 m–1 (B) 1.096 × 107 m–1 (C) 1.096 × 108 m–1 (D) 1.4 × 102 m–1
What transition in the hydrogen spectrum would have same wavelength as the Balmer t r a n s i t i o n , n = 4 to n = 2 of He+ spectrum. (A) n1 = 4, n2 = 2 (B) n1 = 5, n2 = 6 (C) n1 = 1, n2 = 3 (D) n1 = 1, n2 = 2
The wavelength of a certain line in the Paschen series in 1093.6 nm. What is the value of nhigh for this line. [RH = 1.0973 × 10+7 m–1] (A) 10 (B) 5 (C) 6 (D) 8
Sol.
7.
Calculate the total energy emitted when electrons of 1.0g atom of hydrogen undergo transition giving the spectral line of lowest energy in the visible region of its atomic spectrum. (A) 1.827 × 105 J/mol (B) 2.827 × 105 J/mol (C) 18.27 × 105 J/mol (D) 28.27 × 105 J/mol
Sol. 4.
Wavelength of the Balmer H line is 6565 Å. Calculate the wavelength of H, line of same hydrogen like atom. (A) 4863 Å (B) 48.63 Å (C) 4.863 Å (D) 4563 Å : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 62 8.
A photon having = 854 Å causes the ionization of a nitrogen atom. Give the I.E. per mole of nitrogen in kJ. (A) 1503 KJ/mole (B) 1603 KJ/mole (C) 1403 KJ/mole (D) 1504 KJ/mole
ATOMIC STRUCTURE 12.
Sol.
The vapours of Hg absorb some electrons accelerated by a potential difference of 4.5 volt as a result of which light is emitted. If the full energy of single incident e– is supposed to be converted into light emitted by single Hg atom, find the wave number of the light. (A) 3.63 × 106 m–1 (B) 36.3 × 106 m–1 (C) 3.63 × 105 m–1 (D) 3.63 × 107 m–1
Sol.
9.
The radius of the an orbit of hydrogen atom is 0.85 nm. Calculate the velocity of electron in this orbit. (A) 6.44 × 105 m/s (B) 5.44 × 105 m/s (C) 5.44 × 103 m/s (D) 5.44 × 102 m/s
13.
Sol.
If the average life time of an excited state of H atom is of order 10–8 sec, estimate how many orbits an e– makes when it is in the state n = 2 and before it suffers a transition to n = 1 state. (A) 8 × 106 (B) 103 6 (C) 10 (D) 0.8 × 106
Sol.
10.
A doubly ionised lithium atom is hydrogen like with atomic number z = 3. Find the wavelength of the radiation required to excite the electron in Li2+ from the first to the third Bohr orbit. (A) 114.74 Å (B) 113.74 Å (C) 112.74 Å (D) 111.74 Å
14.
Sol.
Calculate the frequency of e– in the first Bohr orbit in a H–atom. (A) 6530 × 1012 Hz (B) 6.530 × 1012 Hz (C) 6530 × 1010 Hz (D) 6530 × 107 Hz
Sol.
11.
Sol.
The energy of an excited H–atom is –3.4 eV. Calculate angular momentum of e– in the given orbit. (A) h/2 (B) h/ (C) 2h/ (D) h/5 15.
A stationary He + ion emitted a photon corresponding to a first line of the Lyman series. The photon liberated a photoelectron from a stationary H atom in ground state. What is the velocity of photoelectron. (A) 3.10 × 108 cm/sec (B) 3.09 × 108 cm/sec (C) 3.10 × 108 cm/sec (D) 3.8 × 108 cm/sec
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ATOMIC STRUCTURE Sol.
16.
Page # 63 19.
Find the number of photons of radiation of frequency 5 × 1013 s–1 that must be absorbed in order to melt one gm ice when the latent heat of fusion of ice is 330 J/g. (A) 1023 (B) 1021 (C) 1040 (D) 1022
Sol.
A proton is accelerated to one–tenth of the velocity of light. If its velocity can be measured with a precision ±1%. What must be its uncertainty in position. (A) 4.05 × 10–4 m (B) 3.05 × 10–4 m (C) 2.05 × 10–3 m (D) 1.05 × 10–3 m
Sol.
20.
To what effective potential a proton beam be subjected to give its protons a wavelength of 1×10–10 m. (A) 1826 volt (B) 082.6 volt (C) 0.0826 volt (D) 826 volt
Sol. 17.
Suppose 10–17J of light energy is needed by the interior of the human eye to see an object. How many photons of green light ( = 550 nm) are needed to generate this minimum amount of energy. (A) 25 photon (B) 20 photon (C) 28 photon (D) 30 photon
21.
Sol.
He atom can be excited to 1s1 2p1 by = 58.44nm. If lowest excited state for He lies 4857 cm–1 below the above. Calculate the energy of the lower excitation state. (A) 6.3×10–15 J (B) 3.3×10–14 J –16 (C) 3.3×10 J (D) 6.3×10–16 J
Sol.
18.
Through what potential difference must an electron pass to have a wavelength of 500 Å. (A) 6.03 × 10–3 volt (B) 6.03 × 10–4 volt (C) 6.03 × 10–2 volt (D) 6.03 × 10–6 volt
22.
Sol.
A certain dye absorbs 4530 Å and fluoresces at 5080 Å these being wavelengths of maxi mum absorption that under given conditions 47% of the absorbed energy is emitted. Calculate the ratio of the number of quanta emitted to the number absorbed. (A) 0.527 (B) 05.27 (C) 052.7 (D) 1.527
Sol.
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Page # 64 23.
The reaction between H2 and Br2 to form HBr in presence of light is initiated by the photo decomposition of Br2 into free Br atoms (free radicals) by absorption of light. The bond dissociation energy of Br2 is 192 kJ/mole. What is the longest wavelength of the photon that would initiate the reaction. (A) 6485 Å (B) 6415 Å (C) 6425 Å (D) 6235 Å
ATOMIC STRUCTURE 27.
Sol.
X–ray emitted from a copper target and a molybdenum target are found to contain a line of wavelength 22.85 nm attributed to the K line of an impurity element. The K lines of copper (Z = 29) and molybdenum (Z = 42) have wavelength 15.42 nm and 7.12 nm respectively. Using Moseley’s law, 1/2 = a(Z – b), calculate the atomic number of the impurity element. (A) 24 (B) 12 (C) 18 (D) 22
Sol.
24.
The quantum yield for decomposition of HI is 0.2. In an experiment 0.01 moles of HI are decomposed. Find the number of photons absorbed. (A) 3 × 1020 (B) 3 × 1022 (C) 5 × 1022 (D) 5 × 1020
Sol.
25.
28.
What is de–Broglie wavelength associated with an e– accelerated through potential difference = 100 KV. (A) 3.88 pm (B) 3.068 pm (C) 4.88 pm (D) 5.68 pm
Sol.
Calculate the wavelength of the radiation that would cause photo dissociation of chlorine molecule if the Cl–Cl bond energy is 243 kJ/ mole. (A) 4.9 × 10–5 m (B) 4.9 × 10–7 m –6 (C) 4.9 × 10 m (D) 5.9 × 10–6 m
29.
Sol.
Cal c ul at e the de –Brogl i e w av el engt h associated with motion of earth (mass 6 × 1024 kg) orbiting around the sun at a speed of 3 × 106 m/s. (A) 3.68 × 10–65 m (B) 5.68 × 10–65 m (C) 2.68 × 10–60 m (D) 6.68 × 10–60 m
Sol.
26.
Sol.
The dissociation energy of H2 is 430.53 KJ/ mole. If H2 is exposed to radiant energy of wavelength 253.7 nm, what % of radiant energy will be converted into K.E.. (A) 8.08% (B) 8.68% (C) 8.18% (D) 7.88%
30.
An electron has a speed of 40 m/s, accurate up to 99.99%. What is the uncertainity in locating its position. (A) 2.0144 m (B) 1.2244 m (C) 1.0144 m (D) 0.0144 m
Sol.
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ATOMIC STRUCTURE 31.
Page # 65
The electrons identified by quantum numbers 'n' and : (a) n = 4, = 1 (b) n = 4, = 0 (c) n = 3, = 2 (d) n = 3, = 1 can be placed in order of increasing energy as (A) (b) < (d) < (a) < (c) (B) (a) < (c) < (b) < (d) (C) (c) < (d) < (b) < (a) (D) (d) < (b) < (c) < (a)
Sol.
Sol.
34.
Mr. Santa has to decode a number “ABCDEF” where each alphabet is represented by a single digit. Suppose an orbital whose radial wave functional is represented as : (r ) k1·e r / k 2
32.
Find the wavelength of the first line of He + ion spectral series whose interval between
From the following information given about each alphabet then write down the answers in the form of “ABCDEF”, for above orbital. Info A = Value of n where “n” is principal quantum number. Info B = No. of angular nodes Info C = Azimuthal quantum number of subshell to orbital belongs Info D = No. of subshell s having energy between (n + 5)s to (n + 5)p where n is principal quantum number. Info E = Orbital angular momentum of given orbital. Info F = Radial distance of the spherical node which is farthest from the nucleus. (Assuming k 3 = 1) (A) 300203 (B) 200103 (C) 300303 (D) 300103
1 1 extreme line is 2.7451 10 4 cm 1 . 1 2
(A) 4559 Å (C) 4689 Å
(B) 5689 Å (D) 5689 Å
Sol.
33.
A cylindrical source of light which emits radiation radially (from curved surface) only, placed at the centre of a hollow, metallic cylindrical surface, as shown in diagram. The power of source is 90 watt and it emits light of wavelength 4000 Å only. The emitted photons strike the metallic cylindrical surface which results in ejection of photoelectrons. All ejected photoelectrons reaches to anode ( l i g h t s o ur c e ). T he m ag ni t u d e o f photocurrent is [Given : h = 6.4 × 10–34 J/ sec] Cathode
(A) 12 amp (C) 11 amp
(r2 – 5k 3r + 6k 32)
Sol.
120º
Anode (light source) (B) 10 amp (D) 15 amp
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EXERCISE – IV
ATOMIC STRUCTURE
PREVIOUS YEARS PROBLEMS
LEVEL – I Q.1
An atom has a mass of 0.02 kg & uncertainity in its velocity is 9.218 × 10 –6 m/s then uncertainity in position is (h = 6.626 × 10–34 J - s) [AIEEE- 2002] –28 (A) 2.86 × 10 m (B) 2.86 × 10–32 cm –27 (C) 1.5 × 10 m (D) 3.9 ×10–10 m
JEE MAIN Q.4
The orbital angular momentum for an electron revolving in an orbit is given by
l l 1.
h . 2
This momentus for an s-electron will be given by [AIEEE- 2003]
Sol. (A)
h 2
(C) +
1 h . 2 2
(B)
2.
h 2
(D) zero
Sol. Q.2
Energy of H–atom in the ground state is –13.6 eV , Hence energy in the second excited state is – [AIEEE- 2002] (A) –6.8 eV (B) –3.4 eV (C) –1.51 eV (D) –4.3 eV
Sol.
Q.5
In Bohr series of lines of hydrogen spectrum, third line from the red end corresponds to where one of the following inter-orbit jumps of electron for Bohr orbits in an atom of hydrogen. [AIEEE- 2003] (A) 4 1 (B) 2 5 (C) 3 2 (D) 5 2
Sol.
Q.3
Unertainty in position of a particle of 25 g in space is 10–5 m. Hence uncertainty in velocity (ms–1) is (Planck’s constant h = 6.6 × 10–34 Js) [AIEEE- 2002] –28 (A) 2.1 × 10 (B) 2.1 × 10–34 –34 (C) 0.5 × 10 (D) 5.0 × 20–24
Q.6
Sol.
The de Broglie wavelength of a tennis ball mass 60 g moving with a velocity of 10 mt. per second is approximately - [AIEEE- 2003] (A) 10–16 metres (B) 10–25 metres –33 (C) 10 metres (D) 10–31metres
Sol.
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ATOMIC STRUCTURE Q.7
Sol.
Q.8
Which of the following sets of quantum numbers is correct for an electron in 4f orbital ? [AIEEE- 2004] 1 (A) n = 4, l = 3 , m = + 4, s = + 2 1 (B) n = 4, l = 4 , m = – 4, s = – 2 1 (C) n = 4, l = 3 , m = + 1, s = + 2 1 (D) n = 4, l = 3 , m = – 2, s = + 2
Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum numbers, l =1 and 2 are, respectively [AIEEE- 2004] (A) 12 and 4 (B) 12 and 5 (C) 16 and 4 (D) 16 and 5
Page # 67 Q.10
Which one of the following sets of ions represents the collection of isoelectronic species ? [AIEEE- 2004] (A) K+, Ca2+, Sc3+, Cl– (B) Na+, Ca2+, Sc3+, F– (C) K+, Cl–, Mg2+, Sc3+ (D) Na+, Mg2+, Al3+, Cl– (Atomic nos.: F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20 , Sc = 21)
Sol.
Q.11
Sol.
In a multi-electron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields ? [AIEEE- 2005] (a) n = 1, l = 0, m = 0 (b) n = 2, l = 0, m = 0 (c) n = 2, l = 1, m = 1 (d) n = 3, l = 2, m = 1 (e) n = 3, l = 2, m = 0 (A) (b) and (c) (B) (a) and (b) (C) (d) and (e) (D) (c) and (d)
Sol.
Q.9
The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×107 m–1) [AIEEE- 2004] (A) 91 nm (B) 192 nm (C) 406 nm (D) 9.1×10–8 nm
Q.12
Sol.
Sol.
Of the following sets which one does NOT contain isoelectronic species ? [AIEEE- 2005] (A) CN–, N2, C22– (B) PO43–, SO42–, ClO4– (C) BO33–, CO32–, NO3– (D) SO32–, CO32–, NO3–
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Page # 68 Q.13
According to Bohr's theory, the angular momentum of an electron in 5th orbit is [AIEEE 2006] (A) 1.0 h/ (B) 10 h/ (C) 2.5 h/ (D) 25 h/
ATOMIC STRUCTURE Q.16
The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol–1. The energy required to excite the electron in the atom from n = 1 to n = 2 is
Sol.
(A) 6.56 ×
[AIEEE 2008] 105
J
mol–1
(B) 7.56 × 105 J mol–1 (C) 9.84 × 105 J mol–1 (D) 8.51 × 105 J mol–1 Sol.
Q.14
Uncertainty in the position of an electron (mass = 9.1 × 10–31 kg) moving with a velocity 300 m/s, accurate upto 0.001 %, will be (h = 6.63 × 10–34 Js) [AIEEE 2006] (A) 5.76 × 10–2 m (B) 1.92 × 10–2 m –2 (C) 3.84 × 10 m (D) 19.2 × 10–2 m
Q.17
Sol.
Sol.
Q.15 Which of the following sets of quantum numbers represents the highest energy of an atom ? [AIEEE 2007] (A) n = 3, = 1, m = 1, s = +½ (B) n = 3, = 2, m = 1, s = +½ (C) n = 4, = 0, m = 0, s = +½ (D) n = 3, = 0, m = 0, s = +½ Sol.
Q.18
Sol.
In an atom, an electron is moving with a speed of 600m/ s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h = 6.6×10–34 kg m 2s–1 ,mass of electron, em = 9.1×10–31 kg ) (A) 1.52×10–4m (B) 5.10×10–3m –3 (C) 1.92×10 m (D) 3.84×10–3m
[AIEEE 2009]
Calculate the wavelength (in nanometer) associated with a proton moving at 1.0×103ms–1 (Mass of proton = 1.67×10 –27 kg and h = 6.63×10–34 Js ) : (A) 0.032 nm (B) 0.40 nm (C) 2.5 nm (D) 14.0 nm
[AIEEE 2009]
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ATOMIC STRUCTURE Q.19
Page # 69
The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ mol–1. The longest wavelength of light capable of breaking a single Cl – Cl bond is (c = 3 x 108 ms–1 and NA = 6.02 x 1023 mol–1) (A) 594 nm (B) 640 nm (C) 700 nm (D) 494 nm
Sol.
[AIEEE 2012]
[AIEEE 2010]
Sol.
Ionisation energy of He is 19.6 x 10 J atom–1. Th e en ergy of the firs t st atio nary sta te 2+ (n = 1) of Li –16 is [AIEEE 2010] –1 (A) 4.41 x 10 –17J atom –1 (B) –4.41 x 10 J atom –15 –1 (C) –2.2 x 10–17 J atom–1 (D) 8.82 x 10 J atom
Q.22 The electrons identified by quantum numbers 'n' and : (a) n = 4, = 1 (b) n = 4, = 0 (c) n = 3, = 2 (d) n = 3, = 1 can be placed in order of increasing energy as : (A) (b) < (d) < (a) < (c) (B) (a) < (c) < (b) < (d) (C) (c) < (d) < (b) < (a) (D) (d) < (b) < (c) < (a) Sol. [AIEEE 2012]
Iron exhibits +2 and +3 oxidation states. Which of the following statements about iron is incorrect ? (A) Ferrous compounds are less volatile than the corresponding ferric compounds. (B) Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds. (C) Ferrous oxide is more basic in nature than the ferric oxide. (D) Ferrous compounds are relatively more ionic than the corresponding ferric compounds.
Q.23 The standard reduction potentials for Zn2+/Zn, Ni2+/Ni, and Fe2+/Fe are –0.76, –0.23 and –0.44 V respectively. The reaction X + Y2+ X2+ + Y will be spontaneous when: (A) X = Fe, Y = Zn (B) X = Zn, Y = Ni (C) X = Ni, Y = Fe (D) X = Ni, Y = Zn
Q.20
+
–18
Sol.
Q.21
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Page # 70 Sol.
ATOMIC STRUCTURE [AIEEE 2012]
Q.26 Which of the following is the wrong statement ? (A) Ozone is diamagnetic gas (B) ONCl and ONO– are not isoelectron (C) O3 molecule is bent (D) Ozone is violet black in solid state Sol. [AIEEE 2013]
Q.24 The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+ will be: (A) – 10.2 eV (B) + 2.55 eV (C) – 2.55 eV (D) – 5.1 eV Sol. [AIEEE 2013]
Q.25 Energy of an electron is given by
Q.27 Which of the following arrangements does not represent the correct order of the property stated against it? (A) Se < Ti < Cr < Mn : number of oxidation states. (B) V2+ < Cr2+ < Mn2+ < Fe2+ : Paramagnetic behaviour. (B) Ni2+< Co2+< Fe2+ < Mn2+ : ionic size (D) Co3+ < Fe3+ < Cr3+ < Sc3+ stability in aqueous solution. Sol. [AIEEE 2013]
Z2 E = –2.178 × 10–18 2 .Wavelength of ligh ren quired to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be (h = 6.62 × 10–34 Js and c = 3.0 × 108 ms–1) (A) 8.500 × 10–7 m (B) 1.214 × 10–7 m (C) 1.816 × 10–7 m (D) 6.500 × 10–7 m Sol. [AIEEE 2013]
28.
In a hydrogen like atom electron make transition from an energy level with quantum number n to another with quantum number (n – 1). If n >> 1, the frequency of radiation emitted it proportional to [AIEEE 2013] (A)
(C)
1 3 /2
n 1 n
(B)
(D)
1 n3 1 n2
Sol.
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ATOMIC STRUCTURE
Page # 71
LEVEL – II 1.
JEE ADVANCED
With what velocity should an –particle travel towards the nucleus of a Cu atom so as to Sol. arrive at a distance 10–13 m. [JEE 1997]
Sol.
5. 2.
A compound of Vanadium has magnetic moment of 1.73 BM work out electronic configuration of Vanadium Ion in the compound. [JEE 1997]
Sol.
Assertion: Zn2+ is diamagnetic. [JEE 1998] Reason: The electrons are lost from 4s orbital to form Zn2+. (A) Both assertion and reason are correct and reason is the correct explanation of the assertion. (B) Both assertion and reason are correct but the reason is not the correct explanation of the assertion. (C) Assertion is correct but reason is incorrect. (D) Assertion is incorrect but reason is correct.
Sol.
3.
The energy of an electron in the first Bohr orbit of H atom is –13.6 eV. The possible energy value(s) of the excited state(s) for electrons in Bohr orbits of hydrogen is/are : [JEE 1998] (A) –3.4 eV (B) –4.2 eV 6. (C) –6.8 eV (D) +6.8 eV
Sol.
The electrons, identified by quantum numbers n and l (i) n = 4, l = 1, (ii) n = 4, l = 0 (iii) n = 3, l = 2 and (iv) n = 3, l = 1 can be placed in order of increasing energy, from the lowest to highest, as [JEE 1999] (A) (iv) < (ii) < (iii) < (i) (B) (ii) < (iv) < (i) < (iii) (C) (i) < (iii) < (ii) < (iv) (D) (iii) < (i) < (iv) < (ii)
Sol. 4.
Which of the following statement(s) is (are) correct ? [JEE 1998] (A) The electronic configuration of Cr is [Ar] 3d54s1 (Atomic Number of Cr = 24) (B) The magnetic quantum number may have a negative value (C) In silver atom, 23 electron have a spin of one type and 24 of the opposite type. (Atomic Number of Ag = 47) (D) The oxidation state of nitrogen in HN3 is -3. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 72 7.
Ground state electronic configuration of nitrogen atom can be represented by [JEE 1999]
ATOMIC STRUCTURE Sol.
(A) (B) (C) (D) 11.
Sol.
Calculate the energy required to excite one litre of hydrogen gas at 1 atm and 298 K to the first excited state of atomic hydrogen. The energy for the dissociation of H–H is 436 kJ mol–1.
Sol.
8.
Electronic configuration of an element is 1s22s22p63s23p63d54s1. This represents its [JEE 2000] (A) excited state (B) ground state (C) cationic form (D) anionic form 12.
Sol.
9.
The wavelength associated with a golf ball weighing 200 g and moving at a speed of 5 m/h is of the order [JEE 2001] (A) 10-10 m (B) 10-20m -30 (C) 10 m (D) 10-40 m
The quantum numbers +1/2 and –1/2 for the electron spin represent : (A) rotation of the electron in clockwise and anticlockwise direction respectively. (B) rotation of the electron in anticlockwise and clockwise direction respectively. (C) magnetic moment of the electron pointing up and down respectively. (D) two quantum mechanical spin states which have no classical analogue. [JEE 2001]
Sol.
Sol.
13.
10.
The number of nodal planes in a px orbital is : [JEE 2000] (A) one (B) two (C) three (D) zero
Rutherfords experiment, which established the nuclear model of atom, used a beam of : (A) –particles, which impinged on a metal foil and get absorbed. (B) –rays, which impinged on a metal foil and ejected electron. (C) Helium atoms, which impinged on a metal foil and got scattered. (D) Helium nuclie, which impinged on a metal foil and got scattered. [JEE 2002]
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ATOMIC STRUCTURE
Page # 73
Sol.
17.
A ball of mass 100 g is moving with 100 m/s. Find its wavelength [JEE 2004]
Sol.
14.
The magnetic moment of cobalt of the compound Hg[Co(SCN)4] is [Given : Co+2] (A)
3
(B)
8
(C)
15
(D)
24
[JEE 2004] Sol. 18.
The number of radial nodes of 3s and 2p orbitals are respectively. [JEE 2004] (A) 2,0 (B) 0,2 (C) 1,2 (D) 2,1
Sol. 15.
The radius of which of the following orbit is same as that of the first Bohr’s orbit of hydrogen atom? (A) He+ (n = 2) (B) Li2+ (n = 2) (C) Li2+ (n = 3) (D) Be3+ (n = 2) [JEE 2004]
Sol.
19.
16.
The Schrodinger wave equation for hydrogen atom is 3 /2
atom in units of
h . 2
r0
1 r 2 0 e a0 2s a0 4 2 a0 Where a0 is Bohr's radius. If the radial node is 2s be at r0, then find r0 in term of a0. [JEE 2004] 1
Find the velocity(m/s) of electron in first Bohr's orbit of radius a0. Also find the de Broglie's wavelength (in m). Find the orbital angular momentum of 2p orbital of hydrogen
[JEE 2004] Sol.
Sol.
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Page # 74 20.
ATOMIC STRUCTURE
Given in hydrogenic atom rn, Vn, E, Kn stand for radius, potential energy, total energy and kinetic energy in nth orbit. Find the value of U, v, x, y. [JEE 2006] Column-I (A) U = Vn / Kn (B)
1 Ex rn
(C)
rn Zy (Z = atomic number)
22.
Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecular B2 is [JEE 2010] (A) 1 and diamagnetic (B) 0 and diamagnetic (C) 1 and paramagnetic (D) 0 and paramagnetic
Sol.
(D) v = (Orbital angular momentum of electron in its lowest energy.) Column-II (P) 1 (Q) –2 (R) –1 (S) 0 Sol.
23.
Paragraph for questions 23 to 25 The hydrogen-like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the hydrogen atom. [JEE 2010] The state S1 is (A) 1 s (B) 2s (C) 2p (D) 3s
Sol. 21.
Match the entries is Column–I with the correctly related quantum number(s) in Column–II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. [JEE 2008] Column–I (A) Orbital angular momentum of the electron in a hydrogen-like atomic orbital (B) A hydrogen-like one-electron wave function function obeying Pauli principle (C) Shape, size and orientation of hydrogen like atomic orbitals (D) Probability density of electron at the nucleus in hydrogen-like atom Column–II (P) Principal quantum number (Q) Azimuthal quantum number (R) Magnetic quantum number (S) Electron spin quantum number
24.
Energy of the state S1 in units of the hydrogen atom ground state energy is (A) 0.75 (B) 1.50 (C) 2.25 (D) 4.50
Sol.
Sol.
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ATOMIC STRUCTURE 25.
Page # 75
The orbital angular momentum quantum number of the state S2 is (A) 0 (B) 1 (C) 2 (D) 3
Sol.
Sol. 29.
The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the sl ope of the stopping potenti al versus frequency plot for Silver to that of Sodium is [JEE 2013]
Sol.
26.
Sol.
The maximum number of electrons that can have principal quantum number, n = 3, and spin quantum number, ms = – 1/2, is: [JEE 2011] 30.
The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0, where a0 is the Bohr radius. Its orbital angular momentum is 3h . It is given that h is Planck constant and R 2 i s R yd be rg c onst ant. T he p os si bl e wavelength(s), when the atom de-excites, is (are) [JEE 2013]
27.
The work function () of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is [JEE 2011] Metal (eV)
Li
Na
K
Mg Cu Ag Fe
Pt
W
(A)
9 32R
(B)
9 16R
(C)
9 5R
(D)
4 3R
Sol.
2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75
Sol.
31.
28.
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius] [JEE 2012] 2 h (A) 42ma2 0 2
h (C) 322ma2 0
The atomic masses of He and Ne are 4 and 20 a.m.u. respectively. The value of the de Broglie wavelength of He gas at –73 ºC is "M" times that of the de Broglie wavelength of Ne at 727 ºC. M is [JEE 2013]
Sol.
2 h (B) 16 2ma2 0 2
h (D) 642ma2 0
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ATOMIC STRUCTURE
ANSWER-KEY OBJECTIVE PROBLEMS (JEE MAIN)
Answer Ex–I 1.
A
2. B
3. A
4.
A
5
7
D
8
9
B
10
D
11 B
12 B
13
A
14 A
15 B
16
D
17 A
18 B
19
C
20 B
21 B
22
C
23 D
24 B
25
B
26 C
27 C
28
B
29 B
30 D
31
A
32 C
33 C
34
A
35 C
36 D
37
D
38 B
39 C
40
B
41 A
42 C
43
A
44 A
45 B
46. D
47. A
48. C
49.
A
50. B
B
D
6
D
OBJECTIVE PROBLEMS (JEE ADVANCED)
Answer Ex–II 1.
ABC
2. ACD
3. BC
4.
AC
5
7
CD
8
9
10
D
11 C
13
D
14 D
15.A-P, B-P,Q,S, C-P,R, D-Q,S
16. A-S, B-R, C-Q, D-P
17
A
18 C
19 A
20
A
21 C
22 B
23
C
24 D
25 C
26
A
27 C
28 8
29
180
30 121
31 16
32
1
33 5
34 4
35
5
36 7
37.A,C
38. B, C
39. B
40. B
41.
A,C,D
42. A, D
43.A,C,D
44. A
45. B
46. A
47.
A,B
48. A,C,D
49. A,C
50. 487.06 nm
51. A
52. B
53.
D
B
A
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AC
6
BCD
12 D
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ATOMIC STRUCTURE
Page # 77
SUBJECTIVE PROBLEMS (JEE ADVANCED)
Answer Ex–III 1.
B
2. B
3. C
4.
A
5
7
A
8
9
B
10
B
11 B
12 A
13
A
14 A
15 B
16
D
17 C
18 B
19
B
20 C
21 C
22
A
23 D
24 B
25
B
26 B
27 A
28
A
29 A
30 D
31
D
32 C
33 B
34
C
C
Answer Ex–IV
B
6
D
PREVIOUS YEARS PROBLEMS
LEVEL – I
JEE MAIN
1.
A
2. C
3. A
4.
D
5
7
CD
8
9
A
10
A
11 C
12 D
13
C
14 B
15 B
16
C
17. C
18. B
19.
D
20. B
21.B
22. D
23. B
24. D
25.
B
26. C
27.B
28. B
B
LEVEL – II 6.3×108 m/s
1. 6.
A
11.
98.2 kJ
17.
2.
+2 23V
: [Ar] 3d1
A
4.
ABC
5.
8.
B
9.
C
10. A
14. C
6.6 × 10–35 s–1
18.A
19. 3.34 × 10–10 m
20.
A-Q, B-P, C-R, D-S
21.A–QR; B–P,Q,RS ; C–P,Q,R; D–P,Q
24.
C
26.0009
30. AC
C
3.
13.D
25. B
6
JEE ADVANCED
7. AD 12. D
B
27. 0004
15. D
22. A
28. C
B
16. 2a0
23. B 29. 1
31. 0005
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Page # 78
STOICHIOMETRY - 1
CLASSIFICATION OF MATTER
LAWS OF CHEMICAL COMBINATION (a)
Law of conservation of mass [Lavoisier] In a chemical change total mass remains conserved i.e. mass before the reaction is always equal to mass after the reaction. H2 (g)
+
1/2 O2
H2O ()
(g)
1 mole 1/2 mole
1 mole
mass before the reaction = 1 × 2 + 1/2 × 32 = 18 gm mass after the reaction = 1 × 18 = 18 gm Ex. A 15.9g sample of sodium carbonate is added to a solution of acetic acid weighing 20.0g. The two substances react, releasing carbon dioxide gas to the atmosphere. After reaction, the contents of the reaction vessel weigh 29.3g. What is the mass of carbon dioxide given off during the reaction? Sol. The total mass of reactants taken = 15.9 + 20.0 = 35.9 gm. From the conservation of mass, the final mass of the contents of the vessel should also be 35.9 gm. But it is only 29.3 gm. The difference is due to the mass of released carbon dioxide gas. Hence, the mass of carbon dioxide gas released = 35.9 – 29.3 = 6.6 gm (b)
Law of constant composition [Proust] All chemical compounds are found to have constant composition irrespective of their method of prepration or sources.
In H2O, Hydrogen & oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is Tap water, river water or seawater or produced by any chemical reaction. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
STOICHIOMETRY - 1 Page # 79 Ex. The following are results of analysis of two samples of the same or two different compounds of phosphorus and chlorine. From these results, decide whether the two samples are from the same or different compounds. Also state the law, which will be obeyed by the given samples. Amount P Amount Cl Compound A
1.156 gm
3.971 gm
Compound B
1.542 gm
5.297 gm
Sol. The mass ratio of phosphorus and chlorine in compound A, mP : mCl = 1.156:3.971 = 0.2911:1.000 The mass ratio of phosphorus and chlorine in compound B, mP : mCl = 1.542:5.297 = 0.2911:1.000 As the mass ratio is same, both the compounds are same and the samples obey the law of definite proportion. (c)
Law of multiple proportions [Dalton] When one element combines with the other element to form two or more different compounds, the mass of one element, which combines with a constant mass of the other bear a simple ratio to one another.
Carbon is found to form two oxides which contain 42.9% & 27.3% of carbon respectively show that these figures shows the law of multiple proportion. First oxide Second oxide Carbon 42.9 % 27.3 % Oxygen 57.1 % 72.7% Given In th first oxide, 57.1 parts by mass of oxygen combine with 42.9 parts of carbon. 1 part of oxygen will combine with
42.9 part of carbon = 0.751 57.1
Similarly in 2nd oxide 27.3 part of carbon = 0.376 72.7 The ratio of carbon that combine with the same mass of oxygen = 0.751 : 0.376 = 2 : 1 This is a simple whole no ratio this means above data shows the law of multiple proportion.
1 part of oxygen will combine with
Ex.
Two oxide samples of lead were heated in the current of hydrogen and were reduced to the metallic lead. The following data were obtained (i) Weight of yellow oxide taken = 3.45 gm; Loss in weight in reduction = 0.24 gm (ii) Weight of brown oxide taken = 1.227 gm; Loss in weight in reduction = 0.16 gm. Show that the data illustrates the law of multiple proportion.
Sol. When the oxide of lead is reduced in the current of hydrogen, metallic lead is formed. Definitely, the loss in weight in reduction is due to removal of the oxygen present in the oxide, to combine with the hydrogen. Therefore, the composition of the yellow oxide is: oxygen = 0.24 gm and lead = 3.45 – 0.24 = 3.21 gm. The mass ratio of lead and oxygen, r1 =
m Pb 3.21 13.375 m O 0.244 1.000
and the compositon of the brown oxide is : oxygen = 0.16 gm and lead = 1.227 – 0.16 = 1.067 gm. The mas ratio of lead and oxygen, r2 =
m Pb 1.067 6.669 mO 0.16 1.000
Now, r1 : r2 = 13.375 : 6.669 = 2.1 (simple ratio) and hence the data illustrates the law of multiple proportion.
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Page # 80 STOICHIOMETRY - 1 (d) Law of reciprocal proportions [Richter] When two elements combine seperately with definite mass of a third element, then the ratio of their masses in which they do so is either the same or some whole number multiple of the ratio in which they combine with each other. This law can be understood easily with the help of the following examples. At. Mass 1 H
At. Mass 32
H2S
S
H
SO
2
2O
O At. Mass 16
Let us consider three elements – hydrogen, sulphur and oxygen. Hydrogen combines with oxygen to form H2O whereas sulphur combines with it to form SO2. Hydrogen and sulphur can also combine together to form H2S. The formation of these compounds is shown in fig. In H2O, the ratio of masses of H and O is 2 : 16. In SO2, the ratio of masses of S and O is 32 : 32. Therefore, the ratio of masses of H and S which combines with a fixed mass of oxygen (say 32 parts) will be 4 : 32 i.e. 1 : 8 ...(1) When H and S combine together, they form H2S in which the ratio of masses of H and S is 2 : 32 i.e., 1 : 16 ...(ii) The two ratios (i) and (ii) are related to each other as 1 1 : 8 16
or
2 :1
i.e., they are whole number multiples of each other. Thus, the ratio masses of H and S which combines with a fixed mass of oxygen is a whole number multiple of the ratio in which H and S combine together. Ex. Methane contains 75 % carbon and 25% hydrogen, by mass. Carbon dioxide contains 27.27 % carbon and 72.73% oxygen, by mass. Water contains 11.11 % hydrogen and 88.89% oxygen, by mass. Show that the data illustrates the law of reciprocal proportion. Sol. Methane and carbon dioxide, both contains carbon and hence, carbon may be considered as the third element. Now, let the fixed mass of carbon = 1 gm. Then, the mass of hydrogen combined with 1 gm carbon in methane =
25 1 gm 75 3
and the mass of oxygen combined with 1 gm carbon in carbon dioxide =
72.73 8 gm 27.27 3
Hence, the mass ratio of hydrogen and oxygen combined with the fixed mass of carbon, r1 =
Now, the mass ratio of hydrogen and oxygen in water, r2 =
1 8 1 : 3 3 8
11.11 1 88.89 8
As r1 and r2 are same , the data is according to the law of reciprocal proportion.
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(e)
STOICHIOMETRY - 1 Page # 81 Gay Lussac law of combining volumes : When two or more gases react with one another, their volumes bear simple whole number ratio with one another and to the volume of products (if they are also gases) provided all volumes are measured under identical conditions of temperature and pressure. When gaseous hydrogen and gaseous chlorine react together to form gaseous hydrogen chloride according to the following equation.
H ( g ) Cl 2 ( g ) 2HCl ( g )
2 one volume
one volume
two volumes
It has been observed experimentally that in this reaction, one volume of hydrogen always reacts with one volume of chlorine to form two volumes of gaseous hydrogen chloride. all reactants and products are in gaseous state and their volumes bear a ratio of 1 : 1 : 2. This ratio is a simple whole number ratio. “These are no longer useful in chemical calculations now but gives an idea of earlier methods of analysing and relating compounds by mass.” Ex. 2.5 ml of a gaseous hydrocarbon exactly requires 12.5 ml oxygen for complete combustion and produces 7.5 ml carbon dioxide and 10.0 ml water vapour. All the volumes are measured at the same pressure and temperature. Show that the data illustrates Gay Lussac’s law of volume combination. Sol. Vhydrocarbon : Voxygen : Vcarbon dioxide : Vwater vapour
= 2.5 : 12.5 : 7.5 : 10.0
= 1 : 5 : 3 : 4 (simple ratio) Hence, the data is according to the law of volume combination. MOLE CONCEPT Definition of mole : One mole is a collection of that many entities as there are number of atoms exactly in 12 gm of C – 12 isotope. or 1 mole = collection of 6.02 × 1023 species 6.02 × 1023 = NA = Avogadro’s No. 1 mole of atoms is also termed as 1 gm-atom, 1 mole of ions is termed as 1 gm-ion and 1 mole of molecule termed as 1 gm-molecule. METHODS OF CALCULATIONS OF MOLE (a) If no. of some species is given, then no. of moles =
Given no. NA
(b) If weight of a given species is given, then no of moles =
or =
Given wt ( for atoms), Atomic wt.
Given wt. (for molecules) Molecular wt.
(c) If volume of a gas is given along with its temperature (T) and pressure (P) use n =
PV RT
where R = 0.0821 lit-atm/mol–K (when P is in atmosphere and V is in litre.) 1 mole of any gas at STP (0°C & 1 bar) occupies 22.7 litre. 1 mole of any gas at STP (0°C & 1 atm) occupies 22.4 litre. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 82 Atom :
STOICHIOMETRY - 1 Atom is smallest particle which can not be divided into its constituents.
Atomic weight : It is the weight of an atom relative to one twelvth of weight of 1 atom of C-12 RELATIONSHIP BETWEEN GRAM AND AMU 1 amu =
1 wt of one C - 12 atom. 12
for C 1 mole C = 12 gm = 6.023 × 1023 atoms wt of 6.023 × 1023 atoms = 12 gm 12 wt of 1 atom of C = N gm (NA Avogadro’s number = 6.23 × 1023) A
1 amu =
1 wt of one C - 12 atom 12
1 12 = 12 N gm A 1 1 amu = N gm A
ELEMENTAL ANALYSIS
Ex.
For n mole of a compound (C3H7O2) Moles of C = 3n Moles of H = 7n Moles of O = 2n Find the wt of water present in 1.61 g of Na2SO4. 10H2O
Sol.
Moles of Na2SO4. 10H2O =
wt. in gram 1.61 = = 0.005 moles molecular wt 322
Moles of water = 10 × moles of Na2SO4. 10H2O = 10 × 0.05 = 0.05 wt of water = 0.5 × 18 = 0.9 gm Ans. AVERAGE ATOMIC WEIGHT = % of isotope X molar mass of isotope. The % obtained by above expression (used in above expression) is by number (i.e. its a mole%) MOLECULAR WEIGHT It is the sum of the atomic weight of all the constituent atom.
n M n i
(a)
Average molecular weight =
i
i
where ni = no. of moles of any compound and mi = molecular mass of any compound. Make yourselves clear in the difference between mole% and mass % in question related to above. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
STOICHIOMETRY - 1 Page # 83 Shortcut for % determination if average atomic weight is given for X having isotopes XA & XB. % of X A
Average atomic wei ght – wt of X B difference in weight of X A & X B
100
Try working out of such a shortcut for XA, XB, XC EMPIRICAL FORMULA, MOLECULAR FORMULA Empirical formula : Formula depicting constituent atom in their simplest ratio. Molecular formula : Formula depicting actual number of atoms in one molecule of the compound. Relation between the two : Molecular formula = Empirical formula × n n
Molecular mass Empirical Formula mass
Ex.
Check out the importance of each step involved in calculations of empirical formula. A molecule of a compound have 13 carbon atoms, 12 hydrogen atom, 3 oxygen atoms and 3.02 × 10–23 gm of other element. Find the molecular wt. of compound.
Sol.
12 1 16 wt. of the 1 molecule of a compound = 13 × N + 12 × N + 3 × N + 3.02 × 10–23 A A A
•
156 12 48 3.02 10 –23 NA = 234.18 / NA = 234 amu. Ans. NA
Density : (a) Absolute density (b) Relative density Absolute density =
Relative density =
Mass volume
density of subs tan ce density of s tan dard subs tan ce
density of subs tan ce Specific gravity = density of H O at 4 C 2
Vapour density : It is defined only for gas. It is a density of gas with respect to H2 gas at same temp & press PMgas / RT Mgas dgas M V.D = dH = PM / RT = M = H2 H2 2 2
V.D =
M 2
Molecular wt of gas V.D = Molecular wt of H gas 2
•
density of Cl2 gas with respect to O2 gas =
Molecular wt of Cl2 gas Molecular wt of O 2 gas
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Page # 84
•
•
STOICHIOMETRY - 1
STOICHIOMETRY : Stoichiometry is the calculations of the quantities of reactants and products involved in a chemical reaction. Following methods can be used for solving problems. (a)
Mole Method
(For Balance reaction)
(b)
POAC method } Balancing not required but common sense ------- use it with slight care.
(c)
Equivalent concept
CONCEPT OF LIMITING REAGENT. Limiting Reagent : It is very important concept in chemical calculation. It refers to reactant which is present in minimum stoichiometry quantity for a chemical reaction. It is reactant consumed fully in a chemical reaction. So all calculations related to various products or in sequence of reactions are made on the basis of limiting reagent. • It comes into picture when reaction involves two or more reactants. For solving any such reactions, first step is to calculate L.R. Calculation of Limiting Reagent. (a) By calculating the required amount by the equation and comparing it with given amount. [Useful when only two reactant are there] (b) By calculating amount of any one product obtained taking each reactant one by one irrespective of other reactants. The one giving least product is limiting reagent. (c) Divide given moles of each reactant by their stoichiometric coefficient, the one with least ratio is limiting reagent. [Useful when number of reactants are more than two.]
•
PERCENTAGE YIELD : The percentage yield of product =
actual yield 100 the theoretical maximum yield
• The actual amount of any limiting reagent consumed in such incomplete reactions is given by [% yield × given moles of limiting reagent] [For reversible reactions] • For irreversible reaction with % yield less than 100, the reactants is converted to product (desired and waste.) Ex.
A compound which contains one atom of X and two atoms of y for each three atoms of z is made of mixing 5 gm of x, 1.15 × 1023 atoms of Y and 0.03 mole of Z atoms. Given that only 4.40 gm of compound results. Calculate the atomic weight of Y if atomic weight of X and Z are 60 and 80 respectively.
Sol.
Moles of x = moles of y =
5 1 = = 0.083 60 12
115 . 10 23
6.023 10 23 moles of z = 0.03
0.19
x + 2y + 3z xy2z3 for limiting reagent, 0.083/1 = 0.083 0.03 0.19 0.01 0.095 , 3 2 Hence z is limiting reagent
wt of xy2z3 = 4.4 gm = moles × molecular wt.
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STOICHIOMETRY - 1
Page # 85
1 moles of xy2z3 = 0.03 = 0.01 3 300 + 2 m = 440 2m = 440 – 300 m = 70 Ans.
P O A C Rule : P O A C is the simple mass conservation. KClO3 KCl + O2 Apply the POAC on K. moles of K in KClO3 = moles of K in KCl 1 × moles of KClO3 = 1 × moles of KCl moles of KClO3 = moles of KCl Apply POAC on O moles O in KClO3 = moles of O in O2 3 × moles of kClO3 = 2 × moles of O2 Ex.
In the gravimetric determination of phosphorous, an aqueous solution of dihydrogen phosphate ion (H2PO4–) is treated with a mix of ammonium & magnesium ions to precipitate magnesium ammonium phosphate MgNH4 PO4.6H2O. This is heated and decomposed to magnesium Pyrophosphate, Mg2P2O7 which is weighted. A solution of H2PO4– yielded 1.054 gm of Mg2P2O7 what weight of NaH2 PO4 was present originally. NaH2PO4
Mg2P2O7
apply POAC on P Let wt of NaH2PO4 = w gm moles of P in NaH2PO4 = moles of P in Mg2P2O7 w 1.054 1 2 120 232
w = 1.054 ×
120 2 = 1.09 gm Ans. 232
SOME EXPERIMENTAL METHODS FOR DETERMINATION OF ATOMIC MASS Dulong’s and Petit’s Law : Atomic weight × specific heat (cal/gm°C) 6.4 Gives approximate atomic weight and is applicable for metals only. Take care of units of specific heat. FOR MOLECULAR MASS DETERMINATION (a)
Victor Maeyer’s process : (for volatile substance) Procedure : Some known weight of a volatile substance (w) is taken, converted to vapour and collected over water. The volume of air displaced over water is given (V) and the following expressions are used. M=
w RT PV
If aq. tension is not given
or
M
w RT (P – P') V
If aq. tension is P
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Page # 86
STOICHIOMETRY - 1 Aqueous tension : Pressure exerted due to water vapours at any given temperature. This comes in picture when any gas is collected over water. Can you guess why ?
(b)
Silver salt method : (for organic acids) Basicity of an acid : No. of replacible H+ atoms in an acid (H bonded to more electronegative atom is acidic) Procedure : Some known amount of silver salt (w1 gm) is heated to obtain w2 gm of white shining residue of silver. Then if the basicity of acid is n, molecular weight of acid would be AgnA nAg+ + A–n Agn A is the salt w 2 1 Msalt w 1 and molecular weight of acid = M – n(108) salt 108 n
This is one good practicle application of POAC. (c)
Chloroplatinate salt method : (for organic bases) Lewis acid : electron pair acceptor Lewis base : electron pair donor Procedure : Some amount of organic base is reacted with H2PtCl6 and forms salt known as chloroplatinate. If base is denoted by B then salt formed. (i) with monoacidic base = B2H2PtCl6 (ii) with diacidic base = B2(H2PtCl6)2 (iii) with triacidic base = B2(H2PtCl6)3 The known amount (w1 gm) of salt is heated and pt residue is measured. (w2 gm). If acidity of base is w 2 1 M – n(410) ‘n’ then 195 n × Msalt = w1 and Mbase = salt 2
•
(a)
For % determination of elements in organic compounds : •
All these methods are applications of POAC
•
Do not remember the formulas, derive them using the concept, its easy.
Liebig’s method : (Carbon and hydrogen) (w) Organic Compound (w1) CO2 + H2O (w2) CuO
w 12 100 % of C = 1 44 w w 2 % of H = 2 100 18 w where w1 = wt. of CO2 produced, w2 = wt. of H2O produced,
w = wt, of organic compound taken (b)
Duma’s method : (for nitrogen) N (P, V, T given) (w) Organic Compound CuO 2
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STOICHIOMETRY - 1
Page # 87
use PV = nRT to calculate moles of N2, n. n 28 100 w w = wt of organic compound taken
% of N =
(c)
Kjeldahl’s method : (for nitrogen) (w) O.C. + H2SO4 (NH4)2SO4 NaOH NH3 + H2SO4 (molarity M and volume (V1) consumed given) MV1 2 14 100 w where M = molarity of H2SO4.
% of N =
• (d)
Some N containing compounds do not give the above set of reaction as in Kjeldahl’s method.
Sulphur : (w) O.C. + HNO3 H2SO4 + BaCl2 (w1) BaSO4
% of S =
w1 233
×1×
32 × 100 w
where w1 = wt. of Ba SO4, w = wt. of organic compound (e)
Phosphorus : O.C+ HNO3 H3PO4 + [NH3 + magnesia mixture ammonium molybdate] MgNH4 PO4 Mg2P2O7
w1 2 31 100 222 w Carius method : (Halogens)
% of P = (f)
O.C. + HNO3 + AgNO3 AgX If X is Cl then colour = white If X is Br then colour = dull yellow If X is I then colour = bright yellow •
Flourine can’t be estimated by this % of X
w1 1 ( At. wt of X ) 100 (M. weight of AgX ) w
Ex.
0.607 g of a silver salt of a tribasic organic acid was quantitatively reduced to 0.370 g of pure silver. Calculate the molecular weight of the acid (Ag = 108)
Sol.
Suppose the tribasic acid is H3A. H3A Ag3A Ag acid
salt
0.607 g 0.37 g Since Ag atoms are conserved, applying POAC for Ag atoms, moles of Ag atoms in Ag3A = moles of Ag atoms in the prduct 3 × moles of Ag3A = moles of Ag in the product
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Page # 88
STOICHIOMETRY - 1 3
0.607 0.37 mol. wt. of Ag3 A 108
(Ag = 108)
mol. wt. of Ag3A = 531. mol. weight of tribasic acid (H3A) = mol wt. of the salt (Ag3A) – 3 × at. wt. of Ag + 3 × at. wt. of H = 531 – 324 + 3 = 210 Ans.
EUDIOMETRY [For reactions involving gaseous reactants and products] •
The stoichiometric coefficient of a balanced chemical reactions also gives that ratio of volumes in which gasesous reactants are reacting and products are formed at same temperature and pressure. The volume of gases produced is often given by mentioning certain solvent which absorb contain gases. Solvent gas(es) absorb KOH CO2, SO2, Cl2 Ammon Cu2Cl2 CO Turpentine oil O3 Alkaline pyrogallol O2 water NH3, HCl CuSO4/CaCl2 H2O Assumption : On cooling the volume of water is negligible Ex.
7.5 mL of a hydrocarbon gas was exploded with excess of oxygen. On cooling, it was found to have undergone a contraction of 15 mL. If the vapour density of the hydrocarbon is 14, determine its molecular formula. (C = 12, H = 1)
Sol.
CxHy + (x +
y y ) O2 X CO2 + HO 4 2 2
7.5 ml
15
on cooling the volume contraction = 15 ml i.e. The volume of H2O (g) = 15 ml V.D. of hydrocarbon = 14 Molecular wt. of CxHy = 28 12x + y = 28 ...(1) From reaction 7.5
y = 15 2
y=4
12 x + 4 = 28 12x = 24 x=2 Hence Hydrocalbon is C2H4.
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STOICHIOMETRY - 1
Page # 89 CONCENTRATION OF SOLUTION
(a)
(b)
(c) (d)
Concentration of solution can be expressed in any of the following ways. % by wt amount of solute dissolved in 100 gm of solution 4.9% H2SO4 by wt. 100 gm of solution contains 4.9 gm of H2SO4 % by volume volume of solute dissolved in 100 ml of solution x% H2SO4 by volume 100 ml of solution contains x ml H2SO4 % wt by volume wt. of solute present in 100 ml of solution % volume by wt. volume of solute present in 100 gm of solution. CONCENTRATION TERMS
•
Molarity (M) : No. of moles of solute present in 1000 ml of solution. moles of solute volume of solution (lit) m.moles of solute M= volume of solution(m l) molarity (M) =
MOLALITY (m) No. of moles of solute present in 1000 gm of solvent m=
moles of solute wt.of solvent in kg
m=
m.moles of solute wt.of solvent in gm
NORMALITY (N) No of gm equivalents of solute present in 1000 ml of solution N=
gm equivalents of solute m. equivalent of solute = volume of solution(lit) volume of solution in (ml)
FORMALITY (f) The formality is the no. of gm -formula weights of the ionic solute present in 1000 ml of solution. wt in gm f = formula wt volume of solution (lit)
MOLE FRACTION The mole fraction of a perticular component in a solution is defined as the number of moles of that component per mole of solution. If a solution has nA mole A & nB mole of B. nA mole fraction of A (XA) = n n A B nB mole fraction of B (XB) = n n A B XA + XB = 1
•
Parts per million (ppm) : =
Mass of solute Mass of solvent
× 106
Mass of solute 10 6 Mass of solution
VOLUME STRENGTH OF H2O2 Strength of H2O2 is represented as 10V, 20V, 30V etc. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 90 STOICHIOMETRY - 1 20V H2O2 means one litre of this sample of H2O2 on decomposition gives 20 It of O2 gas at S.T.P. Decomposition of H2O2 is given as H2O2
H2O +
1 O2 2
•
1 × 22.4 It O2 at S.T.P.. 2 = 34 g = 11.2 It O2 at S.T.P. To obtain 11.2 litre O2 at S.T.P. at lest 34 gm H2O2 must be decomposed
•
for 20 It O2, we should decompose atleast
1 It solution of H2O2 contains
34 20 gm H2O2 112 .
1 It solution of H2O2 contains
34 20 equivalents of H2O2 112 . 17
•
Normality of H2O2 =
1 mole
MH2O 2
NH2O 2 v. f
(EH2O 2
M 34 17) 2 2
34 20 20 112 . 17 5.6 Volume, strength of H2 O 2 5.6
Normality of H2O2(N) =
34 × 20 gm H2O2 112 .
NH2O 2 2
IInd Method : 1 O2 2 From law of equivalence gm eq. of O2 = gm eq. of H2O2
H2O2 H2O +
gm eq. of O2 = moles × n factor of O2, =
20 20 4 = 22.4 5.6
20 5.6 and the volume of H2O2 is 1 lit.
gm. eq. of H2O2 =
this means 1 lit of H2O2 have i.e. Normality N =
20 gm eq. 5.6
20 5.6
NORMALITY OF H2O2 =
volume strength of H2 O 2 5.6
Volume, strength of H2 O 2 11 .2 Strength (in g/) : Denoted by S Strength = molarity × mol. wt. = molarity × 34 strength = Normality × Eq. weight. = Normality × 17
•
Molarity of H2O2(M) =
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Ex. Sol.
STOICHIOMETRY - 1 Page # 91 A bottle labeled with “12V H2O2” contain 700 ml solution. If a sdudent mix 300 ml water in it what is the g/litre strenth & normality and volume strength o final solution. N=
12 5.6
12 700 5.6 let the normality of H2O2 on dilution is N meq. before dilution = meq. after dilution
meq. of H2O2 =
12 700 5.6
N × 1000 =
N=
12 7 × = 1.5 5.6 10
M=
15 . 2
15 . 34 = 25.5 2 84 volume strength = N × 5.6 = = 8.4 V Ans. 10 Strength of Oleum Oleum is SO3 dissolved in 100% H2SO4. Sometimes, oleum is reported as more then 100% by weight, say y% (where y > 100). This means that (y – 100) grams of water, when added to 100 g of given oleum sample, will combine with all the free SO3 in the oleum to give 100% sulphuric acid.
strength gm/lit =
80( y – 100) 18 Calculate the percentage of free SO3 in an oleum (considered as a solution of SO3 in H2SO4) that is labelled '109% H2SO4'. '109% H2SO4' refers to the total mass of pure H2SO4, i.e., 109 g that will be formed when 100 g of oleum is diluted by 9 g of H2O which (H2O) combines with all the free SO3 present in oleum to form H2SO4 H2O + SO3 H2SO4 1 mole of H2O combines with 1 mole of SO3 or 18 g of H2O combines with 80 g of SO3 or 9 g of H2O combines with 40 g of SO3. Thus, 100 g of oleum contains 40 g of SO3 or oleum contains 40% of free SO3.
Hence weight % of free SO3 in oleum = Ex. Sol.
Ex.
A 62% by mass of an aqueous solution of acid has specific gravity 1.8. This solution is diluted such that the specific gravity of solutin became 1.2. Find the % by wt of acid in new solutiuon.
Sol.
density =
mass volume
1.8 =
100 n
volume of sol Let x gm water is added in soluion
then
d= 12 .
12 .
volume of solution =
100 18 .
mass volume 100 x 100 x 18 .
100 12 . x 100 x 18 .
200 12 . x 100 x 3
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Page # 92
STOICHIOMETRY - 1 0.2 x = 100 –
200 100 = 3 3
100 1000 500 = = = 166.67 3 0.2 6 3 mass of new solution = 100 + 166.67 = 266.67 266.67 gm solution contains 62 gm of acid
x=
% by mass =
62 100 = 23.24 % 266.67
RELATION SHIP BETWEEN MOLARITY, MOLALITY & DENSITY OF SOLUTION Let the molarity of solution be 'M', molality be 'm' and the density of solution be d gm/m. Molarity implies that there are M moles of solute in 1000 ml of solution wt of solution = density × volume = 1000 d gm wt of solute = MM1 where M1 is the molecular wt of solute wt of solvent = (1000d – MM1) gm (1000d – MM1) gm of solvent contains M moles of solute M 1000 gm of solvent have = 1000d – MM 1000 mole = Molality 1 1000 M no. of moles of solute present in 1000 gm of solvent = 1000d – MM = Molality 1 M1 1 on simplyfying d M m 1000
RELATION SHIP BETWEEN MOLALITY & MOLE FRACTION consider a binary solution consisting of two components A (Solute) and B (Solvent). Let xA & xB are the mole fraction of A & B respectively. nA nB xA = n n , xb = n n A B A B If molality of solution be m then nA nA 1000 1000 = n mass of solvent B MB where MB is the molecular wt of the solvent B
m
m
x A 1000 xB MB mole fraction of A
1000
molality = mole fraction of B M B m=
mole fraction of solute 1000 mole fraction of solvent molecular wt. of solvent
Ex.
An aqueous solution is 1.33 molal in methanol. Determine the mole fraction of methanol & H2O
Sol.
molality =
mole fraction of solute 1000 mole fraction of solvent mol.wt of solvent xA 1.33 18 xA 23.94 x A 1.33 = x M 1000 , 1000 x , 1000 x B B B B xA = 0.02394 xB, xA + xB = 1
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STOICHIOMETRY - 1 1.02394 xB = 1 xB
Page # 93
1 = 0.98, xA = 0.02 Ans. 102394 .
Second Method : Let wt of solvent = 1000 gm molality = 1.33 = moles of solute m 1.33 moles of solute mole fraction of solute = , m + 1000 1.33 1000 / 18 moles of solute moles of solvent 18
mole fraction of solute = 0.02 mole fraction of solvent = 1 – 0.02 = 0.98 Ex.
Sol.
The density of 3 M solution of sodium thiosulphate (Na2S2O3) is 1.25 g/mL. Calculate (i) amount of sodium thiosulphate (ii) mole fraction of sodium thiosulphate (iii) molality of Na+ and S2O32– ions (i) Let us consider one litre of sodium thiosulphate solution. wt. of the solution = density × volume (mL) = 1.25 × 1000 = 1250 g. wt. of Na2S2O3 present in 1 L of the solution = molarity × mol. wt. = 3 × 158 = 474 g. Ans. 474 100 = 37.92% 1250 (ii) Wt. of solute (Na2S2O3) = 474 g.
wt. % of Na2S2O3 =
474 3 Ans. 158 Wt. of solvent (H2O) = 1250 – 474 = 776 g
Moles of solute =
Moles of solvent =
776 43.11 18
mole fraction of Na2S2O3 =
3 0.063 3 43.11
moles of Na2 S2 O 3 1000 = 3 1000 3.865 wt. of solvent in grams 776 + 1 mole of Na2S2O3 contains 2 moles of Na ions and 1 mole of S2O32– ions. molality of Na+ = 2 × 3.865 = 7.73 m Molality of S2O32– = 3.865 m. Ans. (iii) Molality of Na2S2O3 =
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Page # 94
STOICHIOMETRY - 1
Solved Objective Ex.1
Sol.
Ex.2
Sol.
8 litre of H2 and 6 litre of Cl 2 are allowed to react to maximum possible extent. Find out the final volume of reaction mixture. Suppose P and T remains constant throughout the course of reaction (A) 7 litre (B) 14 litre (C) 2 litre (D) None of these. (B) H2 + Cl 2 2 HCl Volume before reaction 8 lit 6 lit 0 Volume after reaction 2 0 12 Volume after reaction = Volume of H2 left + Volume of HCl formed = 2 + 12 = 14 lit Naturally occurring chlorine is 75.53% Cl35 which has an atomic mass of 34.969 amu and 24.47% Cl 37 which has a mass of 36.966 amu. Calculate the average atomic mass of chlorine(A) 35.5 amu (B) 36.5 amu (C) 71 amu (D) 72 amu (A) Average atomic mass % of I isotope its atomic mass 75.53 x 34.969 24.47 x 36.96 = % of II isotope its atomic mass = 100 100
Ex.3
= 35.5 amu.
Sol.
Calculate the mass in gm of 2g atom of Mg(A) 12 gm (B) 24 gm (C) 6 gm (D) 1 gm atom of Mg has mass = 24 gm 2 gm atom of Mg has mass = 24 x 2 = 48 gm.
Ex.4
In 5 g atom of Ag (At. wt. of Ag = 108), calculate the weight of one atom of Ag -
Sol.
Ex.5 Sol.
Ex.6 Sol.
Ex.7
(D) None of these.
(A) 17.93 × 10–23gm (B) 16.93 × 10–23 gm (C) 17.93 × 1023 gm (D) 36 × 10–23 gm (A) N atoms of Ag weigh 108 gm 108 108 1 atom of Ag weigh = = = 17.93 × 10–23 gm. 6 . 023 x 10 23 N In 5g atom of Ag (at. wt. = 108), calculate the no. of atoms of Ag (A) 1 N (B) 3N (C) 5 N (D) 7 N. (C) 1 gm atom of Ag has atoms = N 5 gm atom of Ag has atoms = 5N. Calculate the mass in gm of 2N molecules of CO2 (A) 22 gm (B) 44 gm (C) 88 gm (D) None of these. (C) N molecules of CO2 has molecular mass = 44. 2N molecules of CO2 has molecular mass = 44 x 2 = 88 gm. How many carbon atoms are present in 0.35 mol of C6H12O6 (A) 6.023 × 1023 carbon atoms (B) 1.26 × 1023 carbon atoms (C) 1.26 × 1024 carbon atoms (D) 6.023 × 1024 carbon atoms
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STOICHIOMETRY - 1 (C) 1 mol of C6H12O6 has = 6 N atoms of C 0.35 mol of C6H12O6 has = 6 × 0.35 N atoms of C = 2.1 N atoms = 2.1 × 6.023 × 1023 = 1.26 × 1024 carbon atoms Ex.8 How many molecules are in 5.23 gm of glucose (C6H12O6) -
Page # 95
Sol.
(A) 1.65 × 1022 (B) 1.75 × 1022 Sol. (B) 180 gm glucose has = N molecules 5.23 gm glucose has =
(C) 1.75 × 1021(
5.23 6.023 10 23 180
D) None of these
= 1.75 × 1022 molecules
What is the weight of 3.01 × 1023 molecules of ammonia (A) 17 gm (B) 8.5 gm (C) 34 gm Sol. (B) 6.023 × 1023 molecules of NH3 has weight = 17 gm Ex.9
(D) None of these
3.01 × 1023 molecules of NH3 has weight =
17 3.01 10 23 6.023 10 23
= 8.50 gm
Ex.10 How many significant figures are in each of the following numbers (a) 4.003 (b) 6.023 × 1023 (c) 5000 (A) 3, 4, 1 (B) 4, 3, 2 (C) 4, 4, 4 (D) 3, 4, 3 Sol. (C) Ex.11 How many molecules are present in one ml of water vapours at STP (A) 1.69 × 1019 (B) 2.69 × 10–19 (C) 1.69 × 10–19 (D) 2.69 × 1019 Sol. (D) 22.4 litre water vapour at STP has = 6.023 × 1023 molecules 1 × 10–3 litre water vapours at STP has =
6.023 10 23 22.4
× 10–3 = 2.69 × 10+19
Ex.12 How many years it would take to spend Avogadro's number of rupees at the rate of 1 million rupees in one second (A) 19.098 × 1019 years (B) 19.098 years Sol.
(C) 19.098 × 109 years (C) 106 rupees are spent in 1sec. 6.023 × 1023 rupees are spent in
=
1 6.023 10 23 6
10 60 60 24 365
(D) None of these
=
1 6.023 10 23 10 6
sec
years , = 19.098 × 109 year
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Page # 96 STOICHIOMETRY - 1 Ex.13 An atom of an element weighs 6.644 × 10–23 g. Calculate g atoms of element in 40 kg(A) 10 gm atom (B) 100 gm atom (C) 1000 gm atom (D) 104 gm atom Sol. (C) weight of 1 atom of element = 6.644 × 10–23 gm weight of 'N' atoms of element = 6.644 × 10–23 × 6.023 × 1023 = 40 gm 40 gm of element has 1 gm atom. 40 x 103 gm of element has
40 10 3 40
, = 103 gm atom.
Ex.14 Calculate the number of Cl – and Ca+2 ions in 222 g anhydrous CaCl 2 (A) 2N ions of Ca+2 4 N ions of Cl – (B) 2N ions of Cl – & 4N ions of Ca+2 +2 – (C) 1N ions of Ca & 1N ions of Cl (D) None of these. Sol. (A) mol. wt. of CaCl 2 = 111 g 111 g CaCl 2 has = N ions of Ca+2 Also
222g of CaCl 2 has
N 222 111
= 2N ions of Ca+2 111 g CaCl2 has = 2N ions of Cl –
222 g CaCl 2 has =
2N 222 ions of Cl – 111
= 4N ions of Cl – . Ex.15 The density of O2 at NTP is 1.429g / litre. Calculate the standard molar volume of gas(A) 22.4 lit. (B) 11.2 lit (C) 33.6 lit (D) 5.6 lit. Sol. (A) 1.429 gm of O2 gas occupies volume = 1 litre. 32 gm of O2 gas occupies =
32 ,= 22.4 litre/mol. 1429 .
Ex.16 Which of the following will weigh maximum amount(A) 40 g iron (B) 1.2 g atom of N (C) 1 × 1023 atoms of carbon (D) 1.12 litre of O2 at STP Sol. (A) (A) Mass of iron = 40 g (B) Mass of 1.2 g atom of N = 14 × 1.2 = 16.8 gm (D) Mass of 1 × 1023 atoms of C (D) Mass of 1.12 litre of O2 at STP
=
12 1 10 23
= 1.99 gm. 6.023 10 23 32 1.2 = = 1.6 g 22 .4
Ex.17 How many moles of potassium chlorate to be heated to produce 11.2 litre oxygen 1 1 1 2 (A) mol (B) mol (C) mol (D) mol. 2 3 4 3 Sol. (B) 2 KClO3 2KCl + 3O2 Mole for reaction 2 2 3 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
STOICHIOMETRY - 1 3 × 22.4 litre O2 is formed by 2 mol KClO3 2 11.2 1 11.2 litre O2 is formed by = mol KClO3 3 22.4 3
Page # 97
Ex.18 Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure lime stone (CaCO3). (A) 104.4 kg (B) 105.4 kg (C) 212.8 kg (D) 106.4 kg Sol. (D) 100 kg impure sample has pure CaCO3 = 95 kg 200 kg impure sample has pure CaCO3 =
95 200 = 190 kg. CaCO3 CaO + CO2 100
100 kg CaCO3 gives CaO = 56 kg.
190 kg CaCO3 gives CaO =
56 190 = 106.4 kg. 100
Ex.19 The chloride of a metal has the formula MCl 3. The formula of its phosphate will be(A) M2PO4 (B) MPO4 (C) M3PO4 (D) M(PO4)2 Sol. (B) AlCl 3 as it is AlPO4 Ex.20 A silver coin weighing 11.34 g was dissolved in nitric acid. When sodium chloride was added to the solution all the silver (present as AgNO3) was precipitated as silver chloride. The weight of the precipitated silver chloride was 14.35 g. Calculate the percentage of silver in the coin (A) 4.8 % (B) 95.2% (C) 90 % (D) 80% Sol. (B) Ag + 2HNO3 AgNO3 + NO2 + H2O 108 AgNO3 + NaCl AgCl + NaNO3 143.5 143.5 gm of silver chloride would be precipitated by 108 g of silver. or 14.35 g of silver chloride would be precipitated 10.8 g of silver. 11.34 g of silver coin contain 10.8 g of pure silver. 100 g of silver coin contain
10.8 × 100 1134 .
= 95.2 %.
SOLVED SUBJECTIVE Ex.1
Sol.
Calculate the following for 49 gm of H2SO4 (a) moles (b) Molecules (e) Total electrons Molecular wt of H2SO4 = 98 (a) moles =
(c) Total H atoms
(d) Total O atoms
wt in gm 49 1 = mole molecular wt 98 2
(b) Since 1 mole = 6.023 × 1023 molecules. 1 1 mole = 6.023 × 1023 × molecules = 3.011 × 1023 molecules 2 2 (c) 1 molecule of H2SO4 Contains 2 H atom 3.011 × 1023 of H2SO4 contain 2 × 3.011 × 1023 atoms = 6.023 × 1023 atoms : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 98
STOICHIOMETRY - 1
(d) 1 molecules of H2SO4 contains 4 O atoms 3.011 × 1023 molecular of H2SO4 contains = 4 × 3.011 × 1023 = 12.044 × 1023 (e) 1 molecule of H2SO4 contains 2H atoms + 1 S atom + 4 O atom this means 1 molecule of H2SO4 Contains (2 + 16 + 4 × 8) e– So 3.011 × 1023 molecules have 3.011 × 1023 × 50 electrons = 1.5055 × 1025 e– Ex.2
Calculate the total ions & charge present in 4.2 gm of N–3
Sol.
mole =
wt in gm 4.2 = = 0.3 Ionic wt 14
total no of ions = 0.3 × NA ions total charge = 0.3 NA × 3 × 1.6 × 10–19 = 0.3 × 6.023 × 1023 × 3 × 1.6 × 10–19 , = 8.67 × 104 C Ans. Ex.3
Find the total number of iron atom present in 224 amu iron.
Sol.
Since
56 amu = 1 atom
therefore 224 amu =
1 × 224 = 4 atom Ans. 56
Ex.4
A compound containing Ca, C, N and S was subjected to quantitative analysis and formula mass determination. A 0.25 g of this compound was mixed with Na2CO3 to convert all Ca into 0.16 g CaCO3. A 0.115 gm sample of compound was carried through a series of reactions until all its S was changed into SO42– and precipitated as 0.344 g of BaSO4. A 0.712 g sample was processed to liberated all of its N as NH3 and 0.155 g NH3 was obtained. The formula mass was found to be 156. Determine the empirical and molecular formula of the compound.
Sol.
Moles of CaCO3 = Wt of Ca =
0.16 = Moles of Ca 100
0.16 × 40 100
0.16 100 40 25.6 100 0.25 0.344 32 100 41 Similarly Mass % of S = 233 0.115
Mass % of Ca =
Similarly Mass % of
N=
0.155 14 100 = 17.9 17 0.712
Mass % of C = 15.48 Now : Elements
Ca
S
N
C
Mass %
25.6
41
17.9
15.48
Mol ratio
0.64
Simple ratio
1
1.28
1.28
1.29
2
2
2
Empirical formula = CaC2N2S2, Molecular formula wt = 156 , n × 156 = 156 n = 1 Hence, molecular formula = CaC2N2S2
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STOICHIOMETRY - 1 Page # 99 A polystyrne having formula Br3C6H3(C3H8)n found to contain 10.46% of bromine by weight. Find the value of n. (At. wt. Br = 80) Sol. Let the wt of compound is 100 gm & molecular wt is M Ex.5
Then moles of compound =
100 M
100 ×3 M 100 wt of Br = × 3 × 80 = 10.46 M M = 2294.45 = 240 + 75 + 44 n , Hence n = 45 Ans.
Moles of Br =
Ex.6 Sol.
A sample of clay was partially dried and then analysed to 50% silica and 7% water. The original clay contained 12% water. Find the percentage of silica in the original sample. In the partially dried clay the total percentage of silica + water = 57%. The rest of 43% must be some 50 impurity. Therefore the ratio of wts. of silica to impurity = . This would be true in the original sample 43 of silica. The total percentage of silica + impurity in the original sample is 88. If x is the percentage of silica, x 50 ; x = 47.3% Ans. 88 – x 43
Ex.7
Sol.
A mixture of CuSO4.5H2O and MgSO4. 7H2O was heated until all the water was driven-off. if 5.0 g of mixture gave 3 g of anhydrous salts, what was the percentage by mass of CuSO4.5H2O in the original mixture ? Let the mixture contain x g CuSO4.5H2O
x 5–x 159.5 120 = 3 x = 3.56 249.5 246 3.56 100 = 71.25 % Ans. Mass percentage of CuSO4. 5H2O = 5 Ex .8 367.5 gm KClO3 (M = 122.5) when heated, How many litre of oxygen gas is proudced at S.T.P.
Sol.
KClO3 KCl + O2 Applying POAC on O, moles of O in KClO3 = moles of O in O2 3 × moles of KClO3 = 2 × moles of O2 367.5 3 367.5 = 2 × n, n = × 122.5 2 122.5 Volume of O2 gas at S.T.P = moles × 22.4
3×
3 367.5 22.4 = 9 × 11.2 = 100.8 lit Ans. 2 122.5 0.532 g of the chloroplatinate of a diacid base on ignition left 0.195 g of residue of Pt. Calculate molecular weight of the base (Pt = 195)
= Ex.9 Sol.
Suppose the diacid base is B. B + H2PtCl6\ diacid acid base
BH2PtCl6 chloroplatinate 0.532 g
Pt 0.195 g
Since Pt atoms are conserved, applying POAC for Pt atoms, moles of Pt atoms in BH2PtCl6 = moles of Pt atoms in the product 1 × moles of BH2PtCl6 = moles of Pt in the product 0.532 0.195 mol. wt. of BH2PtCl6 195 : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 100 mol. wt. of BH2PtCl6 = 532 From the formula BH2PtCl6, we get mol. wt. of B = mol. wt. of BH2PtCl6 – mol. wt. of H2PtCl6 = 532 – 410 = 122. Ans.
STOICHIOMETRY - 1
Ex.10 10 mL of a gaseous organic compound containing. C, H and O only was mixed with 100 mL of oxygen and exploded under conditions which allowed the water formed to condense. The volume of the gas after explosion was 90 mL. On treatment with potash solution, a further contraction of 20 mL in volume was observed. Given that the vapour density of the compound is 23, deduce the molecular formula. All volume measurements were carried out under the same conditions. Sol.
y z y CxHyOz + x – O2 xCO2 + H2O 4 2 2
10 ml after explosion volume of gas = 90 ml 90 = volume of CO2 gas + volume of unreacted O2 on treatment with KOH solution volume reduces by 20 ml. This means the volume of CO2 = 20 ml the volume of unreacted O2 = 70 ml volume of reacted O2 = 30 ml V.D of compoud = 23 molecular wt 12x + y + 16z = 46 ...(1) from equation we can write y z 10 x – 30 , x + y – z = 3 4 2 4 2
4x + y – 2z = 12 ...(2) & 10x = 20 x = 2 from eq. (1) & (2) ; z = 1 & y = 6;
Hence C2H6O
Ans.
Ex.11 A sample of coal gas contained H2, CH4 and CO. 20 mL of this mixture was exploded with 80 mL of oxygen. On cooling, the volume of gases was 68 mL. There was a contraction of 10 mL. When treated with KOH. Find the composition of the original mixture. Sol. H2 + CH4 + CO; at H2 = x ml CH4 = y ml; CO = (20 – x – y) ml H2 + CH4 + CO + O2 CO2 + H2O x y 20 – x – y on cooling the volume of gases = 68 ml = volume of CO2 + unreacted O2 volume contraction due to KOH = 10 ml this means volume of CO2 = 10 ml volume of unreacted O2 = 58 ml volume of reacted O2 = 80 – 58 = 22 ml Applying POAC on C; y + 20 – x – y = volume of CO2, 20 – x = 10 x = 10 Applying POAC on H; 2x + 4y = 2x moles of H2O; moles of H2O = x + 2y Applying POAC on O 1 × moles of CO + 2 × moles of O2 = 2 × moles of CO2 + 1 × moles of H2O 1 × 20 – x – y + 2 × 22 = 2 × 10 + x + 2y 20 – x – y + 44 = 20 + x + 2y; 2x + 3y = 44 3y = 44 – 20 = 24; y = 8 ml; x = 10 ml; volume of CO = 20 – x – y
= 2 ml Ans.
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STOICHIOMETRY - 1
Page # 101
Class Room Problems Problem.1 From the following reaction sequence Cl2 + 2KOH KCl + KClO + H2O 3KClO 2KCl + KClO3 4KClO3 3KClO4 + KCl Calculate the mass of chlorine needed to produce 100 g of KClO4.
Problem.4 5 mL of a gaseous hydrocarbon was exposed to 30 mL of O2. The resultant gas, on cooling is found to measure 25 mL of which 10 mL are absorbed by NaOH and the remainder by pyrogallol. Determine molecular formula of hydrocarbon. All measurements are made at constant pressure and temperature. Sol. C2H4
Sol. 205.04 gm
Problem.2 Calculate the weight of FeO produced from 2 g VO and 5.75 g of Fe2O3. Also report the limiting reagent. VO + Fe2O3 FeO + V2O5 Sol. 5.175 gm
Problem.5 A gaseous alkane is exploded with oxygen. The volume of O2 for complete combustion to CO2 formed is in the ratio of 7 : 4. Deduce molecular formula of alkane. Sol. C2H6
Problem.3 A polystyrene, having formula Br3C6H3 (C 8H8 )n was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. If it was found to contain 10.46% bromine be weight, find the value of n.
Problem.6 A sample of gaseous hydrocarbon occupying 1.12 litre at NTP, when completely burnt in air produced 2.2 g CO2 and 1.8 g H2O. Calculate the weight of hydrocarbon taken and the volume of O2 at NTP required for its combustion.
Sol. 19
Sol. 0.8 gm, 2.24 L
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Page # 102 Problem.7 16 mL of a gaseous aliphatic compound CnH3nOm was mixed with 60 mL O2 and sparked. The gas mixture on cooling occupied 44 mL. After treatment with KOH solution, the volume of gas remaining was 12 mL. Deduce the formul a of compound. All measurements are made at constant pressure and room temperature.
STOICHIOMETRY - 1 Problem.10 What is the purity of conc. H2SO 4 solution (specific gravity 1.8 g/mL), if 5.0 mL of this solution is neutralized by 84.6 mL of 2.0 N NaOH ? Sol. 92.12%
Sol. C2H6O
Problem.11 A sample of H2SO4 (density 1.787 g mL–1) is labelled as 86% by weight. What is molarity of acid ? What volume of acid has to be used to make 1 litre of 0.2 M H2SO4 ? Sol. 15.68 M, 0.013 L Problem.8 In what ratio should you mix 0.2M NaNO3 and 0.1M Ca(MO3)2 solution so that in resulting solution, the concentration of –ve ion is 50% greater than the concentration of +ve ion ? Sol. 1/2
Problem.12 Mole fraction of I2 in C6H6 is 0.2. Calculate molality of I2 in C6H6. Problem.9 How much BaCl2 would be needed to make 250 mL of a solution having same concentration of Cl– as the one containing 3.78 g of NaCl per 100 mL ?
Sol. 3.2
Sol. 16.8 gm
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STOICHIOMETRY - 1 Problem.13 A drop (0.05 mL) of 12 M HCl is spread over a thin sheet of aluminium foil (thickness 0.10 mm and density of Al = 2.70 g/mL). Assuming whole of the HCl is used to dissolve Al, what will be the maximum area of hole produced in foil ?
Page # 103 Problem.16 A mixture of FeO and Fe3O4 when heated in air to constant weight, gains 5% in its weight. Find out composition of mixture. Sol. 20.25
Sol. 5.4×10–3
Problem.14 What would be the molarity of solution obtained by mixing equal volumes of 30% by weight H2SO4 (d = 1.218 g mL–1) and 70% by weight H2SO4 (d = 1.610g mL–1) ? If the resulting solution has density 1.425 g/mL, calculate its molality.
Problem.17 25.4 g of I2 and 14.2 g of Cl2 are made to react completely to yield a mixture of ICl and ICl3. Calculate mole of ICl and ICl3 formed. Sol. 0.1,0.1
Sol. 7.67 M
Problem.15 A mixture of Al and Zn weighing 1.67 g was completley dissolved in acid and evolved 1.69 litre of H2 at NTP. What was the weight of Al in original mixture ? Sol. 1.21 gm
Problem.18 A mixture of HCOOH and H2C2O4 is heated with conc. H 2SO 4. The gas produced is collected and on treating with KOH solution the volume 1 th. Calculate molar ratio 6 of two acids in original mixture.
of the gas decreases by
Sol. 1/4
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Page # 104
STOICHIOMETRY - 1
Problem.19 For the reaction, N2O5(g) 2NO2(g) + 0.5O2(g), calculate the mole fraction of N2O5(g) decomposed at a constant volume and temperature, if the initial pressure is 600 mm Hg and the pressure at any time is 960 mm Hg. Assume ideal gas behaviour. Sol. 0.375 Problem.23 A sample of CaCO3 and MgCO3 weighed 2.21 g is ignited to constant weight of 1.152 g. What is the composition of mixture ? Also calculale the volume of CO2 evolved at 0ºC and 76 cm of pressure. Sol. 46% 1.19 Problem.20 0.22 g sample of volatile compound, containing C, H and Cl only on combustion in O2 gave 0.195 g CO2 and 0.0804 g H2O. If 0.120 g of the compound occupied a volume of 37.24 mL at 105º and 768 mm of pressure, calculate molecular formula of compound. Sol. C2H4Cl2
Problem.24 2.0 g of a mixture of carbonate, bicarbonate and chloride of sodium on heating produced 56 mL of CO2 at NTP. 1.6 g of the same mixture required 25 mL of N HCl solution for neutral i zat i on. Cal cul ate percent age of each component present in mixture. Sol. 69.56%
Problem.21 2.0 g sample containing Na2CO3 and NaHCO3 loses 0.248 g when heated to 300ºC, the temperature at which NaHCO3 decomposes to Na2CO3, and H2O. What is % of Na2CO3 in mixture ? Sol. 66.4% Problem.25 Igni ting MnO 2 i n ai r converts i t quantitatively to Mn3O4. A sample of pyrolusite has MnO2 80%, SiO2 15% and rest having water. The sample is heated in air to constant mass. What is the % of Mn in ignited sample ? Sol. 59.37
Problem.22 10 mL of a solution of KCl containing NaCl gave on evaporation 0.93 g of the mixed salt which gave 1.865 g of AgCl by the reaction with AgNO3. Calculate the quantity of NaCl in 10 mL of solution. Sol. 0.67 gm
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STOICHIOMETRY - 1 Problem.26 A solid mixture 5 g consits of lead nitrate and sodium nitrate was heated below 600ºC until weight of residue was constant. If the loss in weight is 28%, find the amount of lead nitrate and sodium nitrate in mixture.
Page # 105
Sol. 3.32 gm
Problem. 29 0.50 g of a mixture of K2CO3 and Li2CO3 required 30 mL of 0.25 N HCl solution for neturalization. What is % composition of mixture ? Sol.
Problem.27 Determine the formula of ammonia form the folloiwng data : (i) Volume of ammonia = 25 mL. (ii) Volume on addition of O2 after explosion = 71.2 mL. (iii) Volume after explosion and reaction with O2 on cooling = 14.95 mL. (i v) Vol ume after bei ng absorbed by al kali ne pyrogallol = 12.5 mL. Sol.
Problem.30 A mixture in which the mole ratio of H2 and O2 is 2 : 1 is used to prepare water by the reaction, 2H2(g) + O2(g) 2H2O(g) The total pressure in the container is 0.8 atm at 20ºC before the reaction. Determine the final pressure at 120ºC after reaction assuming 80% yield of water. Sol. 0.7865
Problem.28 A mixture of ethane (C2H6) and ethene (C2H4) occupies 40 litre at 1.00 atm and at 400 K. The mixture reacts completely with 130 g of O2 to produce CO2 and H2O. Assuming ideal gas behaviour, calculate the mole fractions of C2H4 and C2H6 in the mixture. Sol. 0.63, 0.37 : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 106
EXERCISE – I
STOICHIOMETRY - 1
OBJECTIVE PROBLEMS (JEE MAIN)
Single correct 1. For the reaction 2x + 3y + 4z 5w Initially if 1 mole of x, 3 mole of y and 4 mole of z is taken. If 1.25 mole of w is obtained then % yield of this reaction is (A) 50% (B) 60% (C) 70% (D) 40% Sol.
Sol.
5.
The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25. Then mole % of gas B in the mixture would be (A) 25% (B) 50% (C) 75% (D) 10%
Sol. 2.
A solution of A (MM = 20) and B (MM = 10), [Mole fraction XB = 0.6] having density 0.7 gm/ ml then molarity and molality of B in this solution will be ________________ and ______________ respectively. (A) 30M,75m (B) 40M,75m (C) 30M,65m (D) 50M,55m
6.
Sol.
For the reaction 2A + 3B + 5C 3D Initially if 2 mole of A, 4 mole of B and 6 mole of C is taken, With 25% yield, moles of D which can be produced are _____________. (A) 0.75 (B) 0.5 (C) 0.25 (D) 0.6
Sol.
3.
125 ml of 8% w/w NaOH solution (sp. gravity 1) is added to 125 ml of 10% w/v HCl solution. The nature of resultant solution would be _________ (A) Acidic (B) Basic (C) Neutral (D) None
7.
Sol.
Two elements X (atomic mass=75) and Y (atomic mass=16) combine to give a compound having 75.8% of X. The formula of the compound is: (A) X2Y3 (B) X2Y (C) X2Y2 (D) XY
Sol.
4.
Ratio of masses of H2SO4 and Al2 (SO4)3 is grams each containing 32 grams of S i s __________. (A) 0.86 (B) 1.72 (C) 0.43 (D) 2.15 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
8.
STOICHIOMETRY - 1 Equal volumes of 10% (v/v) of HCl is mixed with 10% (v/v) NaOH solution. If density of pure NaOH is 1.5 times that of pure HCl then the resultant solution be : (A) basic (B) neutral (C) acidic (D) can’t be predicted.
11.
Sol.
9.
A definite amount of gaseous hydrocarbon was burnt with just sufficient amount of O2 . The volume of all reactants was 600 ml, after the explosion the volume of the products [CO2(g) and H2O(g)] was found to be 700 ml under the similar conditions. The molecular formula of the compound is : (A) C3H8 (B) C3H6 (C) C3H4 (D) C4H10
Page # 107 One mole mixture of CH4 & air (containing 80% N2 20% O2 by volume) of a composition such that when underwent combustion gave maximum heat (assume combustion of only CH4). Then which of the statements are correct, regarding composition of initial mixture. (X presents mole fraction) (A) X CH4
1 2 8 , XO2 = , XN2 11 11 11
(B) X CH4
3 1 1 , X O 2 , X N2 8 8 2
(C) X CH4
1 1 2 , X O 2 , X N2 6 6 3
(D) Data insufficient Sol.
Sol.
12.
C6H5OH(g) + O2(g) CO2(g) + H2O(l) Magnitude of volume change if 30 ml of C6H5OH (g) is burnt with excess amount of oxygen, is (A) 30 ml (B) 60 ml (C) 20 ml (D) 10 ml
Sol. 10.
One gram of the silver salt of an organic dibasic acid yields, on strong heating, 0.5934 g of silver. If the weight percentage of carbon in it 8 times the weight percentage of hydrogen and half the weight percentage of oxygen, determine the molecular formula of the acid. [Atomic weight of Ag = 108] (A) C4H6O4 (B) C4H6O6 (C) C2H6O2 (D) C5H10O5
13.
Sol.
10 ml of a compound containing ‘N’ and ‘O’ is mixed with 30 ml of H2 to produce H2O (l) and 10 ml of N2(g). Molecular formula of compound if both reactants reacts completely, is (A) N2O (B) NO2 (C) N2O3 (D) N2O5
Sol.
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Page # 108 14. Similar to the % labelling of oleum, a mixture of H3PO4 and P4O10 is labelled as (100 + x) % where x is the maximum mass of water which can react with P4O10 present in 100 gm mixture of H3PO4 and P4O10. If such a mixture is labelled as 127% Mass of P4O10 is 100 gm of mixture, is (A) 71 gm (B) 47 gm (C) 83gm (D) 35 gm Sol.
17.
STOICHIOMETRY - 1 In the quantitative determination of nitrogen using Duma’s method, N2 gas liberated from 0.42 gm of a sample of organic compound was collected over water. If the volume of N2 gas 100 collected was ml at total pressure 860 mm 11 Hg at 250 K, % by mass of nitrogen in the organic compound is [Aq. tension at 250K is 24 mm Hg and R = 0.08 L atm mol–1 K–1] (A)
10 % 3
(B)
5 % 3
(C)
20 % 3
(D)
100 % 3
Sol.
15.
Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit. O2 at 1 atm & 273 K according to given reaction, is C(s) + H2(g) + O2(g) C12H22O11(s) (A) 138.5 (B) 155.5 (C) 172.5 (D) 199.5
18.
Sol.
40 gm of a carbonate of an alkali metal or alkaline earth metal containing some inert impurities was made to react with excess HCl solution. The liberated CO2 occupied 12.315 lit. at 1 atm & 300 K. The correct option is (A) Mass of impurity is 1 gm and metal is Be (B) Mass of impurity is 3 gm and metal is Li (C) Mass of impurity is 5 gm and metal is Be (D) Mass of impurity is 2 gm and metal is Mg
Sol. 16.
If 50 gm oleum sample rated as 118% is mixed with 18 gm water, then the correct option is (A) The resulting solution contains 18 gm of water and 118 gm H2SO4 (B) The resulting solution contains 9 gm of water and 59 gm H2SO4 (C) The resulting solution contains only 118 gm pure H2SO4 (D) The resulting solution contains 68 gm of pure H2SO4
19.
Sol.
The percentage by mole of NO2 in a mixture NO2(g) and NO(g) having average molecular mass 34 is : (A) 25% (B) 20% (C) 40% (D) 75%
Sol.
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20.
STOICHIOMETRY - 1 The minimum mass of mixture of A2 and B4 required to produce at least 1 kg of each product is : (Given At. mass of ‘A’ = 10; At mass of ‘B’ = 120) 5A2 + 2B4 2AB2 + 4A2B (A) 2120 gm (B) 1060 gm (C) 560 gm (D) 1660 gm
Page # 109 Sol.
Sol. 24.
74 gm of sample on complete combustion gives 132 gm CO2 and 54 gm of H2O. The molecular formula of the compound may be (A) C5H12 (B) C4H10O (C) C3H6O2 (D) C3H7O2
Sol. 21.
The mass of CO2 produced from 620 gm mixture of C2H4O2 & O2, prepared to produce maximum energy is (A) 413.33 gm (B) 593.04 gm (C) 440 gm (D) 320 gm
Sol.
25.
The % by volume of C4H10 in a gaseous mixture of C4H10, CH4 and CO is 40. When 200 ml of the mixture is burnt in excess of O2. Find volume (in ml) of CO2 produced. (A) 220 (B) 340 (C) 440 (D) 560
Sol. 22.
Assuming complete precipitation of AgCl, calculate the sum of the molar concentration of all the ions if 2 lit of 2M Ag2SO4 is mixed with 4 lit of 1 M NaCl solution is : (A) 4M (B) 2M (C) 3M (D) 2.5 M
26.
Sol.
23.
12.5 gm of fuming H2SO4 (labelled as 112%) is mixed with 100 lit water. Molar concentration of H+ in resultant solution is : [Note : Assume that H2SO4 dissociate completely and there is no change in volume on mixing] (A)
2 700
(B)
2 350
(C)
3 350
(D)
3 700
What volumes should you mix of 0.2 M NaCl and 0.1 M CaCl2 solution so that in resulting solution the concentration of positive ion is 40% lesser than concentration of negative ion. Assuming total volume of solution 1000 ml. (A) 400 ml NaCl, 600 ml CaCl2 (B) 600 ml NaCl, 400 ml CaCl2 (C) 800 ml NaCl, 200 ml CaCl2 (D) None of these
Sol.
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Page # 110 27. An iodized salt contains 0.5% of Nal. A person consumes 3 gm of salt everyday. The number of iodide ions going into his body everyday is (A) 10–4 (B) 6.02 × 10–4 19 (C) 6.02 × 10 (D) 6.02 × 1023 Sol.
31.
Sol.
32. 28.
STOICHIOMETRY - 1 Weight of oxygen in Fe2O3 and FeO is in the simple ratio for the same amount of iron is : (A) 3 : 2 (B) 1 : 2 (C) 2 : 1 (D) 3 : 1
The pair of species having same percentage (mass) of carbon is : (A) CH3COOH and C6H12O6 (B) CH3COOH and C2H5OH (C) HCOOCH3 and C12H22 O11 (D) C6H12O6 and C12H22O11
Sol.
Two elements X (atomic mass 16) and Y (atomic mass 14) combine to form compounds A, B and C. The ratio of different masses of Y which combines with a fixed mass of X in A, B and C is 1:3:5. If 32 parts by mass of X combines with 84 parts by mass of Y in B, then in C, 16 parts by mass of X will combine with___ parts by mass of Y. (A) 14 (B) 42 (C) 70 (D) 84
Sol.
29.
200 ml of a gaseous mixture containing CO, CO2 and N2 on complete combustion in just sufficient amount of O2 showed contraction of 40 ml. When the resulting gases were passed through KOH sol uti on it reduces by 50 % then calculate the volume ratio
33.
of VCO2 : VCO : VN2 in original mixture. (A) 4 : 1 : 5 (C) 1 : 4 : 5
(B) 2 : 3 : 5 (D) 1 : 3 : 5
Sol.
In a textile mill, a double-effect evaporator system concentrates weak liquor containing 4% (by weight) caustic soda to produce a dye containing 25% solids (by weight). Calculate the weight of the water evaporate per 100-kg feed in the evaporator. (A) 125.0 g (B) 50.0 kg (C) 84.0 kg (D) 16.0 kg
Sol.
30.
Sol.
Density of a gas relative to air is 1.17. Find the mol. mass of the gas. [Mair = 29 g/mol] (A) 33.9 (B) 24.7 (C) 29 (D) 22.3
34.
Zinc ore (zinc sulphide) is treated with sulphuric acid, leaving a solution with some undissolved bits of material and releasing hydrogen sulphide gas. If 10.8g of zinc ore is treated with 50.0 ml of sulphuric acid (density 1.153 g/ml), 65.1g of solution and undissolved material remains. In addition, hydrogen sulphide (density 1.393 g/ L) is evolved. What is the volume (in liters) of this gas? (A) 4.3 (B) 3.35 (C) 4.67 (D) 2.40
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STOICHIOMETRY - 1 Sol.
35.
38.
A sample of an ethanol-water solution has a volume of 54.2 cm3 and a mass of 49.6g. What is the percentage of ethanol (by mass) in the solution? (Assume that there is no change in volume when the pure compounds are mixed.) The density of ethanol is 0.80 g/cm3 and that of water is 1.00 g/cm3. (A) 18.4% (B) 37.1% (C) 33.95% (D) 31.2%
Page # 111 A sample of clay contains 40% silica and 15% water. The sample is partially dried by which it loses 5 gm water. If the percentage of water in the partially dried clay is 8, calculate the percentage of silica in the partially dried clay. (A) 21.33% (B) 43.29% (C) 75% (D) 50%
Sol.
39.
Sol.
The density of quartz mineral was determined by adding a weighed piece to a graduated cylinder containing 51.2ml water. After the quartz was submersed, the water level was 65.7 ml. The quartz piece weighed 38.4g. What was the density of quartz? (A) 1.71 gm/ml (B) 1.33 gm/ml (C) 2.65 gm/ml (D) 1.65gm/ml
Sol. 36.
A student gently drops an object weighing 15.8 g into an open vessel that is full of ethanol, so that a volume of ethanol spills out equal to the volume of the object. The experimenter now finds that the vessel and its contents weigh 10.5 g more than the vessel full of ethanol only. The density of ethanol is 0.789 g/cm3. What is the density of the object? (A) 6.717 gm/cm3 (B) 4.182 gm/cm3 3 (C) 1.563 gm/cm (D) 2.352 gm/cm3
40.
Which has maximum number of atoms of oxygen (A) 10 ml H2O(l) (B) 0.1 mole of V2O5 (C) 12 gm O3(g) (D) 12.044 × 1022 molecules of CO2
Sol.
Sol.
41. 37.
A person needs on average of 2.0 mg of riboflavin (vitamin B2) per day. How many gm of butter should be taken by the person per day if it is the only source of riboflavin? Butter contains 5.5 microgram riboflavin per gm. (A) 363.6 gm (B) 2.75 mg (C) 11 gm (D) 19.8 gm
Mass of one atom of the element A is 3.9854 × 10–23. How many atoms are contained in 1g of the element A? (A) 2.509 × 1023 (B) 6.022 × 1023 23 (C) 12.044 × 10 (D) None
Sol.
Sol. 42.
The number of atoms present in 0.5 g-atoms of nitrogen is same as the atoms in (A) 12 g of C (B) 32 g of S (C) 8 g of oxygen (D) 24g of Mg
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Page # 112 Sol.
Sol.
43.
47.
STOICHIOMETRY - 1
A graph is plotted for an element, by putting its weight on X-axis and the corresponding number of number of atoms on Y-axis. Determine the atomic weight of the element for which the graph is plotted.
Two isotopes of an element Q are Q97 (23.4% abundance) and Q94 (76.6% abundance). Q97 is 8.082 times heavier than C12 and Q94 is 7.833 times heavier than C12. What is the average atomic weight of the element Q? (A) 94.702 (B) 78.913 (C) 96.298 (D) 94.695
Sol.
(A) infinite (C) 0.025
(B) 40 (D) 20
48.
Sol.
44.
The O18/O16 ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on earth. The average mass of an atom of oxygen in these meteorites is ___ that of a terrestrial oxygen atom? (A) equal to (B) greater than (C) less than (D) None of these
The element silicon makes up 25.7% of the earth's crust by weight, and is the second most abundant element, with oxygen being the first. Three isotopes of silicon occur in nature: Si28 (92.21%), which has an atomic mass of 27.97693 amu; Si29 (4.70%), with an atomic mass of 28.97649 amu; and Si30 (3.09%), with an atomic mass of 29.97379 amu. What is the atomic weight of silicon? (A) 28.0856 (B) 28.1088 (C) 28.8342 (D) 29.0012
Sol.
Sol.
49. 45.
If isotopic distribution of C12 and C14 is 98.0% and 2.0% respectively, then the number of C14 atoms in 12 gm of carbon is (A) 1.032 ×1022 (B) 1.20 ×1022 23 (C) 5.88 ×10 (D) 6.02 ×1023
The average atomic mass of a mixture containing 79 mol % of 24 Mg and remaining 21 mole % of 25 Mg and 26Mg, is 24.31. % mole of 26Mg is (A) 5 (B) 20 (C) 10 (D) 15
Sol.
Sol.
50. 46.
At one time there was a chemical atomic weight scale based on the assignment of the value 16.0000 to naturally occurring oxygen. What would have been the atomic weight, on such a table, of silver, if current information had been available? The atomic weights of oxygen and silver on the present table are 15.9994 and 107.868. (A) 107.908 (B) 107.864 (C) 107.868 (D) 107.872
The oxide of a metal contains 30% oxygen by weight. If the atomic ratio of metal and oxygen is 2 : 3, determine the atomic weight of metal. (A) 12 (B) 56 (C) 27 (D) 52
Sol.
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STOICHIOMETRY - 1
EXERCISE – II 1.
Page # 113
OBJECTIVE PROBLEMS (JEE ADVANCED)
The average mass of one gold atom in a sample of naturally occuring gold is 3.257 × 10–22 g. Use this to calculate the molar mass of gold.
Sol.
Sol.
2.
A plant virus is found to consist of uniform cylindrical particles of 150 Å in diameter and 5000 Å long. The specific volume of the virus is 0.75 cm3/g. If the virus is considered to be a single particle, find its molecular weight. [V = r2l]
EMPIRICAL & MOLECULAR FORMULA 5. Polychlorinated biphenyls, PCBs, known to be dangerous environmental pollutants, are a group of compounds with the general empirical formula C12HmCl10–m, where m is an integer. What is the value of m, and hence the empirical formula of the PCB that contains 58.9% chlorine by mass ? Sol.
Sol.
6.
MOLE 3. Calculate (a) Number of nitrogen atoms in 160 amu of NH4NO3 (b) Number of gram-atoms of S in 490 kg H2SO4
Given the following empirical formulae and molecular weights, compute the true molecular formulae : Empirical Formula Molecular weight
(a) (b) (c)
CH2 CH2O HO
84 150 34
(d) (e) Sol.
HgCl HF
472 80
(c) Grams of Al2(SO4)3 containing 32 amu of S. Sol.
7. 4.
A chemical compound “dioxin” has been very much in the news in the past few years. (it is the by-product of herbicide manufacture and is through to be quite toxic.) Its formula is C12H4Cl4O2. If you have a sample of dirt (28.3 g) that contains 8.78 × 10–8 moles of dioxin, calculate the percentage of dioxin in the dirt sample ?
What is the empirical formula of a compound 0.2801 gm of which gave on complete combustion 0.9482 gm of carbon dioxide and 0.1939 gm of water ?
Sol.
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Page # 114 8. A 5.5 gm sample of an organic compound gave on quantitative analysis 1.4 gm of N and 3.6 gm of C and 0.5 gm of H. If Molecular mass of the compound is 55 then calculate E.F. and M.F.
STOICHIOMETRY - 1 Sol.
Sol.
12.
PROBLEMS RELATED WITH MIXTURE 9.
One gram of an alloy of aluminium and magnesium when heated with excess of dil. HCl forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen collected over mercury at 0ºC has a volume of 1.12 liters at 1 atm pressure. Calculate the composition of the alloy.
A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.12gm. Calculate % by mass of NaCl in the original mixture.
Sol.
Sol.
LIMITING REACTANT 13. 10.
A sample containing only CaCO3 and MgCO3 is ignited to CaO and MgO. The mixture of oxides produced weight exactly half as much as the original sample. Calculate the percentages of CaCO3 and MgCO3 in the sample.
Titanium, which is used to make air plane engines and frames, can be obtained from titanium tetrachloride, which in turn is obtained from titanium oxide by the following process : 3 TiO2 (s) + 4C(s) + 6Cl2 (g) 3TiCl4(g) + 2CO2(g) + 2Co(g) A vessel contains 4.32 g TiO2, 5.76 g C and; 6.82 g Cl 2 , suppose the reaction goes to completion as written, how many gram of TiCl4 and be produced ? (Ti = 48)
Sol.
Sol.
11.
Determine the percentage composition of a mixture of anhydrous sodium carbonate and sodium bicarbonate from the following data : wt of the mixture taken = 2g Loss in weight on heating = 0.11 gm
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STOICHIOMETRY - 1 More than one correct : 14
16.
Two gases A and B which react according to the equation
aA ( g) bB( g) cC( g) dD( g)
Page # 115 An aqueous solution consisting of 5 M BaCl2, 58.8% w/v NaCl solution & 2m Na2 X has a density of 1.949 gm/ml. Mark the option(s) which represent correct molarity (M) of the specified ion.
to give two gases C and D are taken (amount not known) in an Eudiometer tube (operating at a constant Pressure and temperature) to cause the above.
[Assume 100% dissociation of each salt and molecular mass of X2– is 96]
If on causing the reaction there is no volume change observed then which of the following statement is/are correct.
(C) [Total anions] = 20.5 M
(A) (a + b) = (c + d)
(A) [Cl–] = 20 M (B) [Na+] = 11 M (D) [Total cations]=15 M Sol.
(B) average molecular mass may increase or decrease if either of A or B is present in limited amount. (C) Vapour Density of the mixture will remain same throughout the course of reaction. (D) Total moles of all the component of mixture will change. Sol.
17.
A mixture of 100 ml of CO, CO2 and O2 was sparked. When the resulting gaseous mixture was passed through KOH solution, contraction in volume was found to be 80 ml, the composition of initial mixture may be (in the same order) (A) 30 ml, 60 ml, 10 ml (B) 30 ml, 50 ml, 20 ml (C) 50 ml, 30 ml, 20 ml (D) 30 ml, 40 ml, 30 ml
Sol. 15.
A mixture of C3H8 (g) O2 having total volume 100 ml in an Eudiometry tube is sparked & it is observed that a contraction of 45 ml is observed what can be the composition of reacting mixture. (A) 15 ml C3H8 & 85 ml O2 (B) 25 ml C3H8 & 75 ml O2 (C) 45 ml C3H8 & 55 ml O2 (D) 55 ml C3H8 & 45 ml O2
Sol.
18.
A sample of H2O2 solution labelled as 56 volume has density of 530 gm/L. Mark the correct option(s) representing concentration of same solution in other units. (Solution contains only H2O and H2O2) (A) MH2O2 6 (B) %
w 17 v
(C) Mole fraction of H2O2 = 0.25 (D) mH2O 2 : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 116 Sol.
19.
STOICHIOMETRY - 1 Sol.
Solution(s) containing 30 gm CH3COOH is/are (A) 50 gm of 70% (w/v) CH3COOH [dsol = 1.4 gm/ml] (B) 50 gm of 10 M CH3COOH [dsol = 1 gm/ml] (C) 50 gm of 60% (w/w) CH3COOH
22.
(D) 50 gm of 10 m CH3COOH Sol.
The recommended daily dose is 17.6 milligrams of vitamin C (ascorbic acid) having formula C6H8O6. Match the following. Given : NA = 6 × 1023 Column I
Column II
(A)
O-atoms present
(P) 10–4 mole
(B)
Moles of vitamin C in 1gm (Q) 5.68 × 10–3 of vitamin C
(C)
Moles of vitamin C in 1gm (R) 3.6 × 1020 should be consumed daily
20.
‘2V’ ml of 1 M Na2SO4 is mixed with ‘V’ ml of 2M Ba(NO3)2 solution.
Sol.
(A) Molarity of Na+ ion in final solution can’t be calculated as V is not known. (B) Molarity of BaSO4 in final solution is
(C) Molarity of
NO3–
2 M 3
4 in final solution is M 3
(D) Molarity of NO3– in final solution is
2 M 3
An evacuated glass vessel weighs 50 gm when empty, 148.0 g when completely filled with liquid of density 0.98 gml–1 and 50.5 g when filled with an ideal gas at 760 mm at 300 K. Determine the molecular weight of the gas. [JEE ‘98,3]
Sol.
Sol.
24.
Match the Column 21.
23.
One type of artifical diamond (commonly called YAG for yttrium aluminium garnet) can be represented by the formula Y3Al5O12 [Y = 89, Al = 27]
Column I
Column II
Element
Weight percentage
(A) Y
(P) 22.73 %
(B) Al
(Q) 32.32 %
(C) O
(R) 44.95 %
At 100° C and 1 atmp, if the density of liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules 1 L of steam at that temperature is : [JEE ‘2001 (Scr), 1] (A) 6 cm3
(B) 60 cm3
(C) 0.6 cm3
(D) 0.06 cm3
Sol.
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STOICHIOMETRY - 1 Integer Type 25.
Page # 117 Sol.
If one mole of ethanol (C2H5OH) completely burns to form carbon dioxide and water, the weight of carbon dioxide formed is about -
Sol.
29.
Assuming that petrol is iso-octane (C8H18) and has density 0.8 gm/ml, 1.425 litre of petrol on complete combustion will consume oxygen -
Sol. 26.
The mass of CaCO3 produced when carbon dioxide is passed in excess through 500 ml of 0.5 M Ca(OH)2 will be-
Sol.
27.
The mass of oxygen that would be required to produce enough CO, which completely reduces 1.6 kg Fe2O3 (at. mass Fe = 56) is-
30.
Sol.
A mixture containing 100 gm H2 and 100 gm O2 is ignited so that water is formed according to the reaction, 2H2 + O2 2H2O; How much water will be formed -
Sol.
28.
1.5 gm of divalent metal displaced 4 gm of copper (at. wt. = 64) from a solution of copper sulphate. The atomic weight of the metal is: 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 118
STOICHIOMETRY - 1
EXERCISE – III 1.
A 2 2B 2 A 2B 4
and
SUBJECTIVE PROBLEMS (JEE ADVANCED) 3 A 2 2B 2 A 3B 4 2
Two substance A2 & B2 react in the above manner when A2 is limited it gives A2B4, when in excess gives A3B4. A2B4 can be converted to A3B4 when reacted with A2. Using this information calculate the composition of the final mixture when mentioned amount of A2 & B2 are taken (i) If 4 moles of A2 & 4 moles of B2 is taken in reaction container
4.
Chloride samples are prepared for analysis by using NaCl, KCl and NH4Cl separately or as mixture. What minimum volume of 5% by weight AgNO3 solution (sp gr., 1.04 g ml–1) must be added to a sample of 0.3 g in order to ensure complete precipitation of chloride in every possible case ?
Sol.
1 moles of A2 & 2 moles of B2 is taken in 2 reaction container
(ii) If
(iii) If
5 moles of A2 & 2 moles of B2 is taken 4
Sol.
5.
One litre of milk weighs 1.035 kg. The butter fat is 10% (v/v) of milk has density of 875 kg/m3. The density of fat free skimed milk is ?
Sol.
2.
How much minimum volume of 0.1 M aluminium sulphate solution should be added to excess calcium nitrate to obtain atleast 1 gm of each salt in the reaction. Al2 (SO 4 )3 3Ca(NO3 )2 2Al(NO3 )3 3CaSO 4
Sol.
6. 3.
A sample of fuming sulphuric acid containing H2SO4, SO3 and SO2 weighing 1.00 g is found to require 23.47 ml of 1.00 M alkali (NaOH) for neutralisation. A separate sample shows the presence of 1.50% SO2. Find the percentage of “free” SO3, H2SO4 and “combined” SO3 in the sample.
Sol.
For a hypothetical chemical reaction represented by 3 A(g) C(g) D(g) , the following informations are known. Information (i) At t = 0, only 1 mole of A is present and the gas has V.D. = 60. (ii) At t = 30 min, the gaseous mixture consist of all three gases and has a vapour density = 75. (iii) Molecular Mass of C = 200 Calculate (a) Molecular weight of A and D. (b) Moles of each specie at t = 30 min.
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STOICHIOMETRY - 1 Sol.
7.
Page # 119 Sol.
A mixture of H2, N2 & O2 occupying 100 ml underwent reaction so as to from H2O2(l) and N2H2(g) as the only products, causing the volume to contract by 60 ml. The remaining mixture was passed through pyrogallol causing a contraction of 10ml. To the remaining mixture excess H2 was added and the above reaction was repeated, causing a reduction in volume of 10 ml. Identify the composition of the initial mixture in mol %. (No other products are formed)
9.
A chemist wants to prepare diborane by the reaction 6LiH + 8BF3 6LiBF4 + B2H6 If he starts with 2.0 moles each of LiH & BF3. How many moles of B2H6 can be prepared.
Sol.
Sol.
10. 8.
A mixture of three gases an alkane (general formula Cn H2n + 2), an alkene (general formula CxH2x) and O2 was subjected to sparking to cause combustion of both the hydrocarbon at 127ºC. After the reaction three gases were present and none of the hydrocarbon remained. On passing the gases through KOH (absorb CO 2 ), an increment in mass of KOH solution by 132 gm was observed. The remaining gases were passed over white anhydrous CuSO4 and the weight of blue hydrated CuSO4 crystals was found to be 72 gm more than that of white anhydrous CuSo4. Given that initially total 10 moles of the three gases were taken and moles of alkane and alkene were equal and if molecular mass of alkene molecular mass of alkane = 12 i.e. (Malkene – M alkane = 12), then answer the following questions. (Show calculations) (a) Which three gases are remained after the combustion reactions. (b) What are the number of moles of product gases. (c) What is the molecular formula of the two hydrocarbon. (d) What is the number of moles of each of the two hydrocarbons and O2 gas taken initially.
Carbon reacts with chlorine to form CCl4. 36 gm of carbon was mixed with 142 g of Cl2. Calculate mass of CCl4 produced and the remaining mass of reactant.
Sol.
MISCELLANEOUS PROBLEM 11.
P4S3 + 8O2 P4O10 + 3SO2 Calculate minimum mass of P4S3 is required to produce atleast 1 gm of each product.
Sol.
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Page # 120 12. By the reaction of carbon and oxygen, a mixture of CO and CO 2 is obtained. What i s the composition by mass of the mixture obtained when 20 grams of O2 reacts with 12 grams of carbon ? Sol.
STOICHIOMETRY - 1 Sol.
16. 13.
The chief ore of Zn is the sulphide, ZnS. The ore is concentrated by froth floatation process and then heated in air to convert ZnS to ZnO.
The density of a solution containing 40% by mass of HCl is 1.2 g/mL. Calculate the molarity of the solution.
Sol.
75% 2ZnS 3O 2 2ZnO 2SO2 100%
ZnO H2 SO 4 ZnSO 4 H2 O 80%
2ZnSO 4 2H2O 2Zn 2H2SO 4 O 2 (a) What mass of Zn will be obtained from a sample of ore containing 291 kg of ZnS. (b) Calculate the volume of O2 produced at 1 atm & 273 K in part (a).
17.
Sol.
15g of methyl alcohol is present in 100 mL of solution. If density of solution is 0.90 g mL–1. Calculate the mass percentage of methyl alcohol in solution
Sol.
CONCENTRATION TERMS 14. Calculate the molarity of the following solutions (a) 4g of caustic soda is dissolved in 200 mL of the solution. (b) 5.3 g of anhydrous sodium carbonate is dissolved in 100 mL of solution (c) 0.365 g of pure HCl gas is dissolved in 50 mL of solution. Sol.
18.
Units of parts per million (ppm) or per billion (ppb) are often used to describe the concentrations of solutes in very dilute solutions. The units are defined as the number of grams of solute per million or per billion grams of solvent. Bay of Bengal has 1.9 ppm of lithium ions. What is the molality of Li+ in this water ?
Sol.
15.
Density of a solution containing 13% by mass of sulphuric acid is 1.09 g/mL. Then molarity of solution will be.
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19.
STOICHIOMETRY - 1 A 6.90 M solution of KOH in water contains 30% by mass of KOH. What is density of solution in gm/ml.
Sol.
Page # 121 23. (a) Find molarity of Ca2+ and NO3– in 2 M Ca(NO3)2 aqueous solution of density 1.328 g/mL. (b) Also find mole fraction of solvent in solution. Sol.
24. 20.
A solution of specific gravity 1.6 is 67% by weight. What will be % by weight of the solution of same acid if it is diluted to specific gravity 1.2 ?
Calculate molality (m) of each ion present in the aqueous solution of 2M NH4 Cl assuming 100% dissociation according to reaction. NH4 Cl(aq) NH4 (aq) Cl (aq) Given : Density of solution = 3.107 gm/ml.
Sol. Sol.
25. 21. Find out the volume of 98% w/w H2SO4 (density = 1.8 gm/ml) must be diluted to prepare 12.5 litres of 2.5 M sulphuric acid solution. Sol.
500 ml of 2 M NaCl solution was mixed with 200 ml of 2 M NaCl solution. Calculate the final volume and molarity of NaCl in final solution if final solution has density 1.5 gm/ml.
Sol.
EXPERIMENTAL METHODS 22.
Determine the volume of diluted nitric acid (d = 1.11 gmL–1, 19% w/v HNO3) That can be prepared by diluting with water 50 mL of conc. HNO3 (d = 1.42 g ML–1, 69.8 % w/v).
26.
Sol.
What is the percentage of nitrogen in an organic compound 0.14 gm of which gave by Dumas method 82.1 c.c. of nitrogen collected over water at 27ºC and at a barometric pressure of 774.5 mm ? (aqueous tension of water at 27ºC is 14.5 mm)
Sol.
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Page # 122 27. 0.20 gm of an organic compound was treated by Kjeldahl’s method and the resulting ammonia was passed into 50 cc of M/4 H2 SO4. The residual acid was then found to require 40 cc of M/2 NaOH for neutralisation. What is the percentage of nitrogen in the compound ? Sol.
28. 0.252 gm of an organic compound gave on complete combustion 0.22 gm of carbon dioxide and 0.135 gm of water. 0.252 gm of the same compound gave by Carius method 0.7175 gm of silver chloride. What is the empirical formula of the compound ? Sol.
30.
STOICHIOMETRY - 1 0.80 g of the chloroplatinate of a mono acid base on ignition gave 0.262 g of Pt. Calculate the mol wt of the base.
Sol.
31. The molecular mass of an organic acid was determined by the study of its barium salt. 2.562 g of salt was quantitatively converted to free acid by the reaction 30 ml of 0.2 M H2 SO4, the barium salt was found to have two moles of water of hydration per Ba+2 ion and the acid is mono basic. What is molecular weight of anhydrous acid ? (At mass of Ba = 137) Sol.
SOME TYPICAL CONCENTRATION TERMS 29.
0.6872 gm of an organic compound gave on complete combustion 1.466 gm of carbon dioxide and 0.4283 gm of water. A given weight of the compound when heated with nitric acid and silver nitrate gave an equal weight of silver chloride. 0.3178 gm of the compound gave 26.0 cc of nitrogen at 15ºC and 765 mm pressure. Deduce the empirical formula of the compound ?
32.
Calculate composition of the final solution if 100 gm oleum labelled as 109% is added with (a) 9 gm water (b) 18 gm water (c) 120 gm water
Sol.
Sol.
33.
For ‘44.8 V’ H2O2 solution having d = 1.136 gm/ ml calculate (i) Molarity of H2O2 solution. (ii) Mole fraction of H2O2 solution.
Sol.
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34.
STOICHIOMETRY - 1 An oleum sample is labelled as 118%, Calculate (i) Mass of H2SO4 in 100 gm oleum sample. (ii) Maximum mass of H2SO4 that can be obtained if 30 gm sample is taken. (iii) Composition of mixture (mass of components) if 40 gm water is added to 30 gm given oleum sample.
38.
Page # 123 When a certain quantity of oxygen was ozonised in a suitable apparatus, the volume decreased by 4 ml. On addition of turpentine the volume further decreased by 8 ml. All volumes were measured at the same temperature and pressure. From these data, establish the formula of ozone.
Sol.
Sol.
EUDIOMETRY 35.
10ml of a mixture of CO, CH4 and N2 exploded with excess of oxygen gave a contraction of 6.5 ml. There was a further contraction of 7 ml, when the residual gas treated with KOH. Volume of CO, CH4 and N2 respectively is
39.
Sol.
36. When 100 ml of a O2 – O3 mixture was passed through turpentine, there was reduction of volume by 20 ml. If 100 ml of such a mixture is heated, what will be the increase in volume ? Sol.
Sol.
40.
37.
60 ml of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If 38 ml of N2 was formed, calculate the volume of each gas in the mixture.
10 ml of ammonia were enclosed in an eudiometer and subjected to electric sparks. The sparks were continued till there was no further increase in volume. The volume after sparking measured 20 ml. Now 30 ml of O2 were added and sparking was continued again. The new volume then measured 27.5 ml. All volume were measured under identical conditions of temperature and pressure. V.D. of ammonia is 8.5. Calculate the molecular formula of ammonia. Nitrogen and Hydrogen are diatomic.
10 mL of gaseous organic compound contain C, H and O only was mixed with 100 mL of O2 and exploded under identical conditions and then cooled. The volume left after cooling was 90 mL. On treatment with KOH a contraction of 20 Ml was observed. If vapour density of compound is 23, derive molecular formula of the compound.
Sol.
Sol.
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Page # 124
1.
STOICHIOMETRY - 1
EXERCISE – IV
PREVIOUS YEARS
LEVEL – I
JEE MAIN
The weight of 2.01 × (A) 9.3 gm (C) 1.2 gm
1023
molecules of CO is [AIEEE-2002] (B) 7.2 gm (D) 3 gm
Sol.
Sol.
5. 2.
In an organic compound of molar mass 108 gm mol–1 C, H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular formula can be – [AIEEE-2002] (A) C6H8N2 (B) C7H10N (C) C5H6N3 (D) C4H18N3
How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms ? [AIEEE 2006] (A) 3.125 × 10–2 (B) 1.25 × 10–2 (C) 2.5 × 10–2 (D) 0.02
Sol.
Sol.
6.
3.
Number of atoms in 560 gm of Fe (atomic mass 56 g mol–1) is – [AIEEE-2003] (A) is twice that of 70 gm N (B) is half that of 20 gm H (C) both are correct (D) None is correct
Sol.
In the reaction, [AIEEE 2007] 2Al(s)+6HCl(aq)2Al3+ (aq)+6Cl(aq)+3H2(g), (A) 6L HCl(aq) is consumed for every 3L H2(g) produced (B) 33.6 L H2(g) is produced regardless of temperature and pressure for every mole Al that reacts (C) 67.2 L H2(g) at STP is produced for every mole Al that reacts (D) 11.2 L H2(g) at STP is produced for every mole HCl (aq) consumed
Sol.
4.
6.02 ×1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is [AIEEE-2004] (A) 0.001 M
(B) 0.01 M
(C) 0.02 M (D) 0.1 M (Avogadro constant, NA = 6.02 ×1023 mol–1) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
STOICHIOMETRY - 1
Page # 125
LEVEL – II 1.
JEE ADVANCED
How many moles of e– weigh one Kg (A) 6.023 × 1023
(C)
6.023 10 54 9.108
(B)
1 × 1031 9.108
(D)
1 108 9.108 6.023
[JEE ‘2002 (Scr), 1]
5.
20% surface sites have adsorbed N2. On heating N2 gas evolved from sites and were collected at 0.001 atm and 298 K in a container or volume is 2.46 cm3. Density of surface sites is 6.023 × 1014/cm2 and surface area is 1000 cm2, find out the no. of surface sites occupied per molecule of N2. [JEE 2005]
Sol.
Sol.
6. 2.
Calculate the molarity of pure water using its density to be 1000 kg m–3. [JEE’ 2003]
Sol.
Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is:[JEE 2011] (A) 1.78 M (B) 2.00 M (C) 2.05 M (D) 2.22 M
Sol.
3.
One gm of charcoal absorbs 100 ml 0.5 M CH3COOH to form a monolayer, and there by the molarity of CH3COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/gm. [JEE’ 2003]
7.
Reaction of Br2 with Na2CO3 in aqueous solution gives sodium bromide and sodium bromate with evolution of CO2 gas. The number of sodium bromide molecules involved in the balanced chemical equation is : [JEE 2011]
Sol.
Sol.
8. 4.
Calculate the amount of Calcium oxide required when it reacts with 852 gm of P4O10.[JEE 2005]
Sol.
A decapeptide (Mol. wt. 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight tto the hydrolysed products. The number of gl yci ne uni t s prese nt i n t he decapeptideis : [JEE 2011]
Sol.
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Page # 126
STOICHIOMETRY - 1
ANSWER-KEY OBJECTIVE PROBLEMS (JEE MAIN)
Answer Ex–I 1.
A
2. A
3. A
4. A
5. A
6.
9.
A
10. B
11. A
12. B
13. C
17. A
18. B
19. A
20. A
25. C
26. D
27. C
33. C
34. D
41. D
42. C
49. C
50. B
A
7. A
8. A
14. A
15. B
16. B
21. C
22. B
23. A
24. C
28. A
29. C
30. A
31.A
32.C
35. B
36. D
37. A
38. B
39.C
40. C
43. B
44. B
45. B
46. D
47.D
48. A
OBJECTIVE PROBLEMS (JEE ADVANCED)
Answer Ex–II 1. 196.169 gm
2. 7.092 × 107
3. (a) 4 ; (b) 5000 moles; (c) 1.89 × 10–22gm
4. 1.0 × 10–4%
5. m = 4, C6H2Cl3
6. (a) C6H12, (b) C5H10O5, (c) H2O2, (d) Hg2 Cl2, (e) H4F4
7. CH
8. C3H5N, C3H5N
9. Al = 60%; Mg = 40%
10. CaCO3 = 28.4%; MgCO3 = 71.6%
11. NaHCO3 = 14.9%; Na2CO3 = 85.1%
12. %NaCl = 77.8%
13. 9.12
14. A,C 15. A, B
19. B,C
20. C
21. (A) R, (B) P, (C) Q
23. 123 g/mol
24. C
17. A,B
18. B,D
22. (A) R, (B) Q, (C) P 29. 125
25. 88
26. 25
16. A,B,C
27. 480
28. 24
30. 113
Answer Ex–III
SUBJECTIVE PROBLEMS (JEE ADVANCED)
1. (i) A2 = 1, A3B4 =2; (ii) B2 =1 A2B4=1/2; (iii) A3B4 = 0.5 A2B4 = 0.5
2. 24.51 ml
3. H2SO4 = 35.38%, Free SO3 = 63.1%, combined SO3 = 28.89%
4. 18.38 ml
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STOICHIOMETRY - 1
5. 1.052 gm/ml
Page # 127
6. (a) mA = 120, mD = 160; (b) n A
2 1 , n C 1 , nD 5 5 5
7. N2 = 30 ml, H2 = 40 ml
8.(a) CO2, H2O and O2 ; (b) nCO2 = 3, nH2O = 4 ; (c) C2H4 and CH4 are the H.C; (d) nO2 = 8 9. 0.25 mole
10. wc = 24gm ; WCCl4 = 154 gm
11. 1.1458
12. 21 : 11
13. (a) 117 kg; (b) 20.16 × 103 lit. 14. (a) 0.5 M, (b) 0.5 M, (c) o.2 M 17. 16.66%
18. 2.7 × 10–4
22. 183.68 ml
23. (a) [Ca2+] = 2 molar [NO3–] = 4 molar; (b) 0.965
24. 0.6667, 0.6667
25. 2M
26. 66.67%
30. 92.70
31. 128
28. CH3Cl
20. 29.77
16. 13.15
21. 1.736 litre
27. 35%
19. 1.288
15. 1.445
29. C7H10NCl
32. (a) pure H2SO4 (109 gm); (b) 109 gm H2SO4, 9 gm H2O; (c) 109 gm H2SO4, 111 gm H2O 33. (i) 4M; (ii) 0.06
34. (i) 20 gm H2SO4; (ii) 35.4 gm H2SO4; (iii) H2SO4 = 35.4 gm, H2O = 34.6 gm
35. 5 ml, 2ml, 3ml
36. 10 ml
37. NO = 44 ml; N2O = 16 ml
38. O3
39. NH3
40. C2H6O
Answer Ex–IV
PREVIOUS YEARS PROBLEMS
LEVEL – I 1. A
2. A
JEE MAIN 3. C
4. B
LEVEL – II
5. A
6. D
JEE ADVANCED
1. D
2. 55.5 mol L–1
3. 5 × 10–19 m2
4. 1008 gm
5. 0002
6. C
7. 0005
8. 0006
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Page # 128
PERODIC PROPERTICS
PERIODIC PROPERTIES
PERIODIC TABLE
GROUPS
Old convention IA New convention 1
PERIODS
IIA 2
PERIODIC TABLE
Ionisation energy
IIIB IVB VB 3 4 5
VIB 6
Electron Electro affinity negativity VIIB 7
Radii
VIII 9 10
8
Lanthanide Contraction
IB IIB 11 12
IIIA IVA 13 14
S– Block 1
H
2
Li
Metallic Character VA VIA 15 16
VIIA 0 17 18
p - Block He
Be
B
C
N
O
F
Ne
Al
Si
P
S
Cl
Ar
d - Block 3
Na Mg
4
K
5
Rb Sr
6
Cs Ba La
7
Fr Ra
Ca Sc Y *
Ac
Ti
V
Cr
Zr
Nb Mo Tc
Ru
Hf
Ta
Os
W
Mn Fe
Re
Co
Ni
Cu
Zn Ga
Ge As
Se
Br
Kr
Rh Pd Ag
Cd
Sn Sb
Te
I
Xe
Ir
Hg Tl
Pb
Po
At
Rn
Pt
Au
In
Bi
**
f-Block
Helium
belongs to s-block because last entered electron goes in s-block.
Iridium is the most dense element followed by Osmium.
NOTE :
(1) Last entered electron (According to Aufbau’s Principle) decides the block of the element. (2).
The valence shell determines the period number.
GENERAL PROPERTIES OF PERIODIC TABLE (1)
There are Seventeen non-metals (including hydrogen) in periodic table.
(2)
Five non - metals are solid C, P, S, Se, I.
(3)
One non - metal is liquid i.e. Br.
(4)
Eleven non-metals are gaseous.
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PERODIC PROPERTICS
Page # 129
(5)
Six gases are monoatomic (noble gases) i.e. He, Ne, Ar, Kr, Xe, Rn.
(6)
Five gases are diatomic, they are H, F, N, O, Cl
(7)
There are eight metalloids in periodic table like, B, Si, Ge, As, Sb, Te, Po, At.
(8)
Five elements are liquid at room temperature namely Cs, Fr, Ga, Hg, and Br,
(9)
s - Block and p - Block together are called Representative elements.
(10)
Five elements are radioactive amongst representative elements. They are Po, At, Rn, Fr and Ra
(11)
There are seven periods in long - form of periodic table.
1.DEVELOPEMENT OF MODERN PERIODIC TABLE Dobereiner's Triads : He arranged similar elements in the groups of three elements called as triads, in which the atomic mass of the central element was merely the arithmatic mean of atomic masses of other two elements or all the three elements possessed nearly the same atomic masses Li Na K 7 23 39 7 + 39/2 = 23 Fe Co Ni 55.85 58.93 58.71 nearly same atomic masses It was restricted to few elements, therefore, discarded
NEWLAND'S LAW OF OCTAVES : He was the first to correlate the chemical properties of the elements with their atomic masses. According to him if the elements are arranged in the order of their increasing atomic masses the eighth element starting from given one is similar in properties to the first one. This arrangement of elements is called as Newland's law of Octave. Li Be B C N O F Ns Mg Al Si P S Cl K Ca This classification worked quite well for the ligher elements but it failed in case of heavier elements and, therefore, discarded
LOTHER MEYER'S CLASSIFICATION He determined the atomic volumes by dividing atomic mass with its density in solid states. He ploted a graph between atomic masses against their respective atomic volumes for a number of elements. He found the following observations. (i) Elements with similar properties occupied similar positions on the curve. (ii) Alkali metals having larger atomic volumes occupied the crests. (iii) Transitions elements occupied the troughs. (iv) The halogens occupied the ascending portions of the curve before the inert gases. (v) Alkaline earth metals occupied the positions at about the mid points of the descending portions of the curve. On the basis of these observations he concluded that the atomic volumes (a physical property) of the elements are a periodic function of their atomic masses. It was discarded as it lacks practical utility. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 130
(d)
(i) (ii)
(iii)
(i) (ii) (iii) (iv)
(v) (vi)
PERODIC PROPERTICS
Mendeleev's Periodic Table : Mendeleev's Periodic's Law According to him the physical and chemical properties ofthe elements are a periodic function oftheir atomic masses. He arranged then known elements in order of their increasing atomic masses considering the facts that elements with similar properties should fall in the same vertical columns and leaving out blank spaces where necessary. Merits of Mendeleev's Periodic table : It has simplified and systematised the study of \3lements and their compounds It has helped in predicting the discovery of new elements on the basis of the blank spaces given in its periodic table. Mendeleevs predicted the properties of those missing elements from the known properties of the other elements in the same group. Eka - Aluminium and Eka-silicon names were given for gallium and germanium (not discovered at the time of mendeleevs). Properties predicted by Mendeleevs for these elements and those found experimentally were almost similar.
Atomic weights of elements were corrected. Atomic weight of Be was calculated to be 3 × 4.5 = 13.5 by considering its valency 3, was correctly calculated considering its valency 2 (2 × 4.5 = 9) Demerits In Mendeleev's Periodic Table: Position of hydrogen is uncertain .It has been placed in lAand VilA groups because of its resemblance with both the groups. No separate positions were given to isotopes. Anomalous position of lanthanides and actinides in periodic table. Order of increaseing atomic weights is not strictly followed in the arrangment of elements in the periodic table. For e.g.-Ar(At.wt.39.94) is placed before K(39.08) and Te (127.6) is placed before I (126.9) Similar elements were placed in differents groups(CuIB and Hg IIB) and the elements with different properties were placed in same groups(alkali metals IA and coinage metals IB) It didn't explained the cause of periodicity. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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LONG FORM OF THE PERIODIC TABLE OR MOSELEY'S PERIODIC TABLE He studied (1909) the frequency of the X-ray produecd by the bombardment of a strong beam of electrons on 'metal target. He found that the square root of the frequency of X-rays ( v ) is directly proportional to number of effective nuclear charge (z) of metal i.e. to atomic number and not to atomic mass of the atom of that metal.(as nuclear charge of metal atom is equal to atomic number) i.e. ( v ) = a (z - b) Where 'a' is the proportionality constant and 'b' is a constant for all the lines in a given series of X-rays. Therefore, he, concluded that atomic number was a better fundamental property of an element than its atomic weight He suggested that the atomic number (z) instead of atomic weight should be basis of the classification of the elements. Modern Periodic Law (Moseley's Periodic Law) Physical and chemical properties of elements are the periodic functions of their atomic number.lf the elements are arranged in order of their increasing atomic numper, after a regular interval ,element with similar properties are repeated. Periodicity The repetition of the properties of elements after regular intervals when the elements are arranged in the order of increasing atomic number is called periodicity.
(i) (ii) (iii) (iv) (v) (vi)
Cause of Perlodlcty: The periodic repetition of the properties of the elements is due to the recurrence of similar valence shell electronic configuration after certain regular intervals. For example, alkail metals have same electronic configuration ns1, therefore, have similar properties. The long form of periodic table is the contribution of Range, Werner, Bohr and Bury This table is also referred to as Bohr's table since it follows Bohr's scheme of the arrangements of elements into four types based on electronic configuration of elements The modern periodic table consits of horizontal rows (periods) and vertical column (groups) Periods: There are seven periods numbered as 1, 2,3,4,5,6 and 7. Each period consists of a series of elements haVing same valence shell. Each period corresponds to a particular principal quantum number of the valence shell present in it. Each period starts with an alkali metal having outermost electronic configuration ns1. Each period ends with a noble gas with outermost electronic configuration ns2np6 except helium having outermost electronic configuration 1s2. Each period starts with the filling of new energy level. The number of elements in each period is twice the number of atomic orbitals available in energy level that is being filled. To illustrate 1st period shortest period having only two elements. Filling of electron takes place in the first energy shell, for which, n = 1, . = 0 (s-subshell) and m = O. Only one orbital (1s) is available and thus it contains only two elements. 3rd period short period having only eight elements. Filling of electrons takes place in the third energy level. For which, n = 3, . = 0, 1,2 and no. of orbitals m = 0, 3, 5 no. of orbitals 1 3 5 (3s) (3p) (3d) ________________________ Total no. of orbitals 9 ________________________ : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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But the energy of 3d orbitals are higher than 4s orbitals. Therefore, four orbitals (one 3s and three 3p orbitals) corresponding to n = 3 are filled before filling in 4s orbital (next energy elevel). Hence 3rd period contains eight elements not eighteen elements. Groups: There are eighteen groups numbered as 1,2,3,4,5, ................ 13, 14, 15, 16, 17, 18. Group consists of a series of elements having similar valence shell electronic configuration.
2. CLASSIFICATION OF THE ELEMENTS : It is based on the type of orbitals which receives the differentiating electron (i.e., last electron).
s-block elements When shells upto (n - 1) are completely filled and the last electron enters the s-orbital of the outermost (nth) shell, the elements of this class are called s-block elements. •
Group 1 & 2 elements constitute the s-block.
•
General electronic configuration is [inert gas] ns1-2
•
s-block elements lie on the extreme left of the periodic table.
•
This block includes metals. GROUP – IA OR 1
Solids
Liquids + Radiactive
ALKALI METALS : Li Na K Rb Cs Fr
Metals
¼ y huk dh j c l sQfj ; kn½
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PERODIC PROPERTICS
GROUP – II A OR
Solids
Page # 133 2 , ALKALINE EARTH METALS :
Be Mg Ca Sr Ba Ra
Metals
¼ csVk eka xsdU; k l q Unj cki j kt h½
Radioactive
p-block elements
• • • • •
When shells upto (n -1) are completely filled and differentiating electron entres the p-orbital of the nth orbit, elements of this class are called p-block elements. Group 13 to 18 elements constitute the p-block. General electronic configuration is [inert gas] ns2np1-6 p-block elements lie on the extreme right of the periodic table. This block includes some metals, all non-metals and metalloids. s-block and p-block elements are collectively called normal or representative elements.
GROUP – III A OR 13 , BORON FAMILY :
Solids Liquid
Liquid
Metals
¼ cky xa xk/kj ba fM; u fr y d½
C } Non-metal+Maximum Catenation property in its group. Si Metalloids Ge ¼ dgksfl ; k t h l w uksi zHkq ½ Sn Metals Pb
GROUP – V A OR 15 , NITROGEN FAMILY :
Gas Solids
Metalloids + Highest/Maximum Catenation property in group.
GROUP – IV A OR 14 , CARBON FAMILY :
Solids
B Al Ga In TI
N P As Sb Bi
Maximum Catenation property in its group Metalloids Metals
¼ U; w t hy S . M] i kfdLr ku vkLVªsfy ; k l c fHk[ kkj h½
GROUP – VI A OR 16 , OXYGEN FAMILY :
Gas Solids
O S Se Te Po
Maximum Catenation property in its group Non-metals Metalloids
¼ O style l sVhi ks ½
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d-Block elements When outermost (nth) and penultimate shells (n - 1)th shells are incompletely filled and differentiating electon enters the (n -1) d orbitals (i.e., d-orbital of penultimate shell) then elements of this class are called d-block elements. •
Group 3 to 12 elements constitute the d-block.
•
General electronic configuration is [inert gas] (n – 1)d1–10 ns0–2
•
d-block elements are classified into four series
•
Series
Elements
3d
21
Sc –30Zn
3d
4d
39
Y–
4d
5d
57
La, 72 Hf–80Hg
6d
89
Ac,
48
104
(n–1)d being filled
Cd Rf – 112Uub
5d 6d
(incomplete series)
Those elements which have paratially filled d-orbits in neutral state or in any stable oxidation state are called stransition elements.
f-Block elements When n, (n–1) and (n –2) shells are incompletely filled and last electron enters into f-orbital of antepenultimate i.e., (n–2)th shell, elements of this class are called f-block elements., General electronic configuration is (n – 2)f1–14 (n – 1)d0–1 ns2 •
All f-block elements belong to 3rd group.
•
They are metals
•
Within each series, the properties of the elements are quite similar.
•
They are also called as inner-transition elements as they contain three outer most shell incomplete and were also referred to as rare earth elements since their oxides were rare in earlier days. The elements of f-block have been classfied into two series.
•
The actinides and lanthanides have been placed at the bottom of the periodic table to avoid the undue expansion of the periodic table.
1.
Ist inner transition or 4 f-series, contains 14 elements 58Ce to 70Lu. Filling of electrons takes place in 4f subshell.
2.
IInd inner transition or 5 f-series, contains 14 elements 90Th to 103Lr. Filling of electrons takes placed in 5f subshell.
3.NOMENCLATURE OF THE ELEMENTS WITH ATOMIC NUMBER > 100 (IUPAC) According to IUPAC, elements with atomic number > 100 are represented by three latter symbols. •
These symbols are based on first letter of numbers from 0 to 9. The names of these number are derived from Greek and latin languages.
•
The latin words for various digits of the atomic number are written together in the order of which makes the atomic number and suffix 'ium' is added at the end. In case of bi and tri one 'i' is omitted.
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NOTATION FOR IUPAC NOMENCLATURE OF ELEMENTS Digit
Name
Abbreviation
0
nil
n
1
un
u
2
bi
b
3
tri
t
4
quad
q
5
pent
p
6
hex
h
7
sept
s
8
Oct
o
9
enn
e
•
Official IUPAC name yet to be announced
+
Elements yet to be discovered
••
IUPAC recommended this nomenclature to be followed until their names are officially recognised
4.PREDICTION OF PERIOD, GROUP AND BLOCK
The block of an element corresponds to the type of subshell which receives the last electron The group is predicted from the number of electrons in the valence shell or/and penultimate shell as follows. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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(a) For s-block elements Group number = the no. of valence electrons (b) For p-block elements Group number = 10+ no. of valence electrons (c) For d-block elements Group number = no. of electrons in (n - 1) d sub shell + no. of electrons in valence shell.
5.METALS AND NON-METALS •
The metals are characterised by their nature of readily giving up the electron and from shinning lusture. Metals comprises more than 75% of all known elements and appear on the left hand side of the periodic table. Metals are usually solids at room temperature (except mercury). They have high melting and boiling points and are good conductors of heat and electricity. Oxides of metals are basic in nature. (Some metals in their higher oxidation state form acid oxides e.g. CrO3)
•
Non-metals do not lose electrons but take up electrons to form corresponding anions. Non-metals are located at the top right hand side of the periodic table. Non-metals are usually solids or gases at room temperature with low melting and boiling points. They are poor conductors of heat and electricity. Oxides of non-metals are acidic in nature.
6. METALLOIDS (SEMI METALS)
•
It is very much clear from the periodic table that non-metallic character increases as we move from left to right across a row. It has been found that some elements lying at the border of metallic and non-metallic behaviour, possess the properties that are characteristic of both metals and non-metals. These elements are called semi metals or metalloids.
•
The metalloids comprise of the elements B, Si, Ge, As, Sb and Te.
•
Oxides of metalloids are generally amphoteric in nature.
7.TYPICAL ELEMENTS :
•
Third period elements are called as typical elements. These include Na, Mg, AI, Si, P, S, CI.
•
The properties of all the elements belonging to a particular group resemble the properties of the corresponding typical element of that group. For example, the general properties of alkali metals (IA) can be predicted from the properties of Na, not Li, the first member of the group.
•
The properties of the elements of second period differ in the many respect belonging to the same group due to the smaller atomic size and absence of vacant d-orbitals.
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8. DIAGONAL RELATIONSHIP : Some elements of certain groups of 2nd period resemble much in properties with the elements of third period of next group i.e. elements of second and third period are diagonally related in properties. This phenomenon is known as diagonal relationship.
Diagonal relationship arises because of (i) similar size of atom and ions (Li = 1.23 Å & Mg = 1.36 Å; Li+ = 0.60 Å & Mg2+ = 0.65 Å (ii) similar electropositive characters (iii) similar polarising powers (charge to radius ratio) (iv) similarity in electronegativity values (Li = 1.0 & Mg = 1.2; Be = 1.5 &AI = 1.5) •
Similarities between properties of Li and Mg are as follows.
(a)
Li and Mg both reacts directly with nitrogen to form lithium nitride (Li3N) and magnesium nitride (Mg3N2) whereas other alkali metals of IA group does not form nitride.
(b)
Fluoride, carbonate and phosphate of Li and Mg are insoluble in water whereas these compounds of other alkali metals are soluble.
(c)
Li and Mg both are hard metals, whereas other metals of IA group are soft.
(d)
LiOH and Mg(OH)2 both are weak bases, whereas hydroxides of other elements of IA group are strong base.
(f)
Metallic bond in Li and Mg both are strong compare to other alkali metals.
(g)
Their melting and boiling points are high.
(h)
By thermal disintegration of LiNO3 and Mg (NO3)2 ; Li2O and MgO is obtained respectively.
(I)
Thermal stability of Li2CO3 and Mg CO3 is very less compare to other alkali metals and they liberates CO2 gas easily.
•
Similarly Be shows similarity to AI of IIIA group compare to other elements of IIA group which are as follows.
(a)
These both elements do not provide colour to Bunsen bumer.
(b)
They both are comparatively stable in air.
(c)
Both are insoluble in NH3 therefore do not form blue coloured solution.
(d)
There is no tendency of making peroxide and superoxide in them.
(e)
Reducing power is very less due to low value of standard electrode potential in the form of oxidation potential.
(f)
Be and AI both forms halogen bridge halides.
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9. THE PERIODICITY OF ATOMIC PROPERTIES: EFFECTIVE NUCLEAR CHARGE: Between the outer most valence electrons and the nucleus of an atom, there exists finite number of shells containing electrons. Due to the presence of these intervening electrons, the valence electrons are unable to experience the attractive pull of the actual number of protons in the nucleus. These intervening electrons act as shield between the valence electrons and protons in the nucleus. Thus, the presence of intervening (shielding) electrons reduces the electrostatic attraction between the protons in the nucleus and the valenece electrons because intervening electrons repel the valence electrons. The concept of effective nuclear charge allows us to account for the effects of shielding on periodic properties. The effective nuclear charge (Zeff) is the charge felt by the valence electron. Zeff is given by Zeff = Z – . Where Z is the actual nuclear charge (atomic number of the element) and is the shielding (screening) constant.
ATOMIC RADIUS: Probability of finding the electron is never zero even at large distance from the nucleus. Based on probability concept, an atom does not have well defined boundary. Hence exact value of the atomic radius can't be evaluated. Atomic radius is taken as the effective size which is the distance of the closet approach of one atom to another atom in a given bonding state. Atomic radius can be (A)
Covalent radius:
It is one-half of the distance between the centres of two nuclei (of like atoms) bonded by a single covalent bond. •
Covalent radius is generally used for non-metals.
Single Bond Covalent Radius, SBCR (a) For Homoatomic moleucles rA
•
dA–A = rA + rA or 2rA dA – A 2
(b) For hetrodiatomic molecules in which electro negativity remains approximately same. dA–B = rA + r B
•
For heteronuclear diatomic molecule, A – B, where difference between the electronegativity values of atom A and atom B is relativity larger, dA–B = rA + rB – 0.09 (XA – XB) where XA and XB electronegativity values of high electronegative element A and less electronegative element B, respectively. This formula is given by Stevenson & Schomaker. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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Ex.2
Calculate the bond length of C – X bond, if C – C bond length is 1.54 Å, X – X bond length is 1.00 Å and electronegativity values of C and X are 2.0 and 3.0 respectively
Sol.
(1)
C – C bond length = 1.54 Å 154 . = 0.77 Å 2 100 . rX = = 0.50 Å 2 C – X bond length
rC =
(2)
dC–X = rC + rX – 0.09 (Xx – XC) = 0.77 + 0.50 – 0.09 (3 – 2) = 0.77 + 0.50 – 0.09 × 1 = 1.27 – 0.09 = 1.18 Å Thus C – X bond length is 1.18 Å (B)
Van der Waals radius (Collision radius) :
It is one - half of the internuclear distance between two adjacent atoms in two nearest neighbouring molecules of the substance in solid state. •
van der Waa's radius does not apply to metal. its magnitude depends upon the packing of the atoms when the element is in the solid state.
•
Comparision of convalent radius and van der Waal's radius
(i)
The van der Waal's force of attractions are weak, therefore, their internuclear distances in case of atoms held by van der Waal's forces are much larger than those of between covalently bonded atoms. Therefore van der Waal's radii are always larger than covalent radii.
(ii)
A covalent bond is formed by the overlaping of two half-filled atomic orbitals, a part of the orbital becomes common. Therefore, covalent radii are always smaller than the van der Waals radii. For example,
(C)
Metallic radius (crystal radius) :
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•
PERODIC PROPERTICS
It is one -half of the distance between thenuclei of two adjacent metal atoms in the metallic crystal lattice. Metallic radius of an element is always greater than its covalent radius. It is due to the fact that metallic bond (electrical attraction between positive charge of an atom and mobile electrons) is weaker than covalent bond and hence the hence the internuclear distance between the two adjacent atoms in a metallic crystal is longer than the internuclear distance between the covalently bonded atom. For example : Metallic radius Covalent radius K 231 pm 203 pm Na 186 pm 154 pm rcovalent < rcrystal < rvander Walls
The atomic radius of inert gas (zero group) is given largest in a period because it is represented as van der Waals's radius which is generally larger than the covalent radius. The van der Waal's radius of inert gases also increases from top to bottom in a group. In the transition series (e.g. in first transition series), the covalent radii of the elements decrease from left to right across a row until near the end when the size increases slightly. On moving from left to right, extra protons are placed in the nucleus and the extra electron are added. The orbital electrons shield the nucleus charge incompletely. Thus the nuclear charge attracts all the electrons more strongly, hence a contraction in size occurs. The radii of the elements from Cr to Cu, are very close to one another because the successive addition of d-electrons screen the outer electrons (4s) from the inward pull of the nucleus. As a result of this, the size of the atom does not change much in moving from Cr to Cu. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
PERODIC PROPERTICS Element Atomic radius (A)
Page # 141 Sc 1.44
Ti 1.32
V 1.22
Cr 1.18
Mn 1.17
Fe 1.17
Co 1.16
Ni 1.15
Cu 1.17
Zn 1.25
The lanthanide contraction counter balances almost exactly the normal size increase on descending a group of transition elements. Thus covalent and ionic radii of Nb (5th peroid) and Ta (6th period) are almost same due to poor shielding of f-orbitals electrons.
IONIC RADIUS: The effective distance from the centre of nucleus of the ion up to which it has an influence in the ionic bond is called ionic radius.
For example :
Na
Na+
Number of Electrons
11
10
Number of Protons
11
Electronic Configuration
•
11 1s 2s 2p 3s 2
2
6
1
Number of Electrons
Cl 17
Cl– 18
Number of Proton
17
17
1s2 2s2 2p6
The sizes of ions increases as we go down a group (cosidering the ions of same charge) For example Li+ < Na+ < K+ < Rb+ Be2+ < Mg2+ < Ca2+ < Sr2+ F– < Cl– < Br– < I–
•
The d and f orbitals do not shield the nuclear charge very effectively. Therefore there is significant reduction in the size of the ions, just after d or f orbitals have been filled completely. This is called a lanthanide contraction. Atomic radii of Zr and Hf are almost identical due to lanthanide contraction.
•
The species containing the same number of electrons but differ in the magnitude of their nuclear charges are called as isoelectronic species. For example, N3- , O2-, F-, Ne, Na+ , Mg2+ and Al3+ are all isoelectronic species with same number of electrons (i.e, 10) but different nuclear charges of +7, +8, +9, +10, +11, +12 and +13 respectively. Within a series of isoelectronic species as the nuclear charge increases, the force of attraction by the nucleus on the electrons also increases. As a result, the ionic radii of isoelectronic species decrease with increases in the magnitude of nuclear charges. For example, O2– F– Mg2+ Na+ N2– Al3+ Ionic radii increase As effective nuclear charge decrease.
1 nuclear ch arg e
•
Pauling's empirical formula : Ionic radius
•
Following are the examples of isoelectronic series (i) S2- Cl- K+ Ca+2 Sc+3 (ii) SO2, NO3–, CO32- (iii) N2, CO, CN– : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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IONISATION ENTHALPY lonisation enthalpy/energy (IE) , sometimes also called ionisation potential (IP) , of an element is defined as the amount of energy required to remove an electron from an isolated gaseous atom of that element resulting in the formation of positive ion. (E )
1 M(g) M (g) e –
M IE2 M2 e – M2 IE3 M3 e –
• •
IE2 & IE3 are the IInd & IIlrd ionization energies to remove electron from monovalent and divalent cations respectively. In general: (IE)1 < (IE)2 < (IE)3 < ............. because, as the number of electrons decreases, the attraction between the nucleus'and the remaining electrons increases considerably and hence sUbsequent 1.E.(s) increase. Units of ionisation energy: KJ mol-1 ,K Cal mol-1, eV (electron volt) Factors Influencing lonisation enthalpy (IE) variation in a period and group may or may not be regular and can be influenced by:
(A)
Size of the Atom : lonisation energy decreases with increase in atomic size. As the distance between the outermost electrons and the nucleus increases, the force of attraction between the valence shell electrons and the nucleus decreases. As a result, outer most electrons are held less firmly and lesser amount of energy is required to knock them out. For example, ionisation energy decreases in a group from top to bottom with increase in atomic size.
(B)
Nuclear Charge: The ionisation energy increases with increase in the nuclear charge. This is due to the fact that with increase in the nuclear charge, the electrons of the outer most shell are more firmly held by the nucleus and thus greater amount of energy is required to pull out an electron from the atom. For example, ionisation energy increases as we move from left to right along a period due to increase in nuclear charge.
(C)
Shielding effect: The electrons in the inner shells act as a screen or shield between the nucleus and the electrons in the outermost shell. This is called shielding effect. The larger the number of electrons in the inner shells, greater is the screening effect and smaller the force of attraction and thus (IE) decreases.
(D)
Penetration Effect of the Electron : The ionisation energy increases as the penentration effect of the electrons increases. It is a well known fact that the electrons of the s-orbital has the maximum probability of being found near the nucleus and this probability goes on decreasing in case of p, d and f orbitals of the same energy level. Within the same energy level, the penetration effect decreases in the order s>p>d>f Greater the penetration effect of electron more firmly the electron will be held by the nucleus and thus higher will be the ionisation energy of the atom. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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For example, ionisation energy of aluminium is comparatively less than magnesium as outer most electron is to be removed from p-orbital (having less penetration effect) in aluminium where as in magnesium it will be removed from s-orbital (having large penetration effect) of same energy level. (E)
Electronic Configuration : If an atom has exactly half-filled or completely filled orbitals, then such an arrangement has extrastability. The removal of an electron from such an atom requires more energy than expected. For example,
As noble gases have completely filled electronic configuration, they have highest ionisation energies in their respective periods.
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Metallic or electropositive character of elements increases as the value of ionisation energy decreases. The relative reactivity of the metals increases with the decrease in ionisation energy. The reducing power of elements increases as the value of ionisation energy decreases (Li is exception in Alkali metals group which has highest reducing power) Ex.3
First and second ionisation energies of Mg(g) are 740 and 1450 kJ mor–1. Calculate percentage of Mg+(g) and Mg2+(g), if 1 g of Mg(g) absorbs 50 kJ of energy.
Sol.
Number of moles of 1g of Mg =
1 = 0.0417 24
Energy required to convert Mg(g) to Mg+(g) = 0.0417 x 740 = 30.83 kJ Remaining energy = 50 – 30.83 = 19.17 kJ Number of moles of Mg2+ formed =
19.17 = 0.0132 1450
Thus, remaining Mg+ will be = 0.0417 – 0.0132 = 0.0285 % Mg+ =
0.0285 × 100 = 68.35% 0.0417
% Mg+ = 100 – 68.35 = 31.65%
ELECTRON GAIN ENTHALPY (ELECTRON AFFINITY) : When an electron is added to a neutral gaseous atom (X) to convert it into a negative ion, the enthalpy change accompanying the process is defined as the electron gain enthalpy. Electron gain enthalpy provides a measure of the ease with which an atom adds an electron to form anion. X(g) e – X – ( g); H Heq
...(i)
The negative of the enthalpy change for the process depicted in equation (i) is defined as the electron affinity (EA) of the atom undergoing the change for the formation of.an anion. Depending on the elements, the process of adding an electron to the atom can be either endothermic or exothermic. When an electron is added to the atom and the energy is released, the electron gain enthalpy is negative and when energy is needed to add an electron to the atom, the electron gain enthalpy is positive. The addition of second electron to an anion is opposed by electrostatic repulsion and hence the energy has to be supplied for the addition of second electron. O(g) e – O – (g) + Energy EA (i) O – (g) e – energy O 2 – (g) EA(ii) EA (i) is exothermic whereas EA(ii) is endothermic. Group 17 elements (halogens) have very high negative electron gain enthalpies because they can attain stable noble gas electronic configuration by picking up an electron. Noble gases have large positive electron gain enthalpies because the electron has to enter the next higher energy level leading to a very unstable electronic configuration. Electron gain enthalpy of O or F is less than S or CI. This is due to the fact that when an electron is added to O or F, the added electron goes to the smaller n = 2 energy level and experiences significant repulsion from the other electrons present in this level. In S or CI, the electron goes to the larger n = 3 energy level and consequently occupies a larger region of space leading to much less electronelectron repulsion. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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Electron gain enthalpies of alkaline earth metals are very less or positive because the extra electron is to be added to completely filled s-orbitals in their valence shells. Across a period, with increase in atomic number, electron gain enthalpy becomes more negative because left to right across a period effective nuclear charge increases and consequently it will be easier to add an electron to a small atom. As we move in a group from top to bottom, electron gain entalpy becomes less negative because the size of the atom increases and the added electron would be at larger distance from the nucleus. (i) Electron affinity
1 Atomic size
(ii) Electron affinity Effective nuclear charge (zeff)
1 (iv) Stability of half filled and completely filled orbitals of a subshell screening effect is comparatively more and the addition of an extra electron to such an system is difficult and hence the electron affinity value decreases. (iii) Electron affinity
Ex.4
How many CI atoms can you ionise in the process Cl Cl e the energy liberated for the process Cl e Cl – for one Avogadro number of atoms. Given IP = 13.0 eVand EA= 3.60 eV
Sol.
Let n atoms be ionised. 6.02 × 1023 × EA = n × IP n
Ex.5
6.02 10 23 3.60 = 1.667 × 1023 13
The first ionisation potential of Li is 5.4 eV and the electron affinity of CI is 3.6 eV Calculate H in kcal mol–1 for the reaction. Li(g) Cl(g) Li Cl– Carried out at such low pressures that resulting ions do not combine with each other.
Sol.
The overall reaction is written into two partial equations Li(g) Li e
E1 = 5.4 eV
Cl(g) e Cl–
E2 = –3.6 eV
H = E1 – E2 = 5.4 – 3.6 = 1.8 eV = 1.8 × 23.06 kcal mol–1 = 41.508 kcal mol–1 Ex.6
For the gaseous reaction, K ( g) F( g) K F – ,H was calculated to be 19 kcal under conditions where the cations and anions were prevented by electrosatic separation from combining with each other. The ionisation potential of K is 4.3 eV. What is the electron effinity of F ?
Sol.
K K e F e F –
E1 = 4.3 eV E2 = – E eV
19.0 E1 – E2 4.3 – E 23.06 0.82 = 4.3 – E E = 3.48 : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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PERODIC PROPERTICS
The electron affinity of chlorine is 3.7 eV. How much energy in kcal is released when 2 g of chlorine is completely converted to CI– ion in a gaseous state? (1 eV = 23.06 kcal mol-1)
Sol.
C e Cl – 3.7eV
35.5
3.7 × 23.06 kcal
.'. Energy released for conversion of 2 g gaseous chlorine into CI– ions
3.7 23.06 × 2 = 4.8 kcal 35.5
ELECTRO-NEGATIVITY Electronegativity is a measure of the tendency of an element to attract electrons towards itself in a covalently bonded molecules . X(g) + e– X–(g) The magnitude of electronegativity of an element depends upon its ionisation potential & electron affinity. Higher ionisation potential & electron affinity values indicate higher electronegativity value. With increase in atomic size the distance between nucleus and valence shell electrons incerases, therefore, the force of attraction between the nucleus and the valence shell electrons decreases and hence the electronegativity values also decrease. With increase in nuclear charge force of attraction between nucleus and the valence shell electrons increases and, therefore, electronegativity value increases In higher oxidation state, the element has higher magnitude of positive charge. Thus, due to more positive charge on element, it has higher polarising power. Thus, with increase in the oxidation state of element, its electronegativity also increases. Charge on cation electronegativity of the atom
There is no direct method to measure the value of electronegativity, however, there are some scales to measure its value . (a)
Pauling's Scale: Linus Pauling developed a method for calcUlating relative electronegativities of most elements. According to Pauling A – B 0.208
where A –B – A – A B –B
where 'S refer to bond enthalpies (K Cal mol-1). By assigning F 4.0 , values of other atoms can be computed. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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Page # 147
Mulliken's scale Electronegativity (EN) can be regarded as the average ofthe ionisation energy (IE) and the electron affinity (EA) of an atom. (EN)
IE EA 2
If both (EA) and (IE) are determined in eV units then paulings's electronegativity (EN)p is related to Mulliken's electronegativity. Mulliken's values were about 2.8 times larger than the Pauling's values. (c)
Allred-Rochow's Electronegatlvlty Allred and Rochow defined electronegativity as the force exerted by the nucleus of an atom on its valence electrons: (EN) AR
0.359 Z effective r2
0.744
where Zelleclive is the effective nuclear charge and r the covalent radius (in Å ). The electron negativity of Cs (55) is less than Fr (87). This is due to the increase of + 32 units in nuclear charge of Fr which makes the effective nuclear charge comparatively high. The electroneativity of inert gas elements of zero group is zero. Inert gases exist as monoatomic molecules and the electronegativity is the property of bonded atoms. Ex.8
lonisation potential and electron affinity of fluorine are 17.42 and 3.45 eV respectively. Calculate the electronegativity of fluorine.
Sol.
According to Mulliken equation X
IP EA when both IP and EAare taken in eV.. 5.6
XF
17.42 3.45 3.726 5.6
APPLICATIONS OF ELECTRONEGATIVITY : (I)
Nomenclature: Compounds formed from two nonmetals are called binary compounds. Name of more electronegative element is written at the end and 'ide' is suffixed to it. The name of less electronegative element is written before the name of more electronegative element of the formula.
Ex.9
Write the correct formula and name of the following (a) ICI or CIl (b) FCI or CIF (c) BrCI or CIBr (e)OF2 or F2O
Sol.
Correct formula (a) I+ Cl–
Name Iodine chloride
(b) CI+ F–
Chlorine fluoride
(c) Br CI (d) IBr
Bromine chloride Iodine bromide
(e) OF2
Oxygen difluoride
(f) Cl2O
Dichlorine oxide
+
–
(d) BrI or IBr
(f)Cl2O or OCI2
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PERODIC PROPERTICS
Nature of Bond: If difference of electronegativities of the two elements is 1.7 or more, then ionic bond is formed between them whereas if it is less than 1.7, then covalent bond is formed. (HF is exception in which bond is covalent although difference of electronegativity is 1.9)
(iii)
Metallic and Nonmetallic Nature : Generally values of electronegativity of metallic elements are low, whereas electronegativity values of nonnmetals are high.
(iv)
Partial Ionic Character in Covalent bonds Partial ionic characters are generated in covalent compounds by the difference of electronegativities. Hanny and smith calculated percentage of ionic character from the difference of electronegativity. Percentage of ionic character
=
16(XA – XB) + 3.5 (XA – XB)2
=
16 + 3.52
=
(0.16 + 0.035 2) × 100
XA is electronegativity of element A XB is electronegativity of element B = XA – XB (v)
Bond length When difference of electronegativities of atoms present in a molecule is increased, then bond length decreases. Shoemaker and stephensen determined. Bond length dA – B = rA + rB – 0.09 |XA – XB| or dA
(vi)
–B
=
1 (DA – A + DB – B) – 0.09 |XA – XB| 2
Bond Strength & Stability Bond strength and stability of A – B increases on increase in difference of electronegativities of atoms A and B bonded A – B. Therefore H – F > H – Cl > H – Br > H – I
Ex.10 Electronegativity of which of the following is high ? (1) –CH3(sp3) Ans.
(2) H2C = CH2(sp2) (3) CH CH(sp)
(4) Equal in all
(3)
Ex.11 CF3NH2 is not a base, whereas CH3NH2 is a base. What is the reason ? Sol.
Due to high electronegativity of F tendency of donating the lone pair of electrons present on N will be less
Ex.12 OF2 is called oxygen difluoride, whereas Cl2O is called dichlorine monoxide. Why ? Sol.
Electronegativity of O in OF2 is less than F. Therefore, there will be positive charge on oxygen and negative charge on fluorine. Whereas in Cl and O, electronegativity of Cl is less than that of O therefore there will be positive charge on Cl and negative charge on O. Positive charge is written first followed by negative charge.
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Ex.13 Calculate the electronegativity of fluorine from the following data : EH – H = 104.2 kcal mol–1, EF–F = 36.6 kcal mol–1 EH–F = 134.6 kcal mol–1, XH = 2.1 Sol.
Let the electronegativity of fluorine be XF. Applying Pauling's equation.
XF – XH 0.208[EH–F – (EF –F EH–H )1/ 2 ]1/ 2 In this equation, dissociation energies are taken in kcal mol–1. XF – 2.1 = 0.208 [134.6 – (104.2 × 36.6)1/2]1/2 XF = 3.87 Ex.14 The electron affinity of chlorine is 3.7 eV. How much energy in kcal is released when 2 g of chlorine is completely converted to Cl– ion in a gaseous state ? (1 eV = 23.06 kcal mol–1) Sol.
Cl + e Cl– + 3.7 eV 35.5
3.7 × 23.06 kcal
Energy released for conversion of 2 g gaseous chlorine into Cl– ions
3.7 23.06 × 2 = 4.8 kcal 35.5
Ex.15 Calculate the electronegativity of fluorine from following data : EH–H = 104.2 kcal mol–1 EF–F = 36.6 kcal mol–1 EH–F = 134.6 kcal mol–1 Electronegativity of H is 2.05. Sol.
On Paulling scale : xF – xH = 0.182 H – F (using B.E. in kcal mol–1) H–F = EH–F –
EH–H EF –F = 13.45 –
104.2 36.6 = 72.84 kcal
From (i) xF – xH = 0.182 xF = xH + 1.4434
72.84 = 1.5534
= 2.05 + 1.5534 = 3.6034
UNSOLVED EXERCISE : 1.
The standard enthalpies of formation of gaseous XeF2, XeF4 & XeF5 are –108, –216 & –294 kJ mol–1 respectively and the bond energy in F2 is 159 kJ mol–1. Calculate the average Xe–F bond energy in each of these compounds and use of the value for XeF2 to obtain a value for the electronegativity of xenon on the Pauling scale assuming the electronegativity of fluorine to be 4.
2.
Calculate the electronegativity value of chlorine on Mulliken's scale, given that IP = 13.0 eV and EA = 4.0 eV. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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PERODIC PROPERTICS
(VII) METALLIC PROPERTY Metals have the tendency to form cations by loss of electrons and this property makes the elements as electropositive elements or metals. M(g) M ( g) e –
(IX)
OXIDES Oxygen react with all elements except noble gases, Au, Pd and Pt to form oxides. In general, metallic oxides (O2–), peroxides (O22–) and super oxides (O2–1) are ionic solids. The tendency of group IA metals (alkali metals) to form oxygen rich compounds increases from top to bottom i.e. with increasing cation radii and decreasing charge density on the metal ion. IIA metals also show the similar trend. Except Be, the IIA metals react with oxygen at normal conditions to form normal ionic oxides and at high pressure of O2, they form peroxides (CaO2, SrO2, BaO2). Oxides of metals are called as basic anhydries as most of them combine with water forming hydroxides with no change in oxidation state of metals. Oxides of IA and IIA dissolve in water forming basic solution where as other oxides do not dissolve in water. Na2 O H2 O 2NaOH Oxygen combines with many non-metals to form covalent oxides such as CO, CO2, SO2, P4O10, Cl2O7 etc. Non-metals with limited supply of oxygen usually form oxides in which non-metals are present in lower oxidation states where as with excess of oxygen, oxides with higher oxidatin state are formed. Oxides of non-metals are called as acid anhydrides as most of them dissolve in water forming acids of oxy-acids. P4O10 + 6H2O 4H3 PO4 ; SO3 + H2O H2SO4 : Cl2O7 + H2O 2HClO4 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
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In a group, basic nature of oxides increases or acidic nature decreases. Oxides of the metals are generally basic and oxides of the non-metals are acidic. The oxides of the matalloids are amphotetric in nature. The oxides of Al, Zn, Sn, As and Sb are amphoteric. In a period the nature of the oxides varies from basic to acidic. Na2O
MgO
Strongly basic Basic
Al2O3 amphoteric
SiO2
P4O10
Weakly acidic Acidic
SO3 Acidic
Cl2O7 Strongly acidic
CO, N2O, NO and H2O are neutral oxides.
GRAPHS OF PERIODIC PROPERTIES ALKALI METALS
LANTHANIDES
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PERODIC PROPERTICS
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PERODIC PROPERTICS
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Page # 155
IMPORTANT POINTS TO REMEMBER 1.
The basis of Mendeeleev's periodic table was his periodic law. According to it the physical and chemical properties of the elements are a periodic function of their atomic masses.
2.
According to Moseley, that a plot of v (where v is frequency of X-Rays emitted) against atomic number (Z) gave a straight line and not the plot of ~ vs atomic mass. Therefore, he concluded that atomic number (Z) instead of atomic mass was a better fundamental property of an element and atomic number instead of atomic mass should be basis of the classification of the elements. Long form of periodic table contains seven periods (horizontal rows) and eighteen groups (vertical columns). In the modern periodic table, the period indicates the value of principle quantum number. The number of elements in each period is twice the number of atomic orbitals available in the energy level that is being filled. Group consists of a series of elements having similar valence shell electronic configuration. In modern periodic table each block (s, - p, - d - and f -) contains a number of columns equal to the numbers of electrons that can occupy that sub-shell. The 4f - (i.e., actinides) and 5f - (i.e., lanthanides) inner transition series of element are placed separately in the periodic table to maintain its structure and to preserve the principle of classification. Metals comprise more than 78% of all known elements and appear on the left side of the periodic table. Silicon, germanium, arsenic, antimony, selenium and tellurium all are semi - metals or metalloids. The combined effect of attractive force due to nucleus and repulsive force due to intervening electrons, acting on the valence electrons is that the valence - electron experiences less attraction from the nucleus. This is called shielding or screening effect. Covalent radius < metallic or crystal radius < Van der Waal's radius. Species having same number of electrons but different in the magnitude of their nuclear charges are called as isoelectronic species and their size is inversely proportional to their effective nuclear charge. The smaller the ionisation energy, the easier it is for the neutarl atom to change in to a positive ions in gaseous state.IE1 < IE2 < IE3 .........• The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration. End of valence electrons is marked by a big jump in ionization enthalpy. Electron gain enthalpy provides a measure of the ease with which an atom adds an electron to form anion:
3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
x(g) e – X – ( g); H egH
16. 17. 18. 19.
20.
The negative value of electron gain enthalpy of C > F (similarly S > 0) because there is weak electronelectron repulsion in the bigger 3p subshell of C as compared to compact 2p - subshell of F. Noble gases have larger positive electron gain enthalpies because the electron has to enter the next higher energy level. Addition of 2nd electron to an anion is opposed due to electrostatic repulsion and thus requires the absorption of energy e.g in case of the formation of S2–, O2– etc. The relative reactivity of metals increases with the decrease in their ionisation energies. Similarly the relative reactivity of non-metals increases with increases in the negative value of electron gain enthalpy.
According to Pauling, the electronegativity difference between two atoms is equal to 0.208 where L\ is the extra bond energies in K cal mol-1. The acidic character of oxides increases when electronegativity difference decreases between element and oxygen (E - O).
21.
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EXERCISE – I 1.
PERODIC PROPERTICS
OBJECTIVE PROBLEMS (JEE MAIN)
Which of the following is not a Doeberiner triad (A) Li, Na, K (B) Mg, Ca, Sr (C) Cl, Br, I (D) S, Se, Te
Sol.
Sol. 6.
2.
Which of the following set of elements obeyes Newland’s octave rule (A) Na, K, Rb (B) F, Cl, Br (C) Be, Mg, Ca (D) B, Al, Ga
The places that were left empty by Mendeleef were, for (A) Aluminium & Silicon (B) Gallium and germinium (C) Arsenic and antimony (D) Molybdenum and tungsten
Sol.
Sol.
7. 3.
Which of the following is/are Doeberiners triad (i) P, As, Sb (ii) Cu, Ag, Au (iii) Fe, Co, Ni (iv) S, Se, Te Correct answer is (A) (i) and (ii) (B) (ii) and (iii) (C) (i) and (iv) (D) All
Which of the following pairs of elements do not follow octave rule (A) Na, K (B) Ca, Sr (C) F, Cl (D) O, S
Sol.
Sol. 8.
4.
Which is not anomalous pair of elements in the Medeleeves periodic table (A) Ar and K (B) Co and Ni (C) Te and I (D) Al and Si
Elements which occupied position in the lother meyer curve, on the peaks, were (A) Alkali metals (B) Highly electro positive elements (C) Elements having large atomic volume (D) All
Sol.
Sol.
5.
Which are correct match (i) Eka silicon – Be (ii) Eka aluminium – Ga (iii) Eka mangenese – Tc (iv) Eka scandium – B (A) (ii) & (iii) (B) (i), (ii) & (iv) (C) (i) & (iv) (D) All
9.
Modern periodic table is based on atomic no. experiments which proved importance of at no. was (A) Braggs work on X-ray diffraction (B) Moseleys work on X-ray spectrum (C) Mulliken’s oil drop experiment (D) Lother meyer curve plotted between at vol. & at wt.
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10.
Page # 157 14.
Atomic no. is the base of (i) Lother meyer curve (ii) Newland octave rule (iii) Modern periodic table (iv) Doeberiener triad rule (v) Long form of periodic table (A) (i), (ii), (iv) (B) (iii), (v) (C) (i), (iv) (D) (i), (iii), (v)
The discovery of which of the following group of elements gave a death blow to the Newlands Law (A) Inert gases (B) Alkali metals (C) Transuranic element (D) Halogens
Sol.
15.
Sol.
Which of the following pair of elements follows Newland’s octave rule (A) Ne, Ar (B) C, N (C) Na, K (D) P, S
Sol.
11.
Atomic wt. or Cl = 35.5 and of I = 127. According to doeberiner triad rule, At. wt. of Br will be (A) 80.0 (B) 162.5 (C) 81.25 (D) 91.5
16.
Sol.
238 92 U (IIIB)
234
changes to 90 Th by emission of particle. Daughter element will be in (A) IIIB (B) IB (C) VB (D) IIA
Sol.
12.
The elements of groups, 1, 2, 13, 14, 15, 16 and 17 are collectively called (A) Noble gases (B) Representative or normal elements (C) Transition elements (D) Inner transition elements
17.
Sol.
From the list given below, elements which belongs to the same group or sub-group are (A) Atomic number = 12, 20, 4, 88 (B) Atomic number = 8, 16, 34, 2 (C) Atomic number = 11, 18, 27, 5 (D) Atomic number = 24, 47, 42, 55
Sol.
13.
Justification of putting H in VII A group is (A) H is gas (B) H is non metal (C) It form NaH like salt (D) It has ortho and para allotropes
18.
Sol.
The name ‘Rare earths’ is used for (A) Lanthanides only (B) Actinides only (C) Both lanthanides and actinides (D) Alakaline earth metals
Sol.
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Page # 158 Z/e ratio for N3–, O2– and F– respectively will be (A) 0.7, 0.8 and 0.9 (B) 0.9, 0.8 and 0.7 (C) 7, 8 and 9 (D) 9, 8 and 7
19.
PERODIC PROPERTICS Sol.
Sol. 25.
Atomic number 15, 33, 51 represents the following family (A) Carbon family (B) Nitrogen family (C) Oxygen family (D) None
Sol. 20.
There are 10 neutrons in the nucleus of the element zM19. It belongs to (A) f-block (B) s-block (C) d-block (D) None of these
26.
Sol.
21.
For Rb(Z = 37), points out the number of electrons present in L and N shells respectively (A) 8 and 18 (B) 18 and 8 (C) 8 and 8 (D) 2 and 8
22.
The electronic configuration of an element is 1s2, 2s22p6, 3s23p4 .The atomic number of element present just below the above element in periodic table is (A) 36 (B) 34 (C) 33 (D) 32
The atom having the valence shell electronic configuration 4s2 4p2 would be in (A) Group II A and period 3 (B) Group II B and period 4 (C) Group IV A and period 4 (D) Group IV A and period 3
Sol.
27.
The number of elements know at that time when Mendeleev arranged them in the periodic table was(A) 63 (B) 60 (C) 71 (D) 65
Sol.
Sol.
28.
23.
The number of elements in 5th and 6th period of periodic table are respectively (A) 8, 18 (B) 18, 18 (C) 18, 32 (D) 18, 28
As applied to periodic table, which of the following sets include only magic numbers (A) 2, 8, 20, 28, 50, 82, 126 (B) 2, 8, 8, 18, 18, 32 (C) 2, 2, 8, 8, 18, 32 (D) 2, 8, 18, 18, 32, 32
Sol.
Sol.
29. 24.
Atomic number of Ag is 47. In the same group the atomic number of elements placed above and below Ag will be (A) 37, 67 (B) 29, 79 (C) 39, 69 (D) 29, 65
In the general electronic configuration (n – 2)f1–14 (n – 1)d0–1 ns2, if value of n = 7 the configuration will be (A) Lanthenides (B) Actinides (C) Transition elements (D) None
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30.
Page # 159 34.
The correct order of atomic size of C, N, P, S follows the order (A) N < C < S < P (B) N < C < P < S (C) C < N < S < P (D) C < N < P < S
In w hi ch of the fol l ow i ng comp ound s manganese show maximum radius (A) MnO2 (B) KMnO4 (C) MnO (D) K3[Mn(CN)6]
Sol.
Sol. 35.
31.
Match list-I with list-II and select the correct answer using the codes given below List-I List-II Ion Radius (in pm) (I) Li + (a) 216 + (II) Na (b) 195 (III) Br– (c) 60 – (IV) I (d) 95 Codes : I II III IV (A) a b d c (B) b c a d (C) c d b a (D) d c b a
Arrange in the increasing order of atomic radii of the following elements O, C, F, Cl, Br (A) F < O < C < Cl < Br (B) F < C < O < Cl < Br (C) F < Cl < Br < O < C (D) C < O < F < Cl < Br
Sol.
36.
The correct order of size would be (A) Ni < Pd ~ (B) Pd < Pt < Ni Pt (C) Pt > Ni > Pd (D) Pd > Pt > Ni
Sol.
Sol. 37.
32.
The ioni c radi i of N 3– , O 2– and F – are respectively given by (A) 1.36, 1.40, 1.71 (B) 1.36, 1.71, 1.40 (C) 1.71, 1.40, 1.36 (D) 1.71, 1.36, 1.40
Which group of atoms have nearly same atomic radius (A) Na, K, Rb, Cs (B) Li, Be, B, C (C) Fe, Co, Ni (D) F, Cl, Br, I
Sol.
Sol.
38. 33.
Sol.
The screening effect of d-electrons is (A) Equal to the p-electrons (B) Much more than p-electrons (C) Same as f-electrons (D) Less than p-electrons
In the ions P3– , S2– and Cl – the increasing order of size is (A) Cl– < S2– < P3– (B) P3– < S2– < Cl – (C) S2– < Cl – < P3– (D) S2– < P3– < Cl –
Sol.
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Page # 160 39.
PERODIC PROPERTICS
Atomic radii of Fluorine and Neon in Angstrom units are given by (A) 0.72, 1.60 (B) 1.60, 1.60 (C) 0.72, 0.72 (D) None of these
45.
Sol.
Consider the isoelectronic series : K+, S2– , Cl – and Ca2+, the radii of the ions decrease as (A) Ca2+ > K + > Cl – > S2– (B) Cl– > S2– > K + > Ca2+ (C) S2– > Cl – > K + > Ca2+ (D) K+ > Ca2+ > S2– > Cl –
Sol. 40.
41.
Which of the (A) 1s2, 2s2, (B) 1s2, 2s2, (C) 1s2, 2s2, (D) 1s2, 2s2,
following has largest radius 2p6, 3s2 2p6, 3s2, 3p1 2p6, 3s2, 3p3 2p6, 3s2, 3p5Sol.
Arrange the elements in increasing order of atomic radius Na, Rb, K, Mg (A) Na < K < Mg < Rb (B) K < Na < Mg < Rb (C) Mg < Na < K < Rb (D) Rb < K < Mg < Na
46.
Which of the following is not isoelectronic series(A) Cl –, P3– , Ar (B) N3– , Ne, Mg+2 +3 + (C) B , He, Li (D) N3–, S2– , Cl –
Sol.
47.
Sol.
In the isoelectronic species the ionic radii (Å) of N3–, Ne and Al +3 are respectively given by (A) 1.36, 1.40, 1.71 (B) 1.36, 1.71, 1.40 (C) 1.71, 1.40, 1.36 (D) 1.71, 1.36, 1.40
Sol. 42.
Which of the following sequences is correct for decreasing order of ionic radius (A) Se–2 > I– > Br– > O–2 > F– (B) I– > Se–2 > O–2 > Br– > F– (C) Se–2 > I– > Br– > F– > O–2 (D) I– > Se–2 > Br– > O–2 > F–
48.
Sol.
Correct orders of Ist I.P. are (i) Li < B < Be < C (ii) O < N < F (iii) Be < N < Ne (A) (i), (ii) (B) (ii), (iii) (C) (i), (iii) (D) (i), (ii), (iii)
Sol. 43.
The (A) (B) (C) (D)
order S–2 > Cl – > S–2 > S–2 >
of size is Cl – > O–2 > F– S–2 > O–2 > F– O–2 > Cl – > F– O–2 > F– > Cl –
49.
Sol.
44.
Sol.
Arrange the following in order of increasing atomic radii Na, Si, Al, Ar (A) Na < Si < Al < Ar (B) Si < Al < Na < Ar (C) Ar < Al < Si < Na (D) Na < Al < Si < Ar
The maximum tendency to form unipositive ion is for the elment with the electronic configuration(A) 1s2, 2s2, 2p6, 3s2 (B) 1s2, 2s22p6, 3s23p1 (C) 1s2, 2s22p6, 3s23p2 (D) 1s2, 2s22p6, 3s23p3
Sol.
50.
The second ionisation potentials in electron volts of oxygen and fluorine atoms are respectively given by (A) 35.1., 38.3 (B) 38.3, 38.3 (C) 38.3, 35.1 (D) 35.1, 35.1
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PERODIC PROPERTICS Sol.
51.
56.
A sudden large jump between the values of 2nd and 3 rd IP of an element would be associated with the electronic configuration(A) 1s2, 2s2 2p6, 3s1 (B) 1s2, 2s2 2p6, 3s2 3p5 (C) 1s2, 2s2 2p6, 3s2 3p2 (D) 1s2, 2s2 2p6 3s2
Sol.
52.
Page # 161
The correct order of stability of Al+, Al +2, Al+3 is(A) Al +3 > Al +2 > Al + (B) Al +2 > Al +3 > Al + (C) Al +2 < Al + > Al +3 (D) Al +3 > Al + > Al +2
IP1 and IP2 of Mg are 178 and 348 K.cal mol – 1. The enthalpy required for the reqction Mg Mg2+ + 2e– is (A) + 170 K.cal (B) + 526 K.cal (C) – 170 K.cal (D) – 526 K.cal
Sol.
57.
Highest ionisation potential in a period is shown by (A) Alkali metals (B) Noble gases (C) Halogens (D) Representative elements
Sol.
Sol. 58. 53.
The ionization energy of sodium is 495 kJ mol– 1. How much energy is needed to convert atoms persent in 2.3 mg of sodium into sodium ions (A) 4.95 J (B) 49.5 J (C) 495 J (D) 0.495 J
In which case the energy released is minimum (A) Cl Cl – (B) P P– (C) N N– (D) C C–
Sol.
Sol. 59.
54.
Ionisation energy increases in the order (A) Be < B < C < N (B) B < Be < C < N (C) C < N < Be < B (D) N < C < Be < B
In the formation of a chloride ion, from an isolated gaseous chlorine atom, 3.8 eV energy is released, which would be equal to (A) Electron affinity of Cl – (B) Ionisation potential of Cl (C) Electronegativity of Cl (D) Ionisation potential of Cl –
Sol.
Sol.
55.
Mg forms Mg(II) because of (A) The oxidation state of Mg is +2 (B) Difference between I.P1 and I.P2 is greater than 16.0 eV (C) There are only two electrons in the outermost energy level of Mg (D) Difference between I.P1 and I.P2 is less than 11 eV
60.
The correct order of electron affinity is (A) Be < B < C < N (B) Be < N < B < C (C) N < Be < C < B (D) N < C < B < Be
Sol.
Sol.
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PERODIC PROPERTICS
EXERCISE – II 1.
OBJECTIVE PROBLEMS (JEE ADVANCED)
Which of the following does not reflect the periodicity of element (A) Bonding behaviour (B) Electronegativity (C) Ionisation potential (D) Neutron/Proton ratio
5.
If there were 10 periods in the periodic table then how many elements would this can maximum comprise of ?
Sol.
Sol. 6.
2.
Choose the s-block element in the following : (A) 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s1 (B) 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1 (C) 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 (D) all of the above
The size of the following species increases in the order : (A) Mg2+ < Na+ < F– < Ar (B) F– < Ar < Na+ < Mg2+ (C) Ar < Mg < F– < Na+ (D) Na+ < Ar < F– < Mg2+
Sol.
Sol.
7. 3.
False statement for periodic classification of element is (A) The propeties of the elements are periodic function of their atomic numbers. (B) No. of non-metallic elements is less than the no. of metallic elements. (C) First ionization energy of elements does not change continuously with increasing of atomic no. in a period. (D) d-subshell is filled by final electron with increasing atomic no. of inner transition elements.
8.
The outermost electronic configuration of most electronegative element is : (A) ns2 np (B) ns2np4 2 5 (C) ns np (D) ns2np6
Sol.
Sol.
4.
Element in which maximum ionization energy of following electronic configuration would be (A) [Ne] 3s2 3p1 (B) [Ne] 3s2 3p2 2 3 (C) [Ne] 3s 3p (D) [Ar] 3d10 4s2 4p3 Sol.
Pick out the isoleelctronic structure from the following : I. +CH3 (A) I and II (C) I and III
II. H3O+
III. NH3
9.
IV. CH3–
(B) III and IV (D) II, III and IV
The electron affinity of the member of oxygen of the periodic table, follows the sequence (A) O > S > Se (B) S > O < Se (C) O < S > Se (D) Se > O > S
Sol.
Sol.
10.
The process of requiring absorption of energy is (A) F F– (B) Cl Cl– (C) O– O2– (D) H H–
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PERODIC PROPERTICS Sol.
Page # 163 14.
The correct order of second ionisation potential of C, N, O and F is : (A) C > N > O > F (B) O > N > F > C (C) O > F > N > C (D) F > O > N > C
Sol.
11.
In the following which configuration of element has maximum electronegativity. (A) 1s2, 2s2, 2p5 (B) 1s2, 2s2, 2p6 2 2 4 (C) 1s , 2s , 2p (D) 1s2, 2s2 , 2p6 , 3s2 , 3p3
Sol.
15.
Decreasing ionisation potential for K, Ca & Ba is (A) Ba > K > Ca (B) Ca > Ba > K (C) K > Ba > Ca (D) K > Ca > Ba
Sol.
12.
Highest size will be of (A) Br– (B) I – (C) I (D) I+
Sol.
16.
Element Hg has two oxidation states Hg+1 & Hg+2. the right order of radii of these ions. (A) Hg+1 > Hg+2 (B) Hg+2 > Hg+1 +1 +2 (C) Hg = Hg (D) Hg+2 Hg+1
Sol.
13.
Atomic radii of flourine and neon in Å units are respectively given by (A) 0.72, 1.60 (B) 1.60, 1.60 (C) 0.72, 0.72 (D) 1.60, 0.72
17.
Sol.
The ionization energy will be maximum for the process. (A) Ba Ba++ (B) Be Be++ + (C) Cs Cs (D) Li Li+
Sol.
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PERODIC PROPERTICS
Bond distance C – F in (CF4) & Si–F in (SiF4) are respective 1.33 Å & 1.54 Å. C–Si bond is 1.87 Å. Calculation the covalent radius of F atom ignoring the electronegativity differences. (A) 0.64 Å
(B)
133 . 154 . 18 . Å 3
(C) 0.5 Å
(D)
154 . Å 2
Sol.
19.
20.
Comment which of the following option represent the correct order of true (T) & false (F) statement. I. size of P < size of Q II. size of R < size of S III. size of P < size of R (appreciable difference) IV. size of Q < size of S (appreciable difference) (A) TTTT (B) TTTF (C) FFTT (D) TTFF
Sol.
Two elements A & B are such that B. E of A – A, B – B & A – B are respectively 81 Kcal/mole, 64 Kcal/mole, 76 Kcal/ mole & if electronegativity of B is 2.4. then the electronegativity of A may be approxiamtely (A) 2.81 (B) 1.8 (C) 1.99 (D) 3.0
21.
Order of IE1 values among the following is (A) P > R > S > Q (B) P < R < S < Q (C) R > S > P > Q (D) P > S > R > Q
Sol.
Sol.
Question No. 20 and 21 are based on the following information. Four elements, P,Q,R & S have ground state electronic configuration as : P 1s2 2s2 2p6 3s2 3p3 Q 1s2 2s2 2p6 3s2 3p1 R 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 S 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
PERODIC PROPERTICS
Page # 165
OBJECTIVE PROBLEMS (JEE ADVANCED)
EXERCISE – III 1.
Using the concept of Zeff (from Stater's rule). Explain the following : "In obtaining the electronic configuration of V+ from that of V, an e– is removed from 4s and not from 3d."
Sol.
Sol.
4.
All the values are in kJ/mol If each orbital can hold a maximum of three electrons, the number of elements in 9th period of periodic table (long form) are (A) 48 (B) 162 (C) 50 (D) 75
Sol. 2.
From the following information A–(g) A+2 (g) + 3e– H1 = 1400 kJ A(g) A+2 (aq) + 2e–\ H2 = 700 kJ + HEG[A (g)] = –350 kJ/mol (IE1 + IE2) for A(g) = 950 kJ/mol Find (A) IE1 of A (B) IE2 of A (C) HEG of A (D) HHE of A2+ (g)
5.
Sol.
The Zeff for 3d electron of Cr 4s electron of Cr 3d electron of Cr3+ 3s electron of Cr3+ are in the order respectively. (A) 4.6, 2.95, 4.95, 8.05 (B) 4.95, 2.95, 4.6, 8.05 (C) 4.6, 2.95, 5.3, 12.75 (D) none of these
Sol.
3.
For ionic compound A2+B22– & C22+D2 (a) A4(s) + B2(g) A B2(s) Hf = –500 Find the lattice energy of AB2 (b) Find Hf for according D3(g) + C(s) C2D (s) Given on sublimation of above metal (A4) it dissociates into individual atom. Given : Sublimation energy of A4 is 1600 ; Sublimation energy of C is 100 Dissociation energy of B2 is 200 ; Dissociation energy of D3 is 90
6.
A( g) A ( g)
;
H = 50
7.
A ( g) A 2 (g)
;
H = 150
B – ( g) B(g)
;
H = 260
B – (g) B2– (g)
;
H = 250
D(g) D – (g)
;
H = –50
D – ( g) D 2– (g)
;
H = 100
C (g) C( g)
;
H = – 150
(a) If internuclear distance between Cl atoms in Cl2 is 10Å & between H atoms in H2 is 2Å, then calculate internuclear distance between H & Cl (Electronegativity of H = 2.1 & Cl = 3.0) (b) Compare the following giving reasons Acidic nature of oxides:CaO,CO, CO2, N2O5, SO3
Sol.
With the help of EN values [ENA = 1.8, ENB = 2.6, ENC = 1.6, END = 2.8] answer the following questions for the compounds HAO, HBO, HCO, HDO (a) Compounds whose aqueous solution is acidic and order of their acidic strength (b) Compounds whose aqeous solution is basic and order of their basic strength (c) Comment on the chances of being coloured on the basis of percent ionic character for the compounds CD & AB
Sol.
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PERODIC PROPERTICS
Energy (kJ)
8. 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0
Al+3(g) + 3Cl–(g)
Al+2(g) + 3Cl–(g)
ACl3(s) Al+3(g) + 3Cl–(aq)
Al+3(aq) + 3Cl–(aq)
Na(g)
1 Cl2 (g) 2
Na(s)
1 3 Cl2 (g) Al( s) Cl2 (g) 2 2
With the help of above information answer the following questions. (1) HL.E. of NaAlCl4 will be_____________________ (2) HH.E. of Al+3 (g) ion will be __________________ (3) Hf of NaCl(s) will be _______________________ (4) BECl2 Heg [Cl(g)] will be ___________________
Sol.
11.
(5) Heg1 of [ Al 3 (g)] will be _____________________ Sol.
9.
Why the first ionisation energy of carbon atom is greater than that of boron atom whereas, the reverse is true for the second ionisation energy.
Sol.
If (n + l) rule for energy is not followed, what are the blocks of the following elements if they are filled according to increasing shell number (a) K(19) (b) Fe(26) (c) Ga(31) (d) Sn(50)
12.
On the Pauling's electronegativity scale, which element is next to F.
Sol.
Sol.
10. Use the following system of naming elements is which first alphabets of the digits are written collectively. 0 1 2 3 4 5 6 7 8 9 nil uni bi tri quad pent hex sept oct enn to write three-letter symbols for the elements with atomic number 101 to 109. [Example : 101 is Unn........]
13.
Mg2+, O2–, Na+, F–, N3– (Arrange in decreasing order of ionic size)
Sol.
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PERODIC PROPERTICS 14. Why Ca+ has a smaller ionic radius than K+. Sol.
15.
Which of the ions are paramagnetic Sr2+, Fe3+, Co2+, S2–, Pb2+
Page # 166 20.
Which property will increase and which will decrease for IA group as we go down the group. (a) Atomc size (e) E N (b) Ionic radii (f) At. mass (c) I.E (g) Valance e– (d) Metallic ch (h) Chemical reactivity
Sol.
Sol.
16.
Why do alkaline earth metals always dipositive ions ?
The IE do not follow a regular trend in II & III periods with increasing atomic number. Why ?
Sol.
Sol.
17.
21.
State giving reasons which one have higher value (a) IE1 of F or Cl (b) E A of O or O– (c) ionic radius of K+ or Cl–
22.
Arrange in decreasing order of atomic size. Na, Cs, Mg, Si, Cl.
Sol.
Sol.
23. 18.
Explain why a few elements such as Be(+0.6), N(+0.3) & He (+0.6) have positive electron gain enthalpies while majority of elements do have negative values.
In the ionic compound KF, the K+ and F– ions are found to have practically radii, about 1.34 Å, each. What do you predict about the relative covalent radii of K and F ?
Sol.
Sol.
19.
From among the elements, choose the following : Cl, Br, F, Al, C, Li, Cs & Xe (i) The element with highest electron affinity (ii) The element with lowest ionisation potential (iii) The element whose oxide is amphoteric. (iv) The element which has smallest radii. (v) The element whose atom has 8 electrons in the outermost shell.
24.
Does Na2(g) moleucle exhibit metallic properties ?
Sol.
25.
Sol.
Which will have a higher boiling point, Br2 or ICl, & why ?
Sol.
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Which bond in each pair is more polar (a) P – Cl or P – Br (b) S – Cl or S – O (c) N – O or N – F
Sol.
27.
PERODIC PROPERTICS 32.
Sol.
Arrange noble gases, in the increasing order of b.p.
33.
ENERGY BASED CALCULATIONS Calculate E.N. of chlorine atom on Pauling scale if I.E. of Cl– is 4eV & of E.A. of Cl+ is +13.0 eV,
Sol.
Sol.
28.
How many chlorine atoms will be ionised Cl Cl+ + e–1 by the energy released from the process Cl + e–1 Cl– for 6.02 × 1023 atoms (I.P. for Cl = 1250 kJ mol–1 and E.A. = 350 kJ mole–1)
For the gaseous reactions, K + F K+F–, H was calculated to be 19 kcal/ mol under conditions where the cations and anions were prevented by electrostatic separation from combining with each other. The ionisation potential of K is 4.3 eV atom. What is the electron affinity of F ?
Sol.
34. 29.
Calculate the electronegativity of fluorine from the following data : EH–H = 104.2 kcal mol–1 EF–F = 36.2 kcal mol–1 EH–F = 134.6 kcal mol–1 XH = 2.1
The ionisation potentials of atoms A and B are 400 and 300 kcal mol–1 respectively. The electron affinities of these atoms are 80.0 and 85.0 k cal mol–1 respectively. Prove that which of the atoms has higher electronegativity.
Sol.
Sol.
35. 30.
Calculate the E.N. of Cl from the bond energy of CIF (61 K Cal/mol). Given that bond energies of F2 and Cl2 are 38 and 58 KCal/mol respectively. Sol.
Sol.
31.
The As-Cl bond distance in AsCl3 is 2.20 Å. Estimate the SBCR (single bond covalent radius) of As. (Assume EN of both to be same and radius of Cl = 0.99 Å)
The IE values of Al(g) = Al+ +e is 577.5 kJ mol–1 and H for Al(g) = Al3+ +3e is 5140 kJ mol–1. If second and third IE values are in the ratio 2 : 3. Calculate IE2 and IE3
36.
The Pt-Cl distance has been found to be 2.32 Å in several crystalline compounds. If this value applies to both of the compounds shown in figure. What is Cl – Cl distance in (a) and (b)
Sol.
NH3
(a)
Cl
Pt
NH3
Cl
(b)
Cl
NH3
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Pt
Cl
NH3
PERODIC PROPERTICS Sol.
37.
Page # 169 Sol.
A mixture contais F and Cl atoms. The removal of an electron from each atom of the sample requires 284 kJ while the addition of an electron to each atom of the mixture releases 68.8 kJ. Determine the % composition of the mixture. (IE) per atom (EA) per atom F 27.91 × 10–22 kJ 5.53 × 10–22 kJ –22 Cl 20.77 × 10 kJ 5.78 × 10–22 kJ
42.
Based on location in P.T. which of the following would you expect to be acidic & which basic. (a) CsOH (b) IOH (c) Sr(OH)2 (d) Se(OH)2 (e) FrOH (f) BrOH
Sol.
Sol.
Miscellaneous Properties 38. Arrange the following ions Na+, Mg2+, Al3+ in increasing. (a) extent of hydration (b) hydration energy (c) size of hydrations (d) Ionic mobility (e) size of gaseous ions. Sol.
39.
Arrange following oxides in increasing acidic nature Li2O, BeO, B2O3
Born Haber Cycle 43. Calculate the lattice energy of NaCl crystal from the following data by the use of Born-Haber cycle. Sublimation energy of Na = 26 kcal/g. atom, dissociation energy of Cl2 = 54 kcal/mole, ionisation energy for Na(g) = 117 kcal/mol, electron affinity for Cl(g) = 84 kcal/g atom, heat of formation of NaCl = –99 kcal/mole. Sol.
44.
Sol.
Calculate the electron affinity of iodine with the help of the following data (given in Kcal/mole) (Hfor)Nal = –68.8, (Hsub)Na = 25.9, 1 (Hsub + Hdiss)I 2 2
40. Which oxide is more basic, MgO or BaO ? Why ? Sol.
41.
= 25.5, (IP)Na = 118.4, (U)Nal = –165.4 Sol.
The basic nature of hydroxides of group 13 (IIIA) decreases progressively down the group. Comment. : 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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1.
PERODIC PROPERTICS
EXERCISE – IV
PREVIOUS YEARS PROBLEMS (JEE MAIN)
LEVEL – I
JEE MAIN
The correct order of ionic radius is [AIEEE-2002] (A) Ce > Sm > Tb > Lu (B) Lu > Tb > Sm > Ce (C) Tb > Lu > Sm > Ce (D) Sm > Tb > Lu > Ce
Sol.
Ce3+, La3+, Pm3+ and Yb3+ have ionic radii in the increasing order as – [AIEEE-2002] (A) La3+ < Ce3+ < Pm3+ < Yb3+ (B) Yb3+ < Pm3+ < Ce3+ < La3+ (C) La3+ = Ce3+ < Pm3+ < Yb3+ (D) Yb3+ < Pm3+ < La3+< Ce3+
2.
Sol.
3.
Sol.
6.
Sol.
7.
According to the Periodic Law of elements, the Variation in properties of elements is related to their ? [AIEEE-2003] (A) Nuclear masses (B) Atomic numbers (C) Nuclear neutron-proton number ratio (D) Atomic masses
T he at om i c num b e rs of v anad i um (V). Chromium (Cr), manganese (Mn) and iron (Fe) respectively 23, 24, 25 and 26. Which one of these may be expected to have the higher second ionization enthalpy ? [AIEEE-2003] (A) Cr (B) Mn (C) Fe (D) V
Which one of the following sets of ions represents the collection of isoelectronic species ? [AIEEE-2004] + – 2+ 3+ + (A) K , Cl , Mg , Sc (B) Na , Ca2+, Sc3+, F– (C) K+, Ca2+, Sc3+, Cl– (D) Na+, Mg2+, Al3+, Cl–
Sol.
8.
Sol.
Which one of the following ions has the highest value of ionic radius ? [AIEEE-2004] (A) O2– (C) Li+
(B) B3+ (D) F–
Sol. 4.
The reduction in atomic size with increase in atomic number is a characteristic of elements of [AIEEE-2003] (A) d-block (B) f-block (C) Radioactive series (D) High atomic masses
9.
Sol.
5.
Which one of the following groups represent a collection of isoelectronic species ? (At. no. Cs = 55, Br = 35) [AIEEE-2003] (A) N3–, F–, Na+ (B) Be, Al3+, Cl– (C) Ca2+, Cs+, Br (D) Na+, Ca2+, Mg2+
Among Al2O3, SiO2, P2O3 and SO2 the correct order of acid strength is : [AIEEE-2004] (A) Al2O3 < SiO2 < SO2 < P2O3 (B) SiO2 < SO2 < Al2O3 < P2O3 (C) SO2 < P2O3 < SiO2 < Al2O3 (D) Al2O3 < SiO2 < P2O3 < SO2
Sol.
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PERODIC PROPERTICS 10.
The formation of the oxide ion O 2g requires first an exothermic and then an endothermic step as shown below : [AIEEE-2004] O(g) + e– = O–(g) H° = – 142 kJ mol–1 O–(g) + e– = O2–(g) H° = 844 kJ mol–1 This is because of : (A) O– ion will tend to resist the addition of another electron (B) Oxygen has high electron affinity (C) Oxygen is more electronegative (D) O– ion has comparatively larger size than oxygen atom
Page # 171 14.
The lanthanide contraction is responsible for the fact that [AIEEE-2005] (A) Zr and Y have about the same radius (B) Zr and Nb have similar oxidation state (C) Zr and Hf have about the same radius (D) Zr and Zn have the same oxidation state
Sol.
Sol. 15.
11.
In which of the following arrangements the order is NOT according to the property indicated against it ? [AIEEE-2005] (A) Al3+ < Mg2+ < Na < F– – increasing ionic size (B) B < C < N < O – increasing first ionization enthalpy (C) I < Br < F < Cl – increasing electron gain enthalpy (with negative sign) (D) Li < Na < K < Rb – increasing metallic radius
Which of the following factors may be regarded as the main cause the lanthanide contraction ? [AIEEE-2005] (A)Poor shielding of one of 4f electron by another in the subshell (B) Effective shielding of one of 4f electrons by another in the subshell (C)Poorer shielding of 5d electrons by 4f electrons (D)greater shielding of 5d electrons by 4f electrons
Sol.
Sol.
16. 12.
Which of the following oxides is amphoteric in character ? [AIEEE-2005] (A) SnO2 (B) SiO2 (C) CO2 (D) CaO
The increasing order of the first ionization enthalpies of the elements B, P, S and F (lowest first) is – [AIEEE-2006] (A) F < S < P < B (B) P < S < B < F (C) B < P < S < F (D) B < S < P < F
Sol.
Sol.
13.
Sol.
Pick out the isoelectronic structure from the following : [AIEEE-2005] + + I. CH3 II. H3O III. NH3 IV. CH3– (A) I and II (B) III and IV (C) I and III (D) II, III and IV
17.
Which one of the following sets of ions represents a collection of isoelectronic species ? [AIEEE-2006] (A) N3–, O2–, F–, S2– (B) Li+, Na+, Mg2+, Ca2+ (C) K+, Cl–, Ca2+, Sc3+ (D) Ba2+, Sr2+, K+, Ca2+
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18.
PERODIC PROPERTICS 21.
lanthanoid contraction is caused due to [AIEEE-2006] (A) The same effective nuclear charge from Ce to Lu (B) The imperfect shielding on outer electrons by 4f electrons from the nuclear charge (C) The appreciable shielding on outer electrons by 4f electrons from the nuclear charge (D) The appreciable shielding on outer electrons by 5d electrons from the nuclear charge
The correct sequence which shows decreasing order of the ionic radii of the elements is [AIEEE-2010] (A) A3+ > Mg2+ > Na+ > F– > O2– (B) Na+ > Mg2+ > Al3+ > O2+ > F– (C) Na+ > F– > Mg2+ > O2– > Al3+ (D) O2– > F– > Na+ > Mg2+ > Al3+
Sol.
22.
Sol.
Which one of the following order represents the correct sequence of the increasing basic nature of the given oxides ? [AIEEE-2011] (A) MgO I (D) I > III > II
Sol.
Sol.
4.
Which of the following has the maximum number of unpaired electrons [JEE 1996] (A) Mg2+ (B) Ti3+ (C) V3+ (D) Fe2+
Which of the following oxide is neutral ? [JEE 1996] (A) CO (B) SnO2 (C) ZnO (D) SiO2
10.
Sol.
11.
Which of the following are amphoteric ? [REE 1997] (A) Be(OH)2 (B) Sr(OH)2 (C) Ca(OH)2 (D) Al(OH)3
Li+, Mg2+, K+, Al3+ (Arrange in increasing order of radii) [JEE 1997]
Sol.
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Page # 174 12.
Sol.
13.
Sol.
Anhydrous AlCl3 is covalent. From the data given below predict whether it would remain as a molecule or converts into ions in aqueous solution. [JEE 1997] [L.E. for AlCl3 = 5137 kJ/mol] H hydration for Al3+ = –4665 kJ/mol; Hhydra for Cl– = –381 kJ/mol.
Which one of the following statement (s) is (are) correct ? (A) The electronic configuration of Cr is [Ar] 3d5 4s1. (Atomic No. of Cr = 24) (B) The magnetic quantum number may have a negative value (C) In silver atom, 23 electrons has a spin of one type and 24 of the opposite type. (Atomic No. of Ag = 47) (D) The oxidation state of nitrogen in HN3 is –3.
PERODIC PROPERTICS Sol.
16.
The correct order of radii is : [JEE 2000] (A) N < Be < B (B) F– < O2– < N3– (C) Na < Li < K (D) Fe3+ < Fe2+ < Fe4+
Sol.
17.
Sol.
18.
The correct order of acidic strength is : [JEE 2000] (A) Cl2O7 > SO2 > P4O10 (B) CO2 > N2O5 > SO3 (C) Na2O > MgO > Al2O3 (D) K2O > CaO > MgO
The IE1 of Be is greater than that of B. [T/F] [JEE 2001]
Sol. Directions : The questions below to consist of an 'assertion in column-I and the 'reason' in column-2. Against the specific question number, write in the appropriate space. (A) If both, assertion and reason are CORRECT, and reason is the CORRECT explanation of the assertion. (B) If both assertion and reason are CORRECT, but reason is not the CORRECT explanation of the assertion. (C) If assertion is CORRECT but reason is INCORRECT (D) If assertion is INCORRECT but reason in CORRECT. Sol.
14.
19.
Sol.
20.
Identify the least stable ion amongst the following : [JEE 2002] (A) Li– (B) Be– (C) B– (D) C–
Sol.
Assertion : F atom has less negative electron gain enthalpy than Cl atom. [JEE 2000] Reason : Additional electron is repelled more efficiently by 3p electron in Cl atom than by 2p electron in F atom.
21.
Sol. Sol. 15.
The set representing correct order of IP1 is [JEE 2001] (A) K > Na > Li (B) Be > Mg > Ca (C) B > C > N (D) Fe > Si > C
Identify the correct order of acidic strengths of CO2, CuO, CaO, H2O : [JEE 2002] (A) CaO < CuO < H2O < CO2 (B) H2O < CuO < CaO < CO2 (C) CaO < H2O < CuO < CO2 (D) H2O < CO2 < CaO < CuO
Assertion : Al(OH)3 is amphoteric in nature. Reason : Al – O and O – H bonds can be broken with equal case in Al(OH)3 [JEE 2000] Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
PERODIC PROPERTICS
Page # 175
ANSWER-KEY OBJECTIVE PROBLEMS (JEE MAIN)
Answer Ex–I Q.No. Ans .
1 B
2 C
3 C
4 D
5 A
6 B
7 B
8 D
9 B
10 B
11 C
12 B
13 C
14 A
15 C
16 A
17 A
18 A
19 A
20 D
Q.No.
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Ans . Q.No. Ans .
C 41 C
B 42 D
C 43 A
B 44 B
B 45 C
C 46 D
A 47 C
B 48 D
B 49 B
A 50 C
C 51 D
C 52 C
D 53 B
C 54 B
A 55 D
A 56 B
C 57 B
A 58 C
A 59 D
A 60 B
OBJECTIVE PROBLEMS (JEE ADVANCED)
Answer Ex–II 1. D 9. C 17. B
2. C 10. C 18. C
3. D 11. A 19. A,C
5. 72 13. A 21. A
6. A 14. C
7. C 15. B
8. C 16. A
OBJECTIVE PROBLEMS (JEE ADVANCED)
Answer Ex–III 1.
4. D 12. C 20. B
Zeff = 4.3
2. (a) IE1 of A = 350 kJ/mol, (b) IE2 = 600 kJ/mol, (c) EG H of A = –450 kJ/mol, (d) ME H of (A2+) = –250 kJ/mol. 3. –780, –420, AB > CD 4.
D
5.
C
6. (a) 5.919 Å, (b) CaO < CO < CO2 < N2O5 < SO3 7. (a) Acids : HBO, HDO and order of acidic strength HDO > HBO (b) Bases : AOH, COH
and order of basic strength AOH < COH
(c) % Ionic character
= 16 |XA – XB| + 3.5 (XA – XB)2
for
colourless
CD = 16(1.2) + 3.5 (1.2)2 = 24.24 % AB = 16 × 0.8 + 3.5 (0.8)2 = 15.07 %
coloured
8. (i) –230 kJ, (ii) –40, (iii) 90, (iv) 110, (v) –20 kJ 9. (a) d block, (b) d block, (c) p block, (e) f block 10. 101 Unu
102
103
104
105
106
107
108
109
Unb
Unt
Unq
Unp
Unh
Uns
Uno
Une
11. Zeff & half filled config. 12.
O
13. N3– > O2– > F– > Na+ > Mg2+
14. Isolelectronic Ca+2 (higher)
15. Fe3+, CO2+
16. difference in 1E1 & 1E2 is less than 10ev. 17. (a) F (b) O
(c) Cl–
18. half filled and fully filled orbitals.
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Page # 176
PERODIC PROPERTICS
19. (i) Cl (ii) Cs (iii) Al (iv) F (v) Xe 20. Increases a, b, d, f, h
Decrease c,e
Same g
22. Cs > Na > Mg > Si > Cl
23. rk > 1.34 Å > rF
24. No
26. (a) P – Cl
27. He < Ne < Ar < Kr < Xe < Rn
21. half filled & fully filled orbitals
(b) S – O, (C) N – F
28. 3.03 (Pauling)
29. 3.8752
25. ICl
30. 3.2
31. IE2 = 1825 kJ/mole, IE3 = 2737.5 kJ/mol 32. 1.686 × 1023 atom
33. 3.476 eV
34. EN1 > EN2
35. 1.21 Å
36. 4.64 Å ; b = 3.28 Å
37. F = 37.81%, Cl = 62.19 % 38. (a) Al+3 > Mg+2 > Na+,
(b) Al+3 > Mg+2 > Na+,
(c) Al+3 > Mg+2 > Na+, (d) Na+ > Mg+2 > Al+3,
(e) Na+ > Mg+2 > Al+3 39. Li2O
<
BeO
< B2O3
basic 40. BaO
amphoteric
acidic
41. False
42. (a) basic (b) acidic (c) basic
(d) acidic
43. –185 kcal/mole
44. +73.2 kcal/mole
Answer Ex–IV
(e) basic
PREVIOUS YEARS PROBLEMS (JEE MAIN)
LEVEL – I B
JEE MAIN
1. A
2.
9. D
10. A
11. B
12. A
17. C
18. B
19. B
20. C
Answer Ex–IV
3.
B
4.
D
2. C 10. A,D 17. A
5.
A
6. A
7. C
8.
A
13. D
14. C
15. C
16. D
21. D
22. D
23. C
24. A
PREVIOUS YEARS PROBLEMS (JEE ADVANCED)
LEVEL – II 1. A 9. B 16. B
(f) acidic
JEE ADVANCED 3. C 4. A 11. Al+3 < Mg2+ < Li+ < K+ 18. True 19. B
5. D 12. Ionic 20. B
6. A 13. A,B,C 21. A
7. D 14. C
8. A 15. C
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