Atomic Structure IPE

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Atomic structure

ATOMIC STRUCTURE INTRODUCTION Atom:- The smallest particle which can take part in a chemical reaction with out losing its identity is known as atom. Subatomic particles: Electrons, protons and neutrons are known as subatomic particles. Electron:- The negatively charged fundamental particle present in an atom with negligible mass is called electron. The mass of an electron is 1/1836 of mass of proton or hydrogen atom. In atomic mass units it is equal to 0.000548 amu or 9.1095 X 10-31kg. Its charge is 1.6022 X 10-19coulomb.

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Discovery of Electron : Sir William Crookes designed a cathode ray discharge tube in which cathode rays are observed only at very low pressures and very high voltages. These rays consist of negatively charged particles called electrons. To vacuum pump

+

Anode

High voltage

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Cathode -

Characteristics of cathode rays : 1) Cathode rays are not visible but their behavior can be observed with the help of a fluorescent or a phosphorescent. 2) These rays travel from cathode to anode. 3) These rays travel in straight lines in the absence of electric and magnetic fields. 4) But these rays deflect like negatively charged particles in electric and magnetic fields. Hence the rays constitute negatively charged particles and are known as electrons. 5) These rays are independent of the nature of the cathode material and nature of the gas present in the cathode ray tube. These facts conclude that electrons are the negatively charged fundamental particles present in all the substances. Charge to mass ratio of Electron (e/me) : - Charge to mass ratio of electron is calculated by J.J. Thomson as follows. e  1.75882 x 1011 C kg 1 me

Charge on the Electron :- The charge on the electron was calculated by Millikan in oil drop experiment as 1.60 x 10-19 coulombs. The mass of the electron can be derived as follows. me 

e  9.1094 x 10-31 kg e / me

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Atomic structure

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Proton:- The positively charged fundamental particle present in the atom is called proton. The mass of a proton is 1.007277 amu or 1.67252 X 10-27 kg. Its charge is same as that of electron. Discovery of proton : Protons are discovered in canal ray experiment. These rays are produced in modified cathode ray tube.

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Characteristics of canal rays : 1) Canal rays constitute positively charged particles. The characteristics of these rays depend on the nature of gas present in the cathode ray tube 2. The e/m ratio of the particles depend on the nature of gas taken. 3. Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge. 4. The behavior of these particles in the magnetic or electric field is opposite to that observed for electron. 5. When hydrogen gas is used in the discharge tube, the positively charged particles emitted are found to possess mass of 1 amu and these are called protons.

Relative Mass in amu charge -1 0.000542 amu +1 1.00727 amu 0 1.00867 amu

-1.602 x 10-19 +1.602 x 10-19 0

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Electron Proton Neutron

Charge

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Particle

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Neutron:- The neutral fundamental particle present in the atom is called neutron. The mass of a neutron is 1.00898 amu or 1.67495 X 10-27kg. It has no charge. Discovery of neutrons: Chadwick discovered neutrons by bombarding a thin layer of Be with  particles.

Mass in kg 9.1 x 10-31 kg 1.672 x 10-27 kg 1.674 x 10-27 kg

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Atomic number (Z):-The number of protons or the number of electrons in an atom is called atomic number. It is represented by 'Z'. Moseley discovered a simple relation between the frequencies of the characteristic X-rays of an element and its atomic number. v = a(Z-b)  = frequency of X-rays Z = atomic number a,b are constants which are characteristic of elements Mass number(A):- The total number of protons and neutrons in an atom is called mass number. It is denoted by 'A'. A = no. of protons + no. of neutrons A = Z + no. of neutrons no. of neutrons = A - Z Isotope:- Isotopes are the atoms of an element with same atomic number but differ in their mass numbers i.e., The isotopes of an element have same number of protons but differ in the number of neutrons. eg., Hydrogen has three isotopes - Hydrogen(1H1), Deuterium (1H2) and Tritium (1H3). They have same number of protons (one) but the numbers of neutrons are 1,2 and 3 respectively. Isobars :- The atoms of different elements with same mass number but different atomic numbers are called “Isobars”. Eg., 6C14, 7N14

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Atomic structure

Atomic weight:- The atomic weight of an element is the average weight of all the isotopes of that element. Note:- Atomic number is a whole number but Atomic weight may be a fractional number. ATOMIC MODELS J.J.Thomson’s model: According to this model, an atom has a spherical shape in which the positive charge is uniformly distributed. The electrons are embedded into it in such a manner as to give the most stable electrostatic arrangement. Rutherford’s planetary model :

Gold foil

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Experiment: A narrow beam of  - particles is passed through a thin gold foil which is surrounded by circular screen made up of fluorescent zinc sulphide. Whenever  - particles strike the screen, a tiny flash of light was produced at that point.

Lead plate

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Source of alpha particles

Photographic plate

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Ruther ford's scattering experiment

Observations: 1) Most of the  - particles pass through the foil undeflected. 2) A small fraction of the  - particles were deflected by small angles. 3) A very few  - particles bounced back i.e. were deflected by 180o. Conclusions: 1. Most of the space in the atom is empty. 2. The positive charge in the atom is concentrated in the small dense portion called the nucleus. 3. Electrons revolve around the nucleus in circular paths called orbits. It resembles the solar system. 4. Electrons and the nucleus are held together by electrostatic forces of attraction. Drawbacks: 1) Rutherford’s model could not explain the stability of atom. According to electromagnetic theory, the charged particle under acceleration should continuously emit radiation. Hence the electron moving in the orbits must lose energy and fall into the nucleus. But this is not happening. 2) This model could not explain the electronic structure and energy of electrons. 3) It could not explain the atomic spectra.

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Atomic structure

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NATURE OF LIGHT Light is considered as an electromagnetic radiation. An electromagnetic radiation consists of two components i.e., Electric component and Magnetic component which are perpendicular to each other as well as to the direction of path of radiation. The electromagnetic radiations are produced by the vibrations of a charged particle.

  wavelength E  amplitude of electric field M  amplitude of magnetic field

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The properties of light can be explained by considering it as either wave or particle as follows.

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WAVE NATURE OF LIGHT By considering light as wave, following properties can be defined for it. Wavelength: The distance between two successive similar points on a wave is called as wavelength. It is denoted by  . Units: cm, Angstroms(Ao), nano meters(nm), milli microns(mµ) etc., 1 Ao = 10-8 cm. 1 nm= 10-9m = 10-7cm

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Frequency: The number of vibrations done by a particle in unit time is called frequency. It is denoted by ' '. Units: cycles per second = Hertzs = sec-1. Velocity: Velocity is defined as the distance covered by the wave in unit time. It is denoted by 'c'. Velocity of light = c = 3.0 x 108 m.sec-1 = 3.0 x 1010 cm.sec-1 Note: For all types of electromagnetic radiations, velocity is a constant value. The relation between velocity, wavelength and frequency can be given by following equation. velocity = frequency x wavelength c   Wave number: The number of waves spread in a length of one centimeter is called wave number. It is denoted by  1  -1 units: cm , m-1 Amplitude: The distance from the midline to the peak or the trough is called amplitude of the wave. It is usually denoted by 'A' (a variable). Amplitude is a measure of the intensity or brightness of light radiation.

 

Problems 1) The wave length of a radiation emitted by a sodium lamp is 300 nm. Find its frequency. 2) The frequency of an electromagnetic radiation is 300 Hz. Calculate its wave length. 3) The wave number of a radiation is 9000 cm-1. Calculate its frequency. 4) Calculate the wave number of the yellow light of wave length 600 m  emitted from sodium lamp.

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Atomic structure

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PARTICLE NATURE OF LIGHT Though most of the properties of light can be understood by considering it as a wave, some of the properties of light can only be explained by using particle (corpuscular) nature of it. Newton considered light to possess particle nature. In the year 1900, in order to explain black body radiations, Max Planck proposed Quantum theory by considering light to possess particle nature.

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PLANCK'S QUANTUM THEORY Black body:- The object which absorbs and emits the radiation of energy completely is called a black body. Practically it is not possible to construct a perfect black body. But a hollow metallic sphere coated inside with platinum black with a small aperture in its wall can act as a near black body. When the black body is heated to high temperatures, it emits radiations of different wavelengths. Following curves are obtained when the intensity of radiations are plotted against the wavelengths, at different temperatures.

classical theory

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intensity

7000 K

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5000 K

wavelength

Following conclusions can be drawn from above graphs. 1. At a given temperature, the intensity of radiation increases with wavelength and reaches a maximum value and then starts decreasing. 2. With increase in temperature, the wavelength of maximum intensity ( max ) shifts towards lower wavelengths. According to classical physics, energy should be emitted continuously and the intensity should increase with increase in temperature. The curves should be as shown by dotted line. In order to explain above experimental observations Max Planck proposed following theory. 1. Energy is emitted due to vibrations of charged particles in the black body. 2. The radiation of energy is emitted or absorbed discontinuously in the form of small discrete energy packets called quanta. 3. Each quantum is associated with definite amount of energy which is given by the equation E=h Where h = planck's constant = 6.625 x 10-34 J.sec = 6.625 x10-27 erg.sec  = frequency of radiation

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Atomic structure

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4. The total energy of radiation is quantized i.e., the total energy is an integral multiple of h . It can only have the values of 1 h or 2 h or 3 h . It cannot be the fractional multiple of h . 5. Energy is emitted and absorbed in the form of quanta but propagated in the form of waves. EINSTEIN'S GENERALIZATION OF QUANTUM THEORY Einstein generalized the quantum theory by applying it to all types of electromagnetic radiations. He explained photoelectric effect using this theory.

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Photoelectric Effect:- The ejection of electrons from the surface of a metal, when the metal is exposed to light of certain minimum frequency, is called photoelectric effect. The frequency of light should be equal or greater than a certain minimum value characteristic of the metal. This is called threshold frequency. Photoelectric effect cannot be explained by considering the light as wave. Einstein explained photoelectric effect by applying quantum theory as follows, 1. All electromagnetic radiations consists of small discrete energy packets called photons. These photons are associated with definite amount of energy given by the equation E=h . 2. Energy is emitted, absorbed as well as propagated in the form of photons only. 3. The electron is ejected from the metal, when a photon of sufficient energy strikes the electron. When a photon strikes the electron, some part of the energy of photon is used to free the electron from the attractive forces in the metal and the remaining part is converted into kinetic energy. h = W + K.E where W = energy required to overcome the attractions K.E = kinetic energy of the electron

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Problems 1) Calculate the energy of one photon of radiation whose frequency is 3x1012 Hz. 2) The energy of a electro magnetic radiation is 6.625x10-19 J Calculate The Wavelength of radiation. SPECTRA When electromagnetic radiation is passed through a prism or grating it is splitted and forms a collection of lines representing different wavelengths. This is called spectrum. Spectra can be divided into two types viz., emission and absorption spectra as given below. Emission Spectra Absorption Spectra 1) These are obtained due to emission of radiation These are obtained when substance absorb the from the substances. radiation. 2) White lines are formed on the black back ground. Black lines are formed on the white back ground. 3)These are formed when atoms or molecules are de-These are formed when atoms or molecules are excited from higher energy level to lower energy excited from lower energy level to higher energy level. levels.

Spectra can also be divided into line and band spectra as given below. Line Spectra Band Spectra 1) It consists of sharp and well defined It consists of closely spaced lines lines. called bands. 2) Characteristic of atoms Characteristic of molecules. 3) Formed due to the excitation and de- Formed due to the vibrations and excitation of electrons in the atoms. rotations of atoms in molecules. 4) It is also known as atomic spectra

It is also known as molecular spectra

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Atomic structure

HYDROGEN ATOMIC SPECTRUM A bright light is emitted when a high potential is applied to hydrogen gas at low pressure in a discharge tube. This bright light is dispersed and forms a spectrum upon passing through a prism or grating. The spectrum consists of separate lines corresponding to different wavelengths. This is called Hydrogen atomic spectrum. The spectral lines are formed due to electronic transitions from one energy level to another. These lines are divided into five series according to the range of wavelengths as follows. Spectral region

n1

n2

1. Lyman series

Ultra-violet

1

2,3,4,5,6,7,_ _ _ _

2. Balmer series

Visible

2

3,4,5,6,7,_ _ _ _

3. Paschen series

near infra-red

3

4,5,6,7,_ _ _ _

4. Brackett series

infra-red

4

5,6,7,_ _ _ _

5. Pfund series

far infra-red

5

6,7,_ _ _ _

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Spectral series

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n1and n2 are the principal quantum numbers of the energy levels.

The wave numbers of spectral lines in each series can be calculated using Rydberg's equation as follows.

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1 1   RH Z 2  2 - 2   n1 n2 

n=7 n=6 n=5 n=4

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where n1and n2 are the principal quantum numbers of orbits corresponding to electronic transition. RH = Rydberg's constant = 1,09,677 cm-1 Z = atomic number

n=3

Pfund

Brackette Paschen

n=2 Balmer

n=1 Lyman

Note: Every element has its own characteristic line spectrum. There is regularity in the line spectrum of each element. Hydrogen has the simplest line spectrum among all the elements. The line spectra become com-

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Atomic structure

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plex with increase in atomic number of the element. Problems 1) An electronic transition from n=3 to n=1 shell takes place in a hydrogen atom. Find the wave number and the wave length of radiation emitted. [Given R=1,09,677 cm-1] BOHR'S ATOMIC MODEL In order to explain the Hydrogen atomic spectrum, Bohr proposed following atomic model based on quantum theory.

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MAIN POSTULATES OF BOHR'S ATOMIC THEORY 1. The electrons in an atom revolve around the nucleus in definite closed circular paths called orbits (energy levels or states or shells). 2. The orbits are represented by principal quantum number 'n'. These are numbered 1,2,3,4 _ _ _ (or K,L,M,N,_ _ _) from the nucleus. 3. The energy of an electron in a particular orbit is constant and the electron neither emits nor absorbs energy as long as it revolves in the orbits. Hence these orbits are called stationary orbits.

Nucleus

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n=3 n=2 n=1

4. The angular momentum of electron is quantized. angular momentum = mvr =

Where

nh 2

m = mass of an electron v = velocity of an electron r = radius of orbit h = Planck's constant = 6.625 X 10-34 J.sec = 6.625 X 10-27 erg.sec n = principal quantum number which can have only positive integer values(1,2,3,4 _ _ _). The electron revolves only in those orbitals where the angular momentum of it is the integral multiple h h h h i.e, 1 or 2 or 3 ___ 2 2 2 2 5. Each orbit is associated with definite amount of energy and radius. The energy of the orbit (and of electron in it) increases with increase in the radius of the orbit. Thus farther the orbit from nucleus greater is the energy. 6. The energy of an electron changes when it moves from one energy level to another. Energy is absorbed when the electron jumps from lower orbit to higher orbit. Whereas energy is emitted when the electron jumps from higher orbit to lower orbit. The energy absorbed or emitted during electronic transitions between two orbits is equal to the

of

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Atomic structure

energy difference between these orbits. It is given by  E = E2 - E1 The energy released in the form of radiation will appear as a spectral line in the atomic spectrum.

DERIVATION OF EXPRESSIONS FOR RADIUS OF ORBIT AND ENERGY OF ELECTRON Radius of Orbit In hydrogen atom, there is one proton in the nucleus, and an electron revolving around the nucleus in a circular orbit of radius 'r'. Let the charge on proton = +e the charge on electron = -e.

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Centrifugal force

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+e

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The attraction between the nucleus and electron acts centripetally towards the nucleus. As per Coulomb's law,

e 2 r2 There is also centrifugal force acting away from the nucleus due to the revolving of electron in the

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force of attraction = orbit.

centrifugal force = where

mv 2 r

m = mass of electron, v = velocity of electron During the orbiting of electron in a stationary orbit, these two forces must be equal.

e 2 mv 2 i.e., 2 = r r e2  mv 2 --------- 1 r According to Bohr's theory, Angular momentum of electron in an orbit is given by or

mvr 

or

v

nh 2

nh 2 mr

n2h2 4 2 m 2 r 2 By substituting the value of v2 in equation-1 or

v2 

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Atomic structure

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e2 mn 2 h 2  2 2 2 r 4 m r n2h2 ------------- 2 4 2 me 2 By substituting the values of planck's constant = h = 6.625 x 10-27erg.sec mass of electron = m = 9.1 x 10-28 g charge on electron = e = 4.802 x 10-10 e.s.u radius of nth orbit = r = 0.529 x 10-8 x n2 cm r

or

where

1 1 e2 K.E = mv 2  2 2 r

and

P.E = 

Hence

T .E 

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Energy of Electron The total energy of electron is equal to the sum of Kinetic and Potential energies. i.e., Total Energy (T.E) = Kinetic Energy (K.E) + Potential Energy (P.E)

e2  mv 2 from equation 1) (since r

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e2 e2  2r r

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e2 r

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e2 2r Upon substituting the value of 'r' from equation-2

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T .E  



2 2 me 4 T .E   2 2 nh By substituting the values of constants, the energy of electron (En) in the nth orbit can be written as En  

21.7 x10-12 erg per atom n2

21.7 x10-19  Joule per atom n2 21.7 x10-22  kJ per atom n2 EXPLANATION OF HYDROGEN ATOMIC SPECTRUM Bohr could successfully explain the hydrogen atomic spectrum using his postulates as given below. The energy of an electron in hydrogen atom is given by

Therefore

En  

2 2 me 4 1 . 2      (1) h2 n

E1  

2 2 me 4 1 . 2 h2 n1

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E2  

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2 2 me 4 1 . 2 h2 n2

And

 2 2 me 4 1   2 2 me 4 1  E2  E1    . 2    . 2  h2 n2   h2 n1  

Therefore

E2  E1 

2 2 me 4 h2

1 1   2  2      (2) n2   n1

According to Bohr’s postulates, E2  E1  h Therefore Hence

  c E2  E1  hc 2 2 me4  1 1   2  2   hc 2 h n2   n1 1 2 2 me 4     ch3

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But

 1 1   2  2  n2   n1

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1  RH 

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 

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 1 1   2  2          (3) n2   n1 This equation is similar to Rydberg’s equation or

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2π 2 me 4 Where RH is equal to and its value can be determined by substituting the following values. ch 3 m=9.1 x 10-28 g, e = 4.8 x 10-10 e.s.u.

 = 3.14, c = 3 x 1010 cm, h = 6.626 x 10-27 erg. sec

2π 2me4  1,09,681cm 1 3 ch Above value is almost equal to Rydberg’s constant (RH = 1,09,677 cm-1). The frequencies of spectral lines in hydrogen atomic spectrum can also be determined by using Bohr’s theory. RH =

MERITS AND DEMERITS OF BOHR'S ATOMIC THEORY Merits 1. This theory successfully explained the spectra of mono-electronic species like Hydrogen, He+, Li2+, Be3+ etc., . 2. Bohr successfully calculated the frequencies of spectral lines, radii and energies of orbits which are in excellent agreement with experimental results. Demerits (defects) 1. Bohr's atomic model is a flat model. But atoms are spherical. 2. This theory could not explain the spectra of atoms containing multi-electrons. 3. It was shown later on that the each spectral line in hydrogen atomic spectrum was actually closely spaced group of fine lines. These fine lines were revealed when the spectrum was taken on high resolution spectrometer in later years. Bohr could not explain this fine spectrum.

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4. Splitting of spectral lines when the atoms are placed in strong magnetic field is called Zeeman effect. Splitting of spectral lines when the atoms are placed in strong electric field is called Stark effect. Bohr could not account for these effects. 5. According to Heisenberg's uncertainty principle, it is not possible to calculate the velocity and position of an electron accurately and simultaneously. But Bohr calculated them. This is a contradiction. 6. According to De Broglie's wave concept, electron has wave nature. But Bohr's theory considered electron as a particle.

QUANTUM NUMBERS Quantum numbers are the numbers used to describe the position and energy of an electron in an atom. There are four types of quantum numbers.

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i) Principal quantum number ii) Azimuthal quantum number iii) Magnetic quantum number iv) Spin quantum number

Azimuthal quantum number (l)

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Principal quantum number (n) 1. Principal quantum number was proposed by Bohr to explain the hydrogen atomic spectrum 2. It denotes the main energy level (or shell or orbit). 3. It is denoted by 'n'. It can have positive integral values from 1,2,3,4,_ _ _ (or K,L,M,N_ _ _ ). 4. It also describes the energy and size of the orbit. The energy and the size of orbit increases with increase in 'n' value. 5. The maximum number of electrons that can be accommodated in a given shell is equal to 2n2 .

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1. Azimuthal quantum number was proposed by Sommerfield in order to explain the fine hydrogen atomic spectrum. 2. It denotes the sub-levels (or sub-shells) in the atom. It is denoted by 'l' 3. The number of sub-levels in a given main energy level is equal to the principal quantum number 'n'. These sub-levels are indicated by azimuthal quantum number 'l' which can have values from 0 to n-1. For ex: In an orbit of principal quantum number n=3, there are three sublevels denoted by l =0,1 and 2. 4. Azimuthal quantum number also describes the angular momentum of electron and shape of orbitals present in that sublevel. For ex: l = 0 (s)--- spherical l = 1 (p)--- dumbbell l = 2 (d)--- double dumbbell Magnetic quantum number (m) 1. Magnetic quantum number was proposed by Lande in order to explain the Zeeman and Stark effects. The splitting of spectral lines in strong magnetic field is called Zeeman effect and splitting in strong electric field is called Stark effect. 2. It is denoted by 'm'. It describes the orientation of orbitals. 3. The number of orientations possible for orbitals in a given sublevel 'l' is equal to the (2l+1) and can have values from -l_ _ _ 0 _ _ _+l. For ex: The number of orientations (m values) possible for a sublevel l=2 is (2X2)+1 = 5 and the values of m = -2,-1,0,+1 and +2.

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Prinicipal quantum number (n) n=1 n=2 n=3

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Azimuthal quantum number Magnetic quantum number (l) (m) l=0 m=0 l = 0,1 m = -1,0,+1 l = 0,1,2 m = -2,-1,0,+1,+2

Spin quantum number (s) 1. It was proposed by Goudsmit and Uhlenbeck to explain the double line structure of alkali metal spectra. 2. Spin quantum number denotes the spin of the electron on its own axis. It is denoted by 's'. It can have only two values(+½ and -½). 3. The clockwise spin is indicated by +½ or upwards arrow(  ) and anti-clockwise spin by -½ or downwards arrow(  ). Possible values 1,2,3,4_ _ _ _ _

2. Azimuthal quantum number (l)

0 to n-1

-l _ _ _0_ _ _+l +1/2 or -1/2

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4. Spin quantum bumber (s)

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3. Magnetic quantum number (m)

Significance Denotes the size and energy of orbit. Denotes the sub-level and shapes of orbitals in that sublevel. It indicates the angular momentum of electron. Indicates the spatial orientation of orbital. Denotes the spin of electron.

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Quantum number 1. Prinicipal quantum number (n)

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Problems 1) What is the shape of orbital if the quantum numbers for the electron in it are n = 3, l = 2, m = -1, and s = +1/2 2) Which of the following set of quantum numbers is not possible? a) n = 3, l = 2, m = -2, s = +1/2 b) n = 2, l = 2, m = -1, s = +1/2 c) n = 4, l = 3, m = 0, s = +1/2 d) n = 3, l = 1, m = -2, s = +1/2 3) How many electrons in an atom may have the quantum numbers, n = 3 and ms= -1/2 ? 4) How many sub-shells, orbitals and electrons are present in n = 3 main shell?

DE BROGLIE'S WAVE CONCEPT According to de Broglie, every particle in motion is associated with wave nature. The wavelength (  ) of every particle wave can be expressed by the following equation.



h h (or) p mv

where h = Planck's constant m = mass of the particle v = velocity of the particle p = mv = momentum of the particle

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Derivation According to Einstein, mass and energy equivalence can be expressed as E = mc2 According to Planck's quantum theory, E  h Hence mc 2  h ------------ 1 But c   or c  By substituting the  value in the 1st equation

mc 2 

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

hc 

h h (or) p mc

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or

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where p = momentum of a photon de Broglie proposed that above equation is applicable to every particle in motion and above equation can be written as

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h h  mv p The wavelengths of macro bodies like cricket ball or stones are very small as their masses are large and hence can be ignored. But for micro particles like electrons, the wavelengths are considerable.

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JUSTIFICATION OF BOHR'S CONCEPT

According to de Broglie, electron in an atom is considered as a standing or stationary wave, which revolve around the nucleus in a circular orbit. In order to behave as a stationary and non-energy radiating wave, the electron wave must be in phase i.e., the two ends of the wave must meet at one place and there should be constructive interference of crests and troughs.

Wave in phase - constructive inter ference

Wave out of phase destructive inter ference

To satisfy this condition, the circumference(2  r) of the orbit must be equal to the integral multiple of the wavelength(  ) of the electron wave. i.e.,

n  2 r

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Atomic structure

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2 r where 'n' is an integer n But according to de Broglie's theory

or





h mv

Hence 2 r h  n mv nh 2 This is Bohr's equation. According to Bohr's theory, the angular momentum (mvr) of electron revolv-

or

mvr 

h i.e., angular momentum is quantized. 2 But when the 'n' is not an integer, there is destructive interference and the wave will go out of phase. This results in loss of energy by the electron wave.

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ing in a stationary orbit is an integral multiple of

Problems

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8) What will be the wavelength of a ball of mass 0.1 Kg moving with a velocity of 10 m/sec. 9) The mass of an electron is 9x10-31 kg and its velocity is 930 m/sec.Calculate its wave length. 10) Calculate the momentum of particle whose de Broglie wavelength is 2A0.

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HEISENBERG'S UNCERTAINTY PRINCIPLE According to Heisenberg, it is impossible to calculate the position and momentum of micro particles like electrons accurately and simultaneously. Mathematically, the product of uncertainties in position(  x) and momentum(  p) is always equal to h or greater than n x.p 

h n

where Δx = uncertainty in position Δp = mΔv = uncertainty in momentum m = mass of the particle Δv = uncertainty in velocity h = planck's constant n = 1,2,3,4,...... For an electron n  4 Hence we can write x.p 

or

Δx.Δv 

h 4 m

h 4

for an electron

(  Δp = mΔv )

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Explanation Case-1 When the position is calculated accurately, Δx = 0 then Δp becomes infinity i.e., it is not possible to calculate the momentum accurately Case-2 When the momentum is calculated accurately, Δp = 0 then Δx becomes infinity i.e., it is not possible to calculate the position accurately. Problems

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11) Calculate the uncertainty in the position if the uncertainty in its velocity is 5x105 m/sec for an electron. 12) Calculate the uncertainty in the velocity if the uncertainties in its position is 1A0 for an electron. 13) The uncertainties in the position and velocity of a particle are respectively 1x10-12m and 3x10-24 m/sec. Calculate the mass of the particle.

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QUANTUM MECHANICAL MODEL OF ATOM

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SCHRODINGER'S WAVE EQUATION

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The important consequence of Heisenberg's uncertainty principle is - it is not possible to determine the exact position of an electron in an atom. But it is possible to predict the probability of finding an electron in space around the nucleus. To determine the probability of finding an electron, Erwin Schrodinger proposed wave mechanical model for the motion of electron. He considered electron as a three dimensional wave moving in the electric field of a positively charged nucleus. He derived an equation which describes the motion of an electron wave along the three axes x, y and z as follows

 2  2  2 8 2 m    2  E  V   0 x 2 y 2 z 2 h In the above equation, m = mass of electron E = total energy of electron V = potential energy of electron  = wave function

Meaning and significance of  and  2  :- It is called wave function. It denotes the amplitude or intensity of the electron wave.  2:- It is called probability function. It denotes the probability of finding an electron in space at a given point. When Schrodinger’s wave equation is solved for hydrogen atom, the solutions give possible energy levels and corresponding  values for the electrons. But the accepted solutions to wave functions, which are called eigen wave functions, are obtained by applying following boundary conditions. Boundary conditions 1)  must be continuous. 2)  must be finite. 3)  must be single valued at any point. 4) The probability of finding the electron over the space from +  to -  must be equal to one.

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Atomic structure

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Important features of quantum mechanical model of atom 1) The energy of an electrons in an atom is quantized. 2) The quantized energy levels of electrons are derived from the accepted solutions of Schrodinger’s wave equation by considering electron as a wave. 3. All the information about electron in an atom is contained in its orbital wave function  . 4. The path of the electron can never be determined accurately. Therefore we find only the probability of the electron at different points in space around an atom. 5. The probability of finding an electron at a point within an atom is proportional to the square of the orbital wave function ( 2 ) at that point.  2 is known as probability density and is always positive. It is possible to predict the region around the nucleus in this region  2 has maximum values. Which is called atomic orbital, from the  2 values.

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SHAPES OF ATOMIC ORBITALS Atomic orbital:-The space around the nucleus where the probability of finding electron is maximum ( 2 > 95%) is known as atomic orbital. The shapes of atomic orbitals can be known by solving the Shrodinger's wave equation for  2 values at different points in the space around the nucleus.

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Radial Probability Function (D-function):- The function which denotes the probability of finding an electron in small volume at a radial distance from the nucleus, without any reference to its direction, is known as radial probability function or D-function.

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D  4 r 2 dr. 2

Orbital s p d f

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Nodal Plane:- The plane where the probability of finding the electron is zero ( 2 = 0) is known as nodal plane. The nodal plane passes through the nucleus and hence is also known as angular node. The number of nodal planes for a given orbital = l (azimuthal quantum number) The number of nodal planes for different orbitals are as follows, 'l'

no. of nodal planes

0 1 2 3

0 1 2 3

Nodal region:- The region around the nucleus where the probability of finding the electron is zero ( 2 = 0) is known as nodal region. The nodal region does not pass through the nucleus and hence known as radial node. The number of nodal regions for a given orbital = n - l - 1 For example, the number of nodal regions for 1s orbital = 1-0-1 = 0 for 2s orbital = 2-0-1 = 1 for 2p orbital = 2-1-1 = 0 Note:-The total of no. of nodal planes and regions for a given orbital = n - l Radial Probability Distribution curves:- The graphs plotted between the D-function and the radial

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Atomic structure

distance from the nucleus are known as radial probability distribution curves. These curves give an idea about the variation of electron density with radial distance around the nucleus for a given orbital.

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radial distance

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s-orbital s-orbitals are spherical in shape with spherical symmetry. There are no nodal planes for s-orbitals. The nodal regions for s-orbitals are given as follows orbital no. of nodal regions 1s 0 2s 1 3s 2 The radial probability curves and shapes of 1s and 2s orbitals are given as follows.

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Radial probability distribution curve and shape of 1s orbital

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radial node

radial distance Radial probability distribution curve and shape of 2s orbital

p-orbitals The p-orbitals are double dumbbell in shape. Each p-orbital contains a nodal plane. The shapes and orientations of p-orbitals in a given sub level along with the nodal planes are shown as follows.

radial distance

z-axis

z-axis

z-axis y-axis px

x-axis

x-axis

x-axis

y-axis py

Radial probability distribution curve for 2p orbitals

y-axis pz

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Atomic structure

d-orbitals The d-orbitals are double dumbbell in shape. There are two nodal planes for every d-orbital In a given sub level, there are five degenerate d-orbitals. These are divided into two groups based on their orientation i.e., t2g group and eg group. The lobes of d-orbitals in t2g group are oriented in between the axes by making 450 of angle with them. There are three t2g orbitals i.e., dxy, dxzand dyz. Where as the lobes of d-orbitals in eg group are oriented along the axes. There are two such eg orbitals i.e., d x 2  y2 and d z 2 .

y-axis

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d x2  y2

z-axis

d z2

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dxy

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y-axis

x-axis

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x-axis

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In case of d z 2 orbital, there are only two lobes oriented along the z axis and there is a concentric ring called torus along the xy axes.

ELECTRONIC CONFIGURATION IN MULTI-ELECTRON ATOMS Representation of electronic configuration The electronic configurations of atoms are represented by two methods i.e., nlx method and box method nlx method 'n' is principal quantum number 'l' is azimuthal quantum number, indicated by letters s,p,d,f.... 'x' is the number of electrons in the orbital For example, the electronic configuration of 'He' atom can be represented as 1s2. Here, 'l' is principal quantum number, 's' represents azimuthal quantum number, l = 0 and the superscript 2 represents the number of electrons present. Box method In this method, the orbitals are represented by boxes and the electrons are denoted by upward and downward arrows in the boxes. For example, the electronic configuration of 'He' can be represented as follows.

1s2

Principles and rules: The following principles and rules are used in writing the electronic configurations.

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Pauli's Exclusion Principle: No two electrons in an atom can have same set of quantum numbers. For example, the electronic configuration of 'He' is 1s2 The set of quantum numbers of two electrons in 'He' atom are For first electron, n = 1, l = 0, m = 0, s = +1/2 For second electron, n = 1, l = 0, m = 0, s = -1/2

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Aufbau Rules The electrons in an atom are arranged according to aufbau rules. According to these rules, the differentiating electron in an atom enters into the orbital with lowest energy. The relative energy of an orbital can be decided from following rules. 1. The orbital with least (n + l) value possesses lowest energy. For example, the (n + l) values of 1s and 2s orbitals are for 1s, n + l = 1 + 0 = 1 for 2s, n + l = 2 + 0 = 2 Hence 1s orbital has lower energy than 2s, as it has low n + l value, and the electron will first enter into 1s orbital. 2. If two are more orbitals possess same (n + l) value, the orbital with lower 'n' value has low energy. For example, the (n + l) values of 2p and 3s orbitals are same for 2p, n + l = 2 + 1 = 3 for 3s, n + l = 3 + 0 = 3 But 2p orbital with lower 'n' value (=2) possesses lower energy and hence the electron enters first into this orbital.

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Hund's Rule of Maximum Multiplicity: No pairing of electrons occur until all the degenerate orbitals in an atom are filled with one electron each. For example, the electronic configuration of carbon is 1s2 2s2 2px1 2py1 1s2

2s2

2px12py1

In above case, the differentiating electron is entering into the 2py orbital instead of 2px because both 2px and 2py orbitals are degenerate orbitals. Hence no pairing occured. Anomalous electronic configuration Chromium and copper exhibit anomalous electronic configurations as the atoms with half filled or fullfilled d-orbitals are more stable. Cr (Z =24) has [Ar] 3d5 4s1 configuration instead of [Ar] 3d4 4s2 as the half filled d-sub level is more stable. Cu (Z =29) has [Ar] 3d10 4s1 configuration instead of [Ar] 3d9 4s2 as the full filled d-sub level is more stable. Problems 1) Arrange the electrons, for which the quantum numbers are given below, in their increasing order of energy. a) n = 3, l = 1, m = -1, s = +1/2 b) n = 4, l = 0, m = 0, s = -1/2 c) n = 3, l = 2, m = -2, s = -1/2 d) n = 5, l = 3, m = +2, s = +1/2