Chemistry Form 6 Sem 1 02
Short Description
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Description
4.1 Atomic Spectra. Spectrum ~ display of component of a beam of radiation. When white ray passes through a prism it forms rainbow colour. A continuous spectrum is spectrum composed of visible light of all wavelength while a discontinuous spectrum which the line spectrum represent the atomic emissions Productions of line spectrum depend on what types of element used and the visible light emitted also depend on it. Spectrophotometer is the instrument that is used to split and dispersed into individual wavelength. The study of line spectra can be used to Identify an element Determine the ionisation energy of an element Determine the arrangement of electrons in an atom of an element.
The main components of a typical spectrophotometer Lenses/slits/collimaters narrow and align beam.
Source produces radiation in region of interest. Must be stable and reproducible. In most cases, the source emits many wavelengths.
Sample in compartment absorbs characteristic amount of each incoming wavelength.
Monochromator (wavelength selector) disperses incoming radiation into continuum of component wavelengths that are scanned or individually selected.
Computer converts signal into displayed data.
Detector converts transmitted radiation into amplified electrical signal.
7.3
4.2 The Atomic Emission Spectrum of Hydrogen In hydrogen, the atomic emission spectrum of hydrogen shows a few distinct series of line. These are the ultraviolet region Lyman series – in …………………………… visible Balmer series – in …………………………. region. infrared Paschen series – in ………………………… region The diagram below shows the example of the line emission produced through the spectrophotometer.
4.2.1 Formation of atomic spectra. The absorption spectrum of a substance is a continuous spectrum of radiation produced by examining the substance through spectrophotometer. Hence, there is a absorption and emission process occur in the formation of spectra where Absorption spectroscopy ↑ Occur when electron which is occupying at lower energy level (ground level / state) absorb light at a certain frequency. ↑ When enough energy is absorbed, electron is able to ‘excite’ to a higher energy level. n=3
Emission spectroscopy ↓ Occur when electron which is located at a high energy level (after absorbed energy) released the energy in the form of radiation ↓ When energy is released, electron falls to the original ground state, emitting the radiation and hence give the spectrum n=3
n=2
n=2
n=1
n=1
The following characteristics of any series should be noted : The spectrum consists of discrete lines, each having its own discrete frequency.
1 1 = R H 2 − 2 λ n2 n1 1
λ = wavelength (in m) n1 = series of the spectra line n2 = line where the emission of energy begin. RH = Rydberg constant = 1.097 x 107 m-1
The wavelengths represented by broken lines are called spectrum (plural : spectra) Each spectrum represents the wavelength (or frequency) in a series. The frequency emitted can be related to the spectral line by the Rydberg’s Equation The value of n1 and n2 for the atomic spectrum of hydrogen are given in Table below Series
n1
n2
Lyman
1
2, 3, 4 …….
Balmer
2
3, 4, 5 …….
Paschen
3
4, 5, 6 …….
Example 1 : Calculate the wavelength of the first line of Lyman series
In Lyman series, n1 = 1 First line in series, so n2 = 1 1 1 1 = R H 2 − 2 λ n2 n1
+1=2 1 1 7 1 = 1 . 097 × 10 − 2 2 λ 2 1 λ = 122 nm
Example 2 : Calculate the wavelength of the third line of Balmer series
In Balmer series, n1 = 2 First line in series, so n2 = 2 + 3 = 5 1 1 = R H 2 − 2 λ n2 n1 1
1 1 7 1 = 1 . 097 × 10 − 2 2 λ 5 2 λ = 434 nm
Example 3 : Calculate the wavelength of the forth Example 4 : Calculate the wavelength of the last line of Paschen series line of Balmer series
In Passchen series, n1 = 3 First line in series, so n2 = 3 1 1 1 = RH 2 − 2 λ n2 n1
+4=7
1 1 7 1 = 1 . 097 × 10 − 2 2 λ 7 3 λ = 1005 nm
In Balmer series, n1 = 2 Last line in series, so n2 = 2 + ∞ = ∞ 1 1 = R H 2 − 2 λ n2 n1 1
1 1 7 1 = 1 . 097 × 10 − 2 22 λ ∞ λ = 365 nm
c (speed of light ) 3 .0 × 10 8 m s −1 Since frequency, f = ; f= λ ( wavelength ) λm Example 5 : Calculate the frequency of the second line of Paschen series
In Passchen series, n1 = 3 Second line in series, n2 = 3 + 2 = 5 1 1 = R H 2 − 2 λ n2 n1 1
Example 6 : Calculate the frequency of the last line of Lyman series
In Lyman series, n1 = 1 last line in series, so n2 = 1 + ∞ = ∞
1 1 1 1 1 1 = 1 . 097 × 10 7 2 − 2 = R H 2 − 2 3 n2 5 λ λ n1
1 / λ = 7.80 x 105 m-1 f = c x (1 / λ) = (3.0 x 108)(7.80 x 105) f = 2.34 x 1014 s-1
1 1 7 1 = 1 . 097 × 10 − 2 12 λ ∞
1 / λ = 1.097 x 107 m-1 f =c x (1 / λ) =(3.0 x 108)(1.097 x 107) f = 3.29 x 1015 s-1
• In 1900, Max Planck proposed the quantum theory. He mentioned that radiant energy was emitted in discrete packets or “quantum” Each quantum of energy emitted, E, is proportional to the frequency, f, of the radiation • Where energy, E = Planck constant (h) x frequency (f) • h = 6.63 x 10-34 J s
Energy , E = h ( Planck con tan ts ) ×
c (speed of light ) λ ( wavelength )
Example 7 : Calculate the energy required to excite one electron to the fourth line in Passchen series In Passchen series, n1 = 3 Fourth line in series, n2 = 3 + 4 = 7
Example 8 : Calculate the energy required to excite one electron to the last line in Balmer series. In Balmer series, n1 = 2 Last line in series, n2 = 2 + ∞ = ∞
1 1 = R H 2 − 2 λ n2 n1
1 1 = R H 2 − 2 λ n2 n1
1
1 1 7 1 = 1 . 097 × 10 − 2 2 λ 7 3
1
1 1 7 1 = 1 . 097 × 10 − 2 2 λ ∞ 2
1 / λ = 9.95 x 105 m-1
1 / λ = 2.74 x 106 m-1
f = c x (1 / λ) = (3.0 x 108)(9.95 x 105) f = 2.99 x 1014 s-1 E = hf E = 6.63 x 10-34 x 2.99 x 1014 E = 1.98 x 10-19 J / e
f = c x (1 / λ) = (3.0 x 108)(2.74 x 106) f = 8.23 x 1014 s-1 E = hf E = 6.63 x 10-34 x 8.23 x 1014 E = 5.45 x 10-19 J / e
4.3 Electronic Energy Levels Hydrogen emission spectrums consist of several series of discrete lines in different limited wavelength. This implies that hydrogen atoms do not emit light at of all possible energies but only emit light with certain amount of energy. Bohr made these assumptions to explain line in spectrum. Electron move in orbit around nucleus (like planet and sun) Energy of electron is quantised (only have certain energy). Electron closer to nucleus has lower energy and vice versa. Electron of atom will fill up the lowest energy level first. But when energy is given to the electron, it moves from a lower level to a higher level. This process is called excitation. If the energy loss from the excited level to original level, electron will emit electromagnetic radiation. This is what caused the emission of spectral lines. The convergence of spectral line show the difference between successive energy level become ↓ with increasing distance of the energy levels from the nucleus
Electron
are not randomly distributed but are arranged in a series of shells or orbits which are situated at various distance from the molecules corresponding to differ E-level. Though Bohr theory proves its unsatisfactory when come to spectra of more complicated atoms. Thus Rydberg equation can only used for atoms or ions contain only 1 electron. For atom or ion with many electrons, the analysis of atomic spectra become more difficult.
Wavelength increase
Frequency increase
4.4
Calculate the Ionisation energy of Hydrogen from its Line Spectrum When energy level of the hydrogen spectrum increased, there’s a possibility where it reaches the convergence limit of the energy level where n = When this occur, electron are not able to return to ……………….., ground level the hydrogen atom is …………….., according ionised and eventually to the following equation Equation : H (g) H+ (g) + eFor this to occur, a certain amount of energy is required. The energy required to remove one electron is called ionisation energy Here, a positive charged ion is formed. ……………………………… There are 2 methods of calculating ionisation energy of hydrogen through the line spectrum
Using Rydberg and Planck equations Supposed that electron at energy level n1 = 1 absorbed energy and it is excited. When reaches the convergence limit, n2 = , the wavelength can be calculated Using Rydberg equation n1 = 1 n2 =
∞
∞
1 1 = R H 2 − 2 λ n2 n1 1
1 = 1 .097 x 10 7 λ
For each electron, the energy absorbed can be determined using the Planck equation
E=hf
1 1 7 1 = 1 .097 x 10 − 2 2 λ 1 α
E = h c 1/λ
E = (6.63 x 10-34)(3.0 x 108)(1.097 x 107) E = 2.182 x 10-18 J / e
Since this much energy is required to excite one electron, so for 1 mole of electron, the energy required
Since in 1 mol contain 6.02 x 1023 e, so Ionisation energy, ∆H = E x NA = 2.182 x 10-18 J / e x 6.02 x 1023 = 1.313 x 106 J / mol + 1313 Hence, the ionisation energy of hydrogen is …………. kJ mol-1
4.5 Atomic Orbital Heinsberg uncertainty principle = position & momentum of an electron cannot be known with great precision but probability of finding an electron in a certain position can be calculated. high Orbital - the space within which there’s a ………………… probability of finding an ……………………. Theelectron electron in the orbital is describe as occupying 3-D space around the nucleus. The nucleus is described as being surrounded by
Orbital – boundary within which there is 95% (high) chances to find an electron
Type of orbital
Shape of orbital
Number of electron filled
S
2
P
6
Type of orbital
Shape of orbital
Number of electron filled
10
f
7 orbitals
14
Filling of orbitals in shell Electrons filled in the shell of an atom through occupation of orbitals in each shell The formula used to calculate the number of orbitals that can occupy each shell is No of orbitals in shell = n2 ; (n = the number of shell) Shell, n
No. of orbitals
Type of orbitals
1
n2 = 12 = 1
s
2
n2 = 22 = 4
s,p
3
n2 = 32 = 9
s,p,d
4
n2 = 42 = 16
s,p,d,f
Orbital in shell
1s 2s 2p 3s 3p 3d 4s 4p 4d 4f
4.6 Electronic configuration Electronic configuration of an element is the expression on how the electrons inside the element filled in its orbitals accordingly There are 3 rules for assigning electrons to the orbitals of an atom in the ground state. 1. …………………………………………………………………. Aufbau Principle 2. …………………………………………………………………………. Pauli’s Exclusion Principle 3. …………………………………………………………………………. Hund’s Rule 1.The Aufbau Principle ~ e- occupy orbitals in the order of the energy levels of the orbitals. Orbital with the lowest energy are always occupied first. The arrangement of energy level from the lowest to the highest is shown in the diagram below
Energy level : 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p
2. The Pauli exclusion principle ~ 2 electron may occupy the same orbital but these 2 electrons must have opposite spin.
Same spin (cannot occur)
Opposite spin (Pauli exclusion principle) 3. The Hund’s Rule ~ when electrons are placed in a set of orbitals with equal or degenerated energies, the electrons must occupy them singly with parallel spins before they occupy the orbital in pair In other words, an atom tend to have as many unpaired electrons as possible. Example : when filling in 3 electrons in p orbital
Electronic configuration
Element
No of e-
Hydrogen H
1
____ 1s
1s1
Helium He
2
____ 1s
1s2
Lithium Li
3
____ ____ 1s 2s
1s22s1
Beryllium Be
4
____ ____ 1s 2s
1s22s2
Boron B
5
____ ____ ____ ____ ____ 1s 2s 2p
1s22s22p1
Carbon C
6
____ ____ ____ ____ ____ 1s 2s 2p
1s22s22p2
Nitrogen N
7
____ ____ ____ ____ ____ 1s 2s 2p
1s22s22p3
Orbital diagram
Electronic configuration
Element
No of e-
Oxygen O
8
____ ____ ____ ____ ____ 1s 2s 2p
1s22s22p4
Fluorine F
9
____ ____ ____ ____ ____ 1s 2s 2p
1s22s22p5
Neon Ne
10
____ ____ ____ ____ ____ 1s 2s 2p
1s22s22p6
Sodium Na
11
____ ____ ____ ____ ____ ____ 1s 2s 2p 3s
1s22s22p63s1
12
____ ____ ____ ____ ____ ____ 1s 2s 2p 3s
1s22s22p63s2
Al
13
____ ____ ____ ____ ____ ____ 1s 2s 2p 3s
___ ___ ___ 3p
1s22s22p63s23p1
Silicon Si
14
____ ____ ____ ____ ____ ____ ____ ____ ____ 1s 2s 2p 3s 3p
1s22s22p63s23p2
Magnesium
Mg Aluminium
Orbital diagram
Element
No of e-
Phosphorous, P
15
____ ____ ____ ____ ____ ____ ____ ____ ____ 1s 2s 2p 3s 3p
1s22s22p63s23p3
Sulphur S
16
____ ____ ____ ____ ____ ____ ____ ____ ____ 1s 2s 2p 3s 3p
1s22s22p63s23p4
Chlorine Cl
17
____ ____ ____ ____ ____ ____ ____ ____ ____ 1s 2s 2p 3s 3p
1s22s22p63s23p5
Argon Ar
18
____ ____ ____ ____ ____ ____ ____ ____ ____ 1s 2s 2p 3s 3p
1s22s22p63s23p6
Potassium K
19
____ ____ ____ ____ ____ ____ ____ ____ ____ ___ 1s 2s 2p 3s 3p 4s
1s22s22p63s23p64s1
Calcium Ca
20
____ ____ ____ ____ ____ ____ ____ ____ ____ ___ 1s 2s 2p 3s 3p 4s
1s22s22p63s23p64s2
Orbital diagram
Electronic configuration
1s22s22p63s23p63d14s2
1s22s22p63s23p63d24s2
1s22s22p63s23p63d34s2
1s22s22p63s23p63d54s1
1s22s22p63s23p63d54s2
1s22s22p63s23p63d64s2
1s22s22p63s23p63d74s2
1s22s22p63s23p63d84s2
1s22s22p63s23p63d104s1
1s22s22p63s23p63d104s2
Electron filling in orbitals at the same According to Hund’s rule : ………………………………………………………..… energy level with single electron before pairing ……………………………………… For the case of chromium, it has the valence electronic as 3d54s1 3d44s2 ………………………………... instead of ……………………………. This is due to the extra stability achieved due to the equal / symmetrical distribution of charge around the atom within the half-filled (p3 and d5) orbital and full-filled orbital (p6 and d10) What happened before and after the electron filling in the 4s orbital can be explained by the diagram below
Position of 3 d and 4 s orbital before electron filling in
Position of 3d and 4s orbital after electron filling in
____ ____ ____ ____ ____ 3d
____ 4s ____ ____ ____ ____ ____ 3d
____ 4s
When this occurs, the 3rd electron is expected to be filled in 3d orbital (Aufbau principle). Until the 6th electron is filling in, a degeneration of orbital occur where now the energy level between 3d and 4s orbital are the same, thus one of the electron from the 4s “jumped” to 3d orbital. When this happened, an extra stability occur at the half-filled 3 d orbital has achieved. When the 7th electron is filled, it will be placed in 4s orbital
Position of 3 d and 4 s orbital before the 6th electron filling in
Position of 3d and 4s orbital after the 6th electron filling in
____ 4s ____ ____ ____ ____ ____ 3d
____ 4s ____ ____ ____ ____ ____ 3d
Same
thing happened when comes to the case of copper. An extra stability is achieved when the orbital in 3 d is full-filled with electrons.
Position of 3 d and 4 s orbital before the 11th electron filling in
Position of 3d and 4s orbital after the 11th electron filling in
____ 4s ____ ____ ____ ____ ____ 3d
____ 4s ____ ____ ____ ____ ____ 3d
1
2
2
1
1s B3+ : ……………………………….
2
2
3
1s 2s 2p N3- : ……………………………….
2
2
5
1s 2s 2p F- : ……………………………….
1s 2s : …………………………….
5B
1s 2s 2p : ……………………………..
7N
1s 2s 2p : ……………………………..
9F
1s 2s 2p : ………………………………
12Mg :
2
1s Li+ : ………………………………
2
3Li
2
2
6
2
2
1s 2s 2p 3s ………………………….
2
2
6
2
2
6
2
2
6
1s 2s 2p Mg2+ : …………………………….
22s22p63s23p2 1s 14Si : ………………………….. 22s22p63s23p4 1s S : …………………………….
Si4+
22s22p6 1s : ……………………………….
19K
1s 2s 2p 3s 3p 4s : ……………………………………
22s22p63s23p6 1s : ……………………………… 1s22s22p63s23p6 K+ : …………………………………….
21Sc
1s22s22p63s23p63d14s2 : …………………………………..
1s 2s 2p 3s 3p 3d Sc2+ : ………………………………….
1s22s22p63s23p63d34s2 : ………….……………………….
1s 2s 2p 3s 3p 3d V2+ : …………………………………..
S2-
16
23V
2
2
6
2
6
1
1s22s22p63s23p63d84s2 Ni : …………………………………..
2
2
6
2
6
10
2
6
2
6
1
2
2
6
2
6
3
22s22p63s23p63d84s1 1s : ………………………………….. 22s22p63s23p63d10 1s 2+ Zn : …………………………………..
Ni+
28
30Zn
2
2
1s 2s 2p 3s 3p 3d 4s : ………………………………….
22s22p63s23p63d104s24p5 1s 35Br : ………………………………….
Br-
22s22p63s23p63d104s24p6 1s : ……………………..............
•From the Periodic Table, elements can be classified accordingly to the block. Block of Periodic Table
Group of elements
s-block
Group 1 and 2 (Those with valence electron ns1 and ns2)
p-block
Group 13 , 14 , 15 , 16 , 17 , 18 (Those with valence electrons ns2 np1, 2, 3, 4, 5, 6)
d-block
Group 3 – 12 (Transition element with valence electron (n–1)d1-10 ns2 )
f-block
Lanthanides and Actinides Series
Based on the valence electron, the Group of an element can determined.
Valence electron
Group
ns1
1
ns2
2
ns2np1
13
ns2np2
14
ns2np3
15
ns2np4
16
ns2np5
17
ns2np6
18
be
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