1
4
4.1
The Electromagnetic Spectrum
4.2
Deduction of Electronic Structure from Ionization Enthalpies
4.3
The Wave-mechanical Model of the
Atom
4.4 Atomic Orbitals

Chapter 4 The electronic structure of atoms (SB p.80)
The electronic structure of atoms
Two sources of evidence : -
(a) Study of atomic emission spectra of the elements
(b) Study of ionization enthalpies of the elements
2

3
Spectra  plural of Spectrum
Arises from light emitted from individual atoms

The electromagnetic spectrum
Wavelength (m) 
4
 Frequency (Hz / s  1 )
Speed of light (ms  1 ) = Frequency  Wavelength c =   3  10 8 ms  1

The electromagnetic spectrum violet red
Wavelength (m) 
5
 Frequency (Hz / s  1 )
Visible light

The electromagnetic spectrum violet red
Wavelength (m) 
6
 Frequency (Hz / s  1 )
 Increasing energy
Ultraviolet light

The electromagnetic spectrum violet red
Wavelength (m) 
7
 Frequency (Hz / s  1 )
 Increasing energy
X-rays

The electromagnetic spectrum violet red
Wavelength (m) 
8
 Frequency (Hz / s  1 )
 Increasing energy
Gamma rays

The electromagnetic spectrum violet red
Wavelength (m) 
9
 Frequency (Hz / s  1 )
Decreasing energy 
Infra-red light

The electromagnetic spectrum violet red
Wavelength (m) 
10
 Frequency (Hz / s  1 )
Decreasing energy 
Microwave & radio waves

4.1
The electromagnetic spectrum (SB p.82)
Types of Emission Spectra
1. Continuous spectra
E.g.
Spectra from tungsten filament and sunlight
2. Line Spectra
E.g.
Spectra from excited samples in discharge tubes
11

4.1
The electromagnetic spectrum (SB p.82)
Continuous spectrum of white light
12
Fig.4-5(a)

4.1
The electromagnetic spectrum (SB p.83)
Line spectrum of hydrogen
13
Fig.4-5(b)

14
H
2
(g)
Electric discharge
Or Heating
H(g)
Hydrogen atom in ground state means its electron has the lowest energy

Atoms in excited state
15
H(g)
Ground
Atom in ground state
Electric discharge
Or Heating
H*(g)
Excited

Atoms in excited state
Not Stable h 
H*(g)
Excited
Atom in ground state
Atom returns to ground state h 
16
H(g)
Ground

17
Atoms in excited state
Atom in ground state
Atom returns to ground state
E = h 
Planck’s equation h 

Atoms in excited state h 
18
Atom in ground state
Atom returns to ground state
E = h 
Planck : Nobel laureate Physics, 1918

Atoms in excited state
E
2 h 
19
E
1
Atom in ground state
Atom returns to ground state
E = h 
E = energy of the emitted light = E
2
– E
1

Atoms in excited state
E
2 h 
20
E
1
Atom in ground state
Atom returns to ground state
E = h 
 = Frequency of the emitted light

Atoms in excited state
E
2 h 
21
E
1
Atom in ground state
Atom returns to ground state
E = h  h (Planck’s constant) = 6.63  10  34 Js

Atoms in excited state
E
2 h 
E
1
Atom in ground state
Atom returns to ground state
E = h  h = 6.63  10  34 Js
Energy cannot be absorbed or emitted by an atom in any arbitrary amount.
22

Atoms in excited state
E
2 h 
E
1
Atom in ground state
Atom returns to ground state
E = h  h = 6.63  10  34 Js
Energy can only be absorbed or emitted by an atom in multiples of 6.63
 10  34 J.
23

4.1
The electromagnetic spectrum (SB p.84)
Characteristic Features of the
Hydrogen Emission Line Spectrum
1. The visible region – The Balmer Series green red violet
24

Q.1
E 
 h ν λ  c
ν
E 
 hc
λ

6.63
 10  34 Js
656.3


3.00
10  9
 m
10 8 ms  1

= 3.03
 10  19 J
25

26
E n  3
 E n  2
 3.03
 10  19 J energy of one photon emitted

Q.2
(a) The spectral lines come closer at higher frequency and eventually merge into a continuum
( 連續體 )
(b) n =   n = 2
27

Q.2
e  1
(c) The electron has been removed from the atom.
I.e. the atom has been ionized.
H(g)  H + (g) + e 
28

29 n    n  2 corresponds to the last spectral line of the Balmer series

4.1
The electromagnetic spectrum (SB p.83)
The Complete Hydrogen Emission Spectrum
UV Visible IR
30

Q.3
(a) The spectral lines in each series get closer at higher frequency.
(b) Since the energy levels converge at higher level, the spectral lines also converge at higher frequency.
31

32
Rydberg Equation
1
 R
H
1 a 2

1 b 2
Relates wavelength of the emitted light of hydrogen atom with the electron transition

1

 R
H
1 a 2

1 b 2
1
= wave number of the emitted light
= number of waves in a unit length e.g.
1

 100 m  1  100 waves in 1 meter
1

 
33

1

 R
H

1

 c

1
 
1 a 2

1 b 2
34

35
1
 R
H a
1
2

1 b 2
Electron transition : b  a
, a, b are integers and b > a a represents the lower energy level to which the electron is dropping back b represents the higher energy levels from which the electron is dropping back

36
1

 R
H
1 a 2

1 b 2
Balmer series, a = 2 b = 3, 4, 5,…

37
2 3 4 5 6 
1

 R
H
1 a 2

1 b 2
Lyman series, a = 1 b = 2, 3, 4,…

38
4 5 6 7 
1

 R
H
1 a 2

1 b 2
Paschen series, a = 3 b = 4, 5, 6,…

39
1

 R
H
1 a 2

1 b 2
H

Q.4
1

(10 6 m  1 )
1.52
2.06
2.30
2.44
2.52
2.57
2.60
2.74
1 b 2
0.111
0.063
0.040
0.028
0.020
0.016
0.012
0.000
40

1
10 6 m  1 y-intercept
 2.74
 10 6 m  1  R
H
1
2 2

1
 2
 R
H
R
H
= 1.096
 10 7 m  1
1
4
2
41

4.1
The electromagnetic spectrum (SB p.84)
Interpretation of the atomic hydrogen spectrum
Discrete spectral lines
 energy possessed by electrons within hydrogen atoms cannot be in any arbitrary quantities but only in specified amounts called quanta .
42

4.1
The electromagnetic spectrum (SB p.84)
Interpretation of the atomic hydrogen spectrum
Only certain energy levels are allowed for the electron in a hydrogen atom.
The energy of the electron in a hydrogen atom is quantized .
43

4.1
The electromagnetic spectrum (SB p.84)
Interpretation of the atomic hydrogen spectrum
44

Interpretation of the atomic hydrogen spectrum
Bohr’s Atomic Model of Hydrogen
Nobel Prize Laureate in Physics, 1922
Niels Bohr (1885-1962)
45

Interpretation of the atomic hydrogen spectrum
Nobel Prize Laureate in Physics, 1922 for his services in the investigation of the structure of atoms and of the radiation emanating from them
46

Interpretation of the atomic hydrogen spectrum
47

Interpretation of the atomic hydrogen spectrum
1. The electron can only move around the nucleus of a hydrogen atom in certain circular orbits with fixed radii.
Each allowed orbit is assigned an integer, n, known as the principal quantum number .
48

Interpretation of the atomic hydrogen spectrum
2. Different orbits have different energy levels .
An orbit with higher energy is further away from the nucleus.
49

Interpretation of the atomic hydrogen spectrum
3. Spectral lines arise from electron transitions from higher orbits to lower orbits.
E
2 n = 2 n = 1
50
E
1

Interpretation of the atomic hydrogen spectrum
For the electron transition E
2
 E
1
, the energy emitted is related to the frequency of light emitted by the Plank’s equation :
E
2 n = 2
E
1
 E = E
2
– E
1
= h  n = 1
51

Interpretation of the atomic hydrogen spectrum
4. In a sample containing numerous excited
H * atoms, different H * atoms may undergo different kinds of electron transitions to give a complete emission spectrum .
52

53
Electronic
Transition
2  1
3  1
4  1
3  2
4  2
5  2
4  3
Wavelength Predicted by Bohr (nm)
121.6
Wavelength Determined by Experiment (nm)
121.7
102.6
97.28
102.6
97.32
656.6
486.5
434.3
1876
656.7
486.1
434.1
1876

Interpretation of the atomic hydrogen spectrum
5. The theory failed when applied to elements other than hydrogen ( multielectron systems )
54

Illustrating Bohr’s Theory
First line of Lyman series, n = 2
 n = 1
E
2
By Planck’s equation,
E
 h
  hc

E = E
2
– E
1 n = 2
E
1 n = 1
55

E
 h
  hc

E
2
By Rydberg’s equation,
E = E
2
– E
1
1


R


1 n
1
2

1 n
2
2



E
 hc
1

E
1
 hcR


1 n
1
2

1 n
2
2

 n = 2 n = 1
56

E
 hc

 hcR


1 n
1
2

1 n
2
2
E
2


E = E
2
– E
1

E
 hcR n
1
2
 hcR n
2
2 E
1




 hcR n
2
2 






 hcR n
1
2 



E

2
E
1 n = 2 n = 1
57

All electric potential energies have negative signs except E

E
1
< E
2
< E
3
< E
4
< E
5
< …… < E

= 0
All are negative
58

Balmer series,
Transitions from higher levels to n = 2
E
 hcR



1
2
2

1 n
2 

 n = 3, 4, 5, …
Visible region
59

Lyman series,
Transitions from higher levels to n = 1
E
 hcR


 
2
1
1 1 n
2 

 n = 2, 3, 4,…
More energy released
 Ultraviolet region
60

Paschen series,
Transitions from higher levels to n = 3
E
 hcR
1
3
2

1 n
2
Less energy released
 Infra-red region n = 4, 5, 6,…
61

4.1
The electromagnetic spectrum (SB p.87)
62
UV
 E = E
2
– E
1
= h  visible
IR

63
Q.5 n = 100  n = 99
Energy of the light emitted is extremely small.
1


1 .
096

10
7



1
99
2

1
100
2 

 m

1
  
4 .
5

10

2 m

The electromagnetic spectrum
Wavelength (m) 
64
 Frequency (Hz / s  1 ) Microwave

Q.5 n = 100  n = 99
Energy of the light emitted is extremely small.
1


1 .
096

10
7



1
99
2

1
100
2 

 m

1
  
4 .
5

10

2 m
65
 Microwave region

Q.6
66

67
Energy of the first line

E
3

2

E
3

1

E
2

1 n = 3 n = 2 n = 1

68
Energy of the first line

E
3

2

E
3

1

E
2

1 h

3

2

3

2
 h

 
3

1
3

1
 

2

1 h


2

1

292 .
3

246 .
6


10
13
Hz
= 45.7

10 13 Hz

69
Energy of the second line

E
4

2

E
4

1

E
2

1 n = 4 n = 2 n = 1

70
Energy of the second line

E
4

2

E
4

1

E
2

1 h

4

2

4

2
 h

4

1
 
4

1
 

2

1 h


2

1

308 .
3

246 .
6


10
13
Hz
= 61.7

10 13 Hz

71
Energy of the third line

E
5

2

E
5

1

E
2

1 n = 5 n = 2 n = 1

72
Energy of the third line

E
5

2

E
5

1

E
2

1 h

5

2

5

2
 h

5

1
 
5

1
 

2

1 h


2

1

315 .
7

246 .
6


10
13
Hz
= 69.1

10 13 Hz

Convergence Limits and Ionization Enthalpies
Ionization enthalpy is the energy needed to remove one mole of electrons from one mole of gaseous atoms in ground state to give one moles of gaseous ions (n =  )
X(g)  X + (g) + e 
Units : kJ mol  1
73

Convergence Limits and Ionization Enthalpies
Convergence Limits
74
The frequency at which the spectral lines of a series merge.

Q.7
n = 1 n = 
H(g)  H + (g) + e 
Ionization enthalpy

E
1
 

E
 
1
 h

 
1
Convergence limit of Lyman series
75

76
Atom
Energy Level of e
 to be removed in ground state of atom
Electron transition for ionization of atom
H
He
Li
Na n = 1 n = 1 n = 2 n = 3 n = 1
 n =

Lyman series ?
n = 1
 n =

Balmer series ?
n = 2
 n =

Paschen series ?
n = 3
 n =


Q.9
Ionization enthalpy of helium

E
1
 

E
 
1
 h

 
1
= (6.626

10

34 Js)(5.29

10 15 s

1 )(6.02

10 23 mol

1 )
=
2110 KJ mol  1
> Ionization enthaply of hydrogen
77

H He
E

Relative positions of energy levels depend on the nuclear charge of the atom.
E
2
E
2
’
E
1
E
1
’
Ionization enthalpy : He > H
78

4.1
The electromagnetic spectrum (SB p.89)
The uniqueness of atomic emission spectra
79
No two elements have identical atomic spectra

4.1
The electromagnetic spectrum (SB p.89)
The uniqueness of atomic emission spectra
80 atomic spectra can be used to identify unknown elements.

Flame Test heat
NaCl(s) Na(g) + Cl(g) atomization
Na(g) + Cl(g) Na * (g) + Cl * (g) h 
Na
+ h 
Cl
Na * (g) + Cl * (g) Na(g) + Cl(g)
81

Flame Test
The most intense yellow light is observed.
82

Q.10
E  h ν  hc


(6.626
 10  34 Js)(3.00
450  10  9 m
 10 8 ms  1 )
= 4.42
 10  19 J
83

Emission vs Absorption
Emission spectrum of hydrogen visible
Bright lines against a dark background
Absorption spectrum of hydrogen visible
 (nm)
84
Dark lines against a bright background  (nm)

Absorption spectra are used to determine the distances and chemical compositions of the invisible clouds.
85

The Doppler effect
Lower pitch heard
Higher pitch heard
The frequency of sound waves from a moving object
(a) increases when the object moves towards the observer.
(b) decreases when the object moves away from the observer.
86

Redshift and the Doppler effect
Frequency of light waves emanating from a moving object decreases when the light source moves away from the observer.
 wavelength increases
 spectral lines shift to the red side with longer wavelength
 redshift
87

redshift
Moving at higher speed
Moving at lower speed
88

89
Redshift
Left : -
Absorption spectrum from sunlight.
Right : -
Absorption spectrum from a supercluster of distant galaxies

Only atomic emission spectrum of hydrogen is required in A-level syllabus !!!
90

91
4.2

Evidence of the Existence of Shells
For multi-electron systems,
Two questions need to be answered
92

Evidence of the Existence of Shells
1. How many electrons are allowed to occupy each electron shells ?
Existence of Shells
2. How are the electrons arranged in each electron shell ?
Existence of Subshells
93

Ionization enthalpy
Ionization enthalpy is the energy needed to remove one mole of electrons from one mole of gaseous atoms in ground state to give one moles of gaseous ions (n =  )
X(g)  X + (g) + e 
94

Evidence of the Existence of Shells
Successive ionization enthalpies
Q. 11
95

Q.11(a)
Be(g)  Be + (g) + e 
Be + (g)  Be 2+ (g) + e 
Be 2+ (g)  Be 3+ (g) + e 
Be 3+ (g)  Be 4+ (g) + e 
IE
1
IE
2
IE
3
IE
4
96

Q.11(b)
IE
1
< IE
2
< IE
3
< IE
4
Positive ions with higher charges attract electrons more strongly.
Thus, more energy is needed to remove an electron from positive ions with higher charges.
97

Q.11(c)
IE
1
(kJ mol

1 )
900
IE
2
(kJ mol

1 )
1758
IE
3
(kJ mol

1 )
14905
IE
4
(kJ mol

1 )
21060
98

Q.11(c)
(i) The first two electrons are relatively easy to be removed.
 they experience less attraction from the nucleus,
 they are further away from the nucleus and occupy the n = 2 electron shell .
99

Q.11(c)
(ii) The last two electrons are very difficult to be removed.
 they experience stronger attraction from the nucleus,
 they are close to the nucleus and occupy the n = 1 electron shell .
100

101
Electron Diagram of Beryllium
Second Shell
First Shell

Energy Level Diagram of Beryllium n =2 n= 1
102

E
2
E
1
E

Be
IE
1
= E

- E
2
103

E

Be Be +
IE
1
E
2 E
2
’
E
1 E
1
’
IE
1
< IE
2
104
IE
2

E

Be Be +
IE
1
IE
2
E
2
E
1
E
2
’
E
2
’’
IE
1
E
1
’
E
1
’’
< IE
2
<< IE
3
105
Be 2+
IE
3

E

Be Be + Be 2+ Be 3+
IE
1
IE
2
IE
3
E
2
E
1
E
2
’
E
2
’’
E
2
’’’
IE
1
E
1
’
E
1
’’
< IE
2
<< IE
3
< IE
4
E
1
’’’
106
IE
4

The Concept of Spin(
自旋
)
Spin is the angular momentum intrinsic to a body.
E.g. Earth’s spin is the angular momentum associated with
Earth’s rotation about its own axis.
自轉
107

On the other hand, orbital angular momentum of the
Earth is associated with its annual motion around the Sun (
公轉
)
108

Subatomic particles like protons and electrons possess spin properties.
i.e. they have spin angular momentum .
But their spins cannot be associated with rotation since they display both particle-like and wave-like behaviours.
109

Paired electrons in an energy level should have opposite spins.
Electrons with opposite spins are represented by arrows in opposite directions.
Q.12
110

111
Q.12
4 groups of electrons

112
Which group of electrons is in the first shell ?
n = 1 n = 3 n = 2
Q.12
n = 4

113
2
8
8
2
2, 8, 8, 2
Q.12

114 n = 4 n = 3 n = 2 n = 1
Q.12

4.2
Deduction of electronic structure from ionization enthalpies (p.91)
Evidence of the Existence of Shells
115

116 n = 3 n = 2 n = 1
Na

Variation of IE
1 with Atomic Number
Evidence for Subshell
117

Only patterns across periods are discussed
Refer to pp.27-29 for further discussion
118

1. A general  in IE
1 with atomic number across Periods 2 and 3.
13.(a)
119

120
2 – 3 - 3
13.(a)

2. IE
1
3. IE
1 value : Group 2 > Group 3 value : Group 5 > Group 6 13.(a)
121

2. Peaks appear at Groups 2 & 5 13.(a)
3. Troughs appear at Groups 3 & 6
122

On moving across a period from left to right,
1. the nuclear charge of the atoms  (from +3 to +10 or +11 to +18)
2. electrons are being removed from the same shell
13.(b)
 e  s removed experience stronger attraction from the nucleus.
2, 8
2, 1
2, 2
2, 3
2,
2, 5
2, 7
4
2, 6
2,8, 1
2,8, 5
2,8, 7
2,8, 2
2,8, 3
2,8, 4
2,8, 6
2,8, 8
123

IE
1
: B(2,3) < Be(2,2) 13.(b)
The electron removed from B occupies a subshell of higher energy within the n = 2 quantum shell
124

n = 
2p
2s
125
1s
IE
2s
IE
2p
13.(b)
IE
2s
(Be) > IE
2p
(B)
IE
1
: Be > B

IE
1
: Al(2,8,3) < Mg(2,8,2) 13.(b)
The electron removed from Al occupies a subshell of higher energy within the n = 3 quantum shell
126

n = 
3p
3s
127
2p
IE
3s
IE
3p
13.(b)
IE
3s
(Mg) > IE
3p
(Al)
IE
1
: Mg > Al

n = 
3p
3s
128
2p
Mg(12)
IE
3s
Al(13)
IE
3p

IE
1
: N(2,5) > O(2,6) ; P(2,8,5) > S(2,8,6)
It is more difficult to remove an electron from a half-filled p subshell
13.(b)
129

n = 
2p
2s
First IE N(7)
IE
N
>
O(8)
IE
O
130

n = 
IE
N
2p
2s
First IE N(7)
>
O(8)
The three electrons in the half-filled 2p subshell occupy three different orbitals (2p x
, 2p y
, 2p z
).
 repulsion between electrons is minimized .
131
IE
O

n = 
IE
P
3p
3s
First IE P(15)
>
S(16)
The three electrons in the half-filled 3p subshell occupy three different orbitals (3p x
, 3p y
, 3p z
).
 repulsion between electrons is minimized .
132
IE
S

The removal of an electron from O or S results in a half-filled p subshell with extra stability Misleading !!!
133

O(2,6)  O + (2,5)
134
2p
Is the 2p energy level of O + lower or higher than that of O ?
Electrons in O + experience a stronger attractive force from the nucleus.
Not because of the half-filled 2p subshell

O(2,6)  O + (2,5)
2p
As a whole, O + is always less stable than O.
It is because O(g) has one more electron than O + (g) and this extra electron has a negative potential energy.
135

Conclusions : -
(a) Each electron in an atom is described by a set of four quantum numbers.
(b) No two electrons in the same atom can have the same set of quantum numbers.
The quantum numbers can be obtained by solving the Schrodinger equation (p.16).
136

Quantum Numbers
1. Principal quantum number (n) n = 1, 2, 3,… related to the size and energy of the principal quantum shell .
E.g. n = 1 shell has the smallest size and electrons in it possess the lowest energy
137

138
Quantum Numbers
 = 0, 1, 2,…,n-1 related to the shape of the subshells .

139
Quantum Numbers





= 0, 1, 2,…,n-1
= 0  spherical s subshell
= 1  dumb-bell p subshell
= 2  d subshell
= 3  f subshell complicated shapes

Each principal quantum shell can have one or more subshells depending on the value of n.
If n = 1,
No. of subshell : 1
 name of subshell : 1s
0 to (1-1)  0
140

Each principal quantum shell can have one or more subshells depending on the value of n.
If n = 2,

No. of subshells : 2
0 to (2-1)  0, 1 names of subshells : 2s, 2p
141

Each principal quantum shell can have one or more subshells depending on the value of n.
If n = 3,

No. of subshells : 3
0 to (3-1)  0, 1, 2 names of subshells : 3s, 3p, 3d
142

Each principal quantum shell can have one or more subshells depending on the value of n.
If n = 4,

No. of subshells : 4
0 to (4-1)  0, 1, 2, 3 names of subshells : 4s, 4p, 4d, 4f
143

Quantum Numbers
3. Magnetic quantum number (m) m =   ,…0,…+  related to the spatial orientation of the orbitals in a magnetic field.
144

Each subshell consists of one or more orbitals depending on the value of 
Possible range of m :  0 to +0  0
No. of orbital : 1
Name of orbital : s
145

Each subshell consists of one or more orbitals depending on the value of 
Possible range of m :  1 to +1   1 ,0,+1
No. of orbitals : 3
Names of orbitals : p x
, p y
, p z
146

Each subshell consists of one or more orbitals depending on the value of 
Possible range of m :
 2 to +2   2,  1, 0, +1, +2
No. of orbitals : 5
Names of orbitals : d xy
, d xz
, d yz
, d x 2  y 2
, d z 2
147

Each subshell consists of one or more orbitals depending on the value of 
Possible range of m :
 3 to +3   3,  2,  1, 0, +1, +2, +3
No. of orbitals : 7
Names of orbitals : Not required in AL
148

Total no. of orbitals in a subshell
Energy of subshells : s < p < d < f
149

4. Spin quantum number (m s
) m s
=

1
2 or

1
2
They describe the spin property of the electron, either clockwise, or anti-clockwise
150

4. Spin quantum number (m s
)
Each orbital can accommodate a maximum of two electrons with opposite spins
151

Q.14(a)
Principal quantum shell n = 1 n = 2 n = 3 n = 4
Subshells
Total no. of orbitals
1s 1(1s)
2s, 2p 1+3=4
3s, 3p, 3d 1+3+5=9
4s, 4p,
4d, 4f
1+3+5+7=16
Total no. of electrons
2
8
18
32
152

14(b)
The total number of electrons in a principal quantum shell = 2n 2
153

Q.15(a)
The two electrons of helium are in the
1s orbital of the
1s subshell of the first principal quantum shell .
1 st electron : n = 1 , l = 0 , m = 0 , m s
2 nd electron : n = 1 , l = 0 , m = 0 , m s
=
=


1
2
1
2
154

Q.15(b)
There are only 2 electrons in the 3 rd quantum shell.
In the ground state, these two electrons should occupy the 3s subshell since electrons in it have the lowest energy.
155

Q.15(b)
The two outermost electrons of magnesium are in the
3s orbital of the
3s subshell of the third principal quantum shell .
1 st electron : n = 3 , l = 0 , m = 0 , m s
2 nd electron : n = 3 , l = 0 , m = 0 , m s
= 
= 
1
2
1
2
156

157
4.3

Models of the Atoms
1. Plum-pudding model by J.J. Thomson (1899)
2. Planetary(orbit)model by Niels Bohr (1913)
3. Orbital model by E. Schrodinger (1920s)
158

Electrons display both particle nature and wave nature.
Particle nature : mass, momemtum,…
Wave nature : frequency, wavelength, diffraction,…
159

(1) E
 h
 By Planck
(2) E
 mc
2 By Einstein mc
2  h

(3) mc
 h
 

 c
Particle property : momentum of a photon h
 Wave property
160

Evidence : photoelectric effect by
Albert Einstein (1905)
161

L. De Broglie (1924)
Nobel Prize Laureate in Physics,
1929 mc
 h


ν c
 h

(3) mv
 h

(4)
162

163 mc
 h
 mv
 h

(3)
(4)
Any particle (not only photon) in motion (with a momentum, mv) is associated with a wavelength

Q.16
mv
 h
λ
  h mv
Electron
λ 
6 .
63

10

34
9 .
1

10

31 
5 .
0

10
6

1 .
5

10

10 m
164

Q.16
mv
 h
λ
λ  h mv
Helium atom
λ 
6 .
63

10

34
6 .
6

10

27 
1 .
4

10
3

7 .
2

10

11 m
165

Q.16
mv
 h
λ
λ  h mv
λ 
100m world record holder
100
10.5
m m s  1
 9.52
s
6.63
86 Kg


10  34
10.5
Js ms 1
 7.3
 10  37 m
166

4.3
The Wave-mechanical model of the atom (p.95)
Wave nature of electrons (1927)
Evidence
 of electron
 inter-atomic spacing in metallic crystals (10  10 m)
167

For very massive ‘particles’,
The wavelength associated with them
(10  37 m) are much smaller than the dimensions of any physical system.
Wave properties cannot be observed.
168

 = L
 = L
 = 2L
Standing waves in a cavity
L
Standing waves only have certain allowable modes of vibration
Similarly, electrons as waves only have certain allowable energies.
169

The uncertainty principle
Heinsenberg
Nobel Prize Laureate in
Physics, 1932
170

171
The uncertainty principle
It is impossible to simultaneously determine the exact position and the exact momentum of an electron.
(
 x )(
 p ) uncertainty in position measurement
 h
4
 uncertainty in momentum measurement

The uncertainty principle
Small and light electron high energy photon
The momentum of electron would change greatly after collision
172

The uncertainty principle
Small and light electron low energy photon
The position of electron cannot be located accurately
173

The uncertainty principle high energy photon
No problem in macroscopic world !
174

Implications : -
The concept of well-defined orbits in Bohr’s model has to be abandoned.
We can only consider the probability of finding an electron of a ‘certain’ energy and momentum within a given space .
175

Schrodinger Equation
Nobel Prize Laureate in
Physics, 1933 de Broglie : electrons as waves
 Use wave functions (  ) to describe electrons
Pronounced as psi
176

Schrodinger Equation
Nobel Prize Laureate in
Physics, 1933
Heisenberg : Uncertainty principle
 Probability (  2 ) of finding electron at a certain position < 1 .
177

Schrodinger Equation
 2

 x
2

 2

 y
2

 2

 z
2

8

2 m
( E h
2

V )
 
0

: wave function m : mass of electron h : Planck’s constant
E : Total energy of electron
V : Potential energy of electron
178

Schrodinger Equation
 2

 x
2

 2

 y
2

 2

 z
2

8

2 m
( E h
2

V )
 
0
The equation can only be solved for certain
 i and E i d 2 sinx dx 2
  sinx
179

Schrodinger Equation
The wave function of an 1s electron is

( 1 s )


1
1
2


Z a
0


3
2 e

Zr a
0
Radius of 1s orbit of Bohr’s model
Z : nuclear charge ( Z = 1 for Hydrogen) a
0
: Bohr radius =
0.529Å (1Å = 10 
10 m) r : distance of electron from the nucleus
180

Schrodinger Equation
The allowed energies of H atom are given by
E

 me
4
Z
2
 2 n
2
8 h
0
2 n = 1, 2, 3,… n is the principal quantum number,
All other terms in the expression are constants
181

182
E

 me
4
Z
2
 2 n
2
8 h
0
2
E

0 n = 1, 2, 3,… refer to p.6 of notes
E

1 n
2

E
 hc

 hcR


1 n
1
2

1 n
2
2
E
2


E = E
2
– E
1

E
 hcR n
1
2
 hcR n
2
2 E
1




 hcR n
2
2 






 hcR n
1
2 



E

2
E
1 n = 2 n = 1
183

4.4
Atomic orbitals (p.98)
 2 is the probability of finding an electron at a particular point in space. (electron density)
 2 (1s)
Probability never becomes zero
 There is no limit to the size of an atom
184


2  as r  at the nucleus relative probability of finding the electron contour diagram
185

In practice, a boundary surface is chosen such that within which there is a high probability
(e.g. 90%) of finding the electron.
186

The electron spends
90% of time within the boundary surface
187

A 3-dimensional time exposure diagram.
The density of the dots represents the probability of finding the electron at that position.
188

The 3-dimensional region within which there is a high probability of finding an electron in an atom is called an atomic orbital .
Each atomic orbital is represented by a specific wave function(  ) .
The wave function of a specific atomic orbital describes the behaviour of the electron in the orbital.
189

dr is infinitesimally small
Total probability of finding the electron within the ‘shell’ of thickness dr
=

2 4
 r 2 dr electron density within the ‘shell’ total volume of the ‘shell’
 2 is the probability of finding the electron per unit volume
190


2
4
 r
2
 2  as r  , 4  r 2  as r 
 a maximum at 0.529 Å
Orbital Model
191

192
1
Bohr’s Orbit Model
0.529
Å r (Å)

Total probability of finding the electron within the ‘shell’ of thickness dr
=

2 4
 r 2 dr
193
The sum of the probabilities of finding the electron within all ‘shells’ = 1
 r r  
 0
2
r
2



2
4
 r
2
 r r  
 0
2
r
2

The total area bounded by the curve and the x-axis = 1
Check Point 4-3
194

4.4
Atomic orbitals (p.98)
s Orbitals
195

nodal surface
There is no chance of finding the electron on the nodal surface.
196

197

198
Probabilities of finding the electron at
A or B > 0
Probability of finding the electron at C = 0
A
C
B
How can the electron move between
A & B ?

199
 can be considered as the amplitude of the wave.
 2 is always  0

200

2 = 0

4.4
Atomic orbitals (p.100)
p Orbitals
201
Two lobes along an axis

4.4
Atomic orbitals (p.100)
For each 2p orbital, there is a nodal plane on which the probability of finding the electron is zero.
yz plane xz plane xy plane
202

4.4
Atomic orbitals (p.101)
d Orbitals Four lobes between two axes
Four lobes along two axes
203
Two lobes & one belt

http://www.orbitals.com/orb/orbtable.htm#table1
204

205
The END

4.1 The electromagnetic spectrum (SB p.82)
206
Some insects, such as bees, can see light of shorter wavelengths than humans can. What kind of radiation do you think a bee sees?
Answer
Ultraviolet radiation
Back

4.1 The electromagnetic spectrum (SB p.87)
207
What does the convergence limit in the Balmer series correspond to?
Answer
The convergence limit in the Balmer series corresponds to the electronic transition from n =
 to n = 2.
Back

4.1 The electromagnetic spectrum (SB p.88)
Given the frequency of the convergence limit of the Lyman series of hydrogen, find the ionization enthalpy of hydrogen.
Frequency of the convergence limit = 3.29  10 15 Hz
Planck constant = 6.626  10 -34 J s
Avogadro constant = 6.02  10 23 mol -1 Answer
208

4.1 The electromagnetic spectrum (SB p.88)
Back
For one hydrogen atom,
E = h

= 6.626

10 -34 J s

3.29

10 15 s -1
= 2.18

10 -18 J
For one mole of hydrogen atoms,
E = 2.18

10 -18 J

6.02

10 23 mol -1
= 1312360 J mol -1
= 1312 kJ mol -1
The ionization enthalpy of hydrogen is 1312 kJ mol -1 .
209

4.1 The electromagnetic spectrum (SB p.88)
The emission spectrum of atomic sodium is studied. The wavelength of the convergence limit corresponding to the ionization of a sodium atom is found. Based on this wavelength, find the ionization enthalpy of sodium.
Wavelength of the convergence limit = 242 nm
Planck constant = 6.626  10 -34 J s
Avogadro constant = 6.02  10 23 mol -1
Speed of light = 3  108 m s -1
1 nm = 10 -9 m
Answer
210

4.1 The electromagnetic spectrum (SB p.88)
Back
For one mole of sodium atoms,
E = h

L h c L
=
=

6.626

10

34 J s

3

10 8 m s -1

242

10 9 m
= 494486 J mol -1
6.02

10 23 mol

1
= 494 kJ mol -1
The ionization enthalpy of sodium of 494 kJ mol -1 .
211

4.2
Deduction of electronic structure from ionization enthalpies (p.94)
(a) Given the successive ionization enthalpies of boron, plot a graph of the logarithm of successive ionization enthalpies of boron against the number of electrons removed. Comment on the graph obtained.
Successive I.E. (in kJ mol-1): 800, 2400, 3700, 25000, 32800
Answer
212

4.2
Deduction of electronic structure from ionization enthalpies (p.94)
(a)
213
The first three electrons of boron are easier to be removed because they are in the outermost shell of the atom. As the fourth and fifth electrons are in the inner shell, a larger amount of energy is required to remove them.

4.2
Deduction of electronic structure from ionization enthalpies (p.94)
(b) Give a sketch of the logarithm of successive ionization enthalpies of potassium against no. of electrons removed.
Explain your sketch.
Answer
214

4.2
Deduction of electronic structure from ionization enthalpies (p.94)
(b)
There are altogether 19 electrons in a potassium atom. They are in four different energy levels. The first electron is removed from the shell of the highest energy level which is the farthest from the nucleus, I.e. the fourth (outermost) shell. It is the most easiest to be removed. The second to ninth electrons are removed from the third shell, and the next eight electrons are removed from the second shell.
The last two electrons with highest ionization enthalpy are removed from the first (innermost) shell of the atom. They are the most difficult to be removed.
215

4.2
Deduction of electronic structure from ionization enthalpies (p.94)
(c) There is always a drastic increase in ionization enthalpy whenever electrons are removed from a completely filled electron shell. Explain briefly.
Answer
(c) A completely filled electron shell has extra stability. Once an electron is removed, the stable electronic configuration will be destroyed.
Therefore, a larger amount of energy is required to remove an electron from such a stable electronic configuration.
Back
216

4.3
The Wave-mechanical model of the atom (p.97)
Back
(a) What are the limitations of Bohr’s atomic model?
(a) It cannot explain the more complicated spectral lines observed in emission spectra other than that of atomic hydrogen. There is no experimental evidence to prove that electrons are moving around the nucleus in fixed orbits.
(b) Explain the term “dual nature of electrons”.
(b) Electrons can behave either as particles or a wave.
(c) For principal quantum number 4, how many sub-shells are present? What are their symbols?
(c) When n = 4, l = 0, 1, 2 and 3, there are 4 sub-shells. The symbols are 4 s , 4 p , 4 d and 4 f respectively.
Answer
217

4.4
Atomic orbitals (p.101)
(a) Distinguish between the terms orbit and orbital.
(a) “Orbit” is the track or path where an electron is revolving around the nucleus. “Orbital” is a region of space in which the probability of finding an electron is very high (about 90 %).
(b) Sketch the pictorial representations of an s orbital and a p orbital. What shapes are they?
(b) s orbital is spherical in shape whereas p orbital is dumb-bell in shape.
Answer
218

4.4
Atomic orbitals (p.101)
Back
(c) How do the 1 s and 2 s
orbitals differ from each other?
(c) Both 1s and 2s orbitals are spherical in shape, but the 2s orbital consists of a region of zero probability of finding the electron known as a nodal surface.
(d) How do the 2 p orbitals differ from each other?
(d) There are three types of p orbitals. All are dumb-bell in shape. They are aligned in three different spatial orientations designated as x , y and z . Hence, the 2 p orbitals are designated as 2 p x
, 2 p y and 2 p z
.
Answer
219