Transcript The Electronic Structure of Atoms

4
1
The Electronic
Structure of Atoms
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
Atomic Emission Spectra
原子放射光譜
Spectra  plural of Spectrum
Arises from light emitted from
individual atoms
3
The electromagnetic spectrum
Wavelength (m) 
 Frequency (Hz / s1)
Speed of light (ms1) = Frequency  Wavelength
c =   3108 ms1
4
The electromagnetic spectrum
red
violet
 Frequency (Hz / s1)
5
Wavelength (m) 
Visible light
The electromagnetic spectrum
red
violet
 Frequency (Hz / s1)
 Increasing energy
6
Wavelength (m) 
Ultraviolet light
The electromagnetic spectrum
red
violet
 Frequency (Hz / s1)
 Increasing energy
7
Wavelength (m) 
X-rays
The electromagnetic spectrum
red
violet
 Frequency (Hz / s1)
 Increasing energy
8
Wavelength (m) 
Gamma rays
The electromagnetic spectrum
red
violet
 Frequency (Hz / s1)
Decreasing energy 
9
Wavelength (m) 
Infra-red light
The electromagnetic spectrum
red
violet
 Frequency (Hz / s1)
Decreasing energy 
10
Wavelength (m) 
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.
11
Spectra from excited samples in
discharge tubes
4.1 The electromagnetic spectrum (SB p.82)
Continuous spectrum of white light
Fig.4-5(a)
12
4.1 The electromagnetic spectrum (SB p.83)
Line spectrum of hydrogen
Fig.4-5(b)
13
How Do Atoms Emit Light ?
H2(g)
Electric discharge
Or Heating
H(g)
Hydrogen atom in ground state
means
its electron has the lowest energy
14
Atoms in excited state
Energy
Atom in
ground state
H(g)
Ground
15
Electric discharge
Or Heating
H*(g)
Excited
Atoms in excited state
Not Stable
Energy
h
Atom in
Atom returns to
ground state ground state
h
H*(g)
H(g)
Excited
Ground
16
Atoms in excited state
Energy
h
Atom in
Atom returns to
ground state ground state
E = h
Planck’s equation
17
Atoms in excited state
Energy
h
Atom in
Atom returns to
ground state ground state
E = h
Planck : Nobel laureate Physics, 1918
18
Atoms in excited state
E2
Energy
E1
h
Atom in
Atom returns to
ground state ground state
E = h
E = energy of the emitted light = E2 – E1
19
Atoms in excited state
E2
Energy
E1
h
Atom in
Atom returns to
ground state ground state
E = h
 = Frequency of the emitted light
20
Atoms in excited state
E2
Energy
E1
h
Atom in
Atom returns to
ground state ground state
E = h
h (Planck’s constant) = 6.63  1034 Js
21
Atoms in excited state
E2
Energy
E1
h
Atom in
Atom returns to
ground state 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
E2
Energy
E1
h
Atom in
Atom returns to
ground state 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
Convergence
ultraviolet
limit
24
violet
red
Q.1
E  hν
hc
E
λ
c
λ
ν

6.63 10


Js 3.00 10 ms
9
656.3 10 m
34
= 3.031019 J
25
8
1

En 3  En 2
 3.03 10
19
J
energy of one
photon emitted
26
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
n  n 2
corresponds to
the last spectral
line of the Balmer
series
29
4.1 The electromagnetic spectrum (SB p.83)
The Complete Hydrogen Emission Spectrum
UV
30
Visible
IR
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
Rydberg Equation
1
1 
 1
 RH  2  2 
λ
a b 
Relates wavelength of the emitted light of
hydrogen atom with the electron transition
32
1
1 
 1
 RH  2  2 

a b 
1
λ
= wave number of the emitted light
= number of waves in a unit length
1
e.g.  100 m 1  100 waves in 1 meter

1
 ν
λ
33
1
1 
 1
 RH  2  2 

a b 
1 
 
λ c
1
 
λ
34
1
1 
 1
 RH  2  2 
λ
a b 
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
35
1
1 
 1
 RH  2  2 

a b 
Balmer series,
a=2
b = 3, 4, 5,…
36
1
1 
 1
 RH  2  2 

a b 
Lyman series,
a=1
b = 2, 3, 4,…
2
37
3
4 5 6
1
1 
 1
 RH  2  2 

a b 
Paschen series,
a=3
b = 4, 5, 6,…
4
38
5
6 7 
1
1 
 1
 RH  2  2 

a b 
RH is the Rydberg constant
39
Q.4
1

1.52
2.06
2.30
2.44
2.52
2.57
2.60
2.74
(106 m1)
1
2
b
40
0.111 0.063 0.040 0.028 0.020 0.016 0.012 0.000
1
λ
106 m1
y-intercept
1 
 1
1
 2.74 10 m  RH  2  2   RH  
2  
4
6
1
RH = 1.096107 m1
1
2
b
41
4.1 The electromagnetic spectrum (SB p.84)
Interpretation of the atomic hydrogen
spectrum
Discrete spectral lines

42
energy possessed by electrons
within hydrogen atoms cannot be
in any arbitrary quantities but
only in specified amounts called
quanta.
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
Bohr’s model of H atom
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.
50
E2
n = 2
E1
n = 1
Interpretation of the atomic hydrogen
spectrum
For the electron transition E2  E1,
the energy emitted is related to the frequency
of light emitted by the Plank’s equation :
E2
n = 2
E = E2 – E1 = h
E1
51
n = 1
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
Electronic
Transition
53
Wavelength Predicted Wavelength Determined
by Bohr (nm)
by Experiment (nm)
2  1
121.6
121.7
3  1
102.6
102.6
4  1
97.28
97.32
3  2
656.6
656.7
4  2
486.5
486.1
5  2
434.3
434.1
4  3
1876
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
E2
By Planck’s equation,
E  h 
hc
E = E2 – E1

E1
55
n = 2
n = 1
E  h 
hc

E2
By Rydberg’s equation,
 1
1 
 R 2  2 

 n1 n2 
n = 2
E = E2 – E1
1
E1
 1
1 
1
 E  hc   hcR 2  2 

 n1 n2 
56
n = 1
E2
 1
1 
E
 hcR 2  2 

 n1 n2 
n = 2
hc
hcR hcR
E 2  2
n1
n2
E = E2 – E1
E1
 hcR   hcR 
   2     2   E2  E1
 n2   n1 
57
n = 1
All electric potential energies have
negative signs except E
E1 < E2 < E3 < E4 < E5 < …… < E = 0
All are negative
58
Balmer series,
Transitions from higher levels to n = 2
1 1
E  hcR 2  2 
2 n 
Visible region
59
n = 3, 4, 5, …
Lyman series,
Transitions from higher levels to n = 1
1 1 
E  hcR 2  2 
1 n 
n = 2, 3, 4,…
More energy released
 Ultraviolet region
60
Paschen series,
Transitions from higher levels to n = 3
1 1
E  hcR 2  2 
3 n 
n = 4, 5, 6,…
Less energy released
 Infra-red region
61
4.1 The electromagnetic spectrum (SB p.87)
E = E2 – E1 = h
IR
UV
62
visible
Q.5 n = 100  n = 99
Energy of the light emitted is
extremely small.
1  1
 1
 1.09610  2 
m
2 

 99 100 
1
7
   4.5 10 m
2
63
The electromagnetic spectrum
Wavelength (m) 
 Frequency (Hz / s1)
64
Microwave
Q.5 n = 100  n = 99
Energy of the light emitted is
extremely small.
1  1
 1
 1.09610  2 
m
2 

 99 100 
1
7
   4.5 10 m
2
 Microwave region
65
Q.6
66
Energy of the first line
 E32  E31  E21
n=3
n=2
n=1
67
Energy of the first line
 E32  E31  E21
h 32  h 31  h 21
 32   31  21  292.3  246.61013 Hz
= 45.71013 Hz
68
Energy of the second line
 E42  E41  E21
n=4
n=2
n=1
69
Energy of the second line
 E42  E41  E21
h 42  h 41  h 21
 42   41  21  308.3  246.61013 Hz
= 61.71013 Hz
70
Energy of the third line
 E52  E51  E21
n=5
n=2
n=1
71
Energy of the third line
 E52  E51  E21
h 52  h 51  h 21
 52   51  21  315.7  246.61013 Hz
= 69.11013 Hz
72
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
The frequency at which the
spectral lines of a series merge.
74
Q.7
n=1
n=
H(g)  H+(g) + e
Ionization enthalpy
 E1  E1  h 1
Convergence limit of Lyman series
75
Atom
Energy Level of e to be
removed in ground state of
atom
Electron transition for
ionization of atom
H
n=1
n=1n=
He
Li
Na
76
n=1
n=2
n=3
Lyman series ?
n=1n=
Balmer series ?
n=2n=
Paschen series ?
n=3n=
Q.9
Ionization enthalpy of helium
 E1  E1  h 1
= (6.6261034 Js)(5.291015 s1)(6.021023 mol1)
= 2110 KJ mol1
> Ionization enthaply of hydrogen
77
E
H
He
Relative positions of energy levels depend on
the nuclear charge of the atom.
E2
E2’
E1
E1’
Ionization enthalpy : He > H
78
4.1 The electromagnetic spectrum (SB p.89)
The uniqueness of atomic emission spectra
No two elements have identical atomic spectra
79
4.1 The electromagnetic spectrum (SB p.89)
The uniqueness of atomic emission spectra
atomic spectra can be used to identify unknown
elements.
80
Flame Test
NaCl(s)
heat
atomization
Na(g) + Cl(g)
Na*(g) + Cl*(g)
81
heat
Na(g) + Cl(g)
Na*(g) + Cl*(g)
hNa + hCl
Na(g) + Cl(g)
Flame Test
The most intense yellow light is observed.
82
Q.10
hc
E  hν 

(6.626 10 Js)(3.00 10 ms )

9
450  10 m
34
= 4.421019 J
83
8
1
Emission vs Absorption
Emission spectrum of hydrogen
visible
(nm)
Bright lines against a dark background
Absorption spectrum of hydrogen
Dark lines against a bright background
84
visible
(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
Atomic Absorption Spectra
redshift
Moving at
higher speed
Moving at
lower speed
88
Redshift
Left : Absorption spectrum
from sunlight.
Right : Absorption spectrum
from a supercluster of
distant galaxies
89
Atomic Absorption Spectra
Only
atomic emission spectrum of hydrogen
is required in A-level syllabus !!!
90
4.2
Deduction of
Electronic Structure
from Ionization
Enthalpies
91
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)
+ e
IE1
Be+(g)  Be2+(g) + e
IE2
Be2+(g)  Be3+(g) + e
IE3
Be3+(g)  Be4+(g) + e
IE4
96
 Be+(g)
Q.11(b)
IE1 < IE2 < IE3 < IE4
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)
IE1
IE2
IE3
IE4
(kJ mol1) (kJ mol1) (kJ mol1) (kJ mol1)
900
98
1758
14905
21060
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
Electron Diagram of Beryllium
Second Shell
First Shell
101
Energy Level Diagram of Beryllium
n =2
n= 1
102
E
Be
IE1 = E - E2
E2
E1
103
E
Be
Be+
IE1
IE2
E2
E2’
E1
E 1’
IE1 < IE2
104
E
Be
Be+
IE1
Be2+
IE2
E2
E2’
E2’’
E1
E 1’
E1’’
IE1 < IE2 << IE3
105
IE3
E
Be
Be+
IE1
Be2+
IE2
Be3+
IE3
IE4
E2
E2’
E2’’
E2’’’
E1
E 1’
E1’’
E1’’’
IE1 < IE2 << IE3 < IE4
106
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
Q.12
4 groups of electrons
111
Which group of electrons
is in the first shell ?
n=3
n=4
112
n=1
n=2
Q.12
2
8
8
2
Q.12
2, 8, 8, 2
113
n=4
n=3
n=2
n=1
114
Q.12
4.2 Deduction of electronic structure from ionization enthalpies (p.91)
Evidence of the Existence of Shells
2, 8, 1
115
Na
n=3
n=2
n=1
116
Variation of IE1 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 IE1 with atomic number
across Periods 2 and 3.
13.(a)
119
2–3-3
120
13.(a)
2. IE1 value : Group 2 > Group 3
3. IE1 value : Group 5 > Group 6
121
13.(a)
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
2,5
2,7
2,8,7
2,8,5
2,2
2,8,2
2,4 2,6
2,3
2,1
123
2,8,1
2,8,4
2,8,3
2,8,6
2,8,8
IE1 : 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
13.(b)
n=
2p
IE2s
IE2p
2s
IE2s(Be) > IE2p(B)
1s
125
IE1 : Be > B
IE1 : 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
13.(b)
n=
3p
3s
IE3s
IE3p
IE3s(Mg) > IE3p(Al)
IE1 : Mg > Al
2p
127
n=
IE3p
IE3s
3p
3s
2p
Mg(12)
128
Al(13)
IE1 : 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=
IEO
IEN
2p
2s
First IE
130
N(7)
>
O(8)
n=
IEO
IEN
2p
2s
First IE
N(7)
>
O(8)
The three electrons in the half-filled 2p subshell
occupy three different orbitals (2px , 2py , 2pz).
 repulsion between electrons is minimized.
131
n=
IES
IEP
3p
3s
First IE
P(15)
>
S(16)
The three electrons in the half-filled 3p subshell
occupy three different orbitals (3px , 3py , 3pz).
 repulsion between electrons is minimized.
132
The removal of an electron from O or S
results in a half-filled p subshell with extra
stability.(p.28, part 3)
Misleading !!!
133
O(2,6)

O+(2,5)
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
134
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
Quantum Numbers
2. Subsidiary quantum number ()
 = 0, 1, 2,…,n-1
related to the shape of the subshells.
138
Quantum Numbers
2.

Subsidiary quantum number ( )
 = 0, 1, 2,…,n-1
 = 0  spherical s subshell
 = 1  dumb-bell p subshell
 = 2  d subshell complicated
shapes
 = 3  f subshell
139
Each principal quantum shell can have
one or more subshells depending on
the value of n.
If n = 1,

possible range of : 0 to (1-1)  0
No. of subshell : 1
name of subshell : 1s
140
Each principal quantum shell can have
one or more subshells depending on
the value of n.
If n = 2,
possible range of
 : 0 to (2-1)  0, 1
No. of subshells : 2
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,
possible range of
 : 0 to (3-1)  0, 1, 2
No. of subshells : 3
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,
possible range of
 : 0 to (4-1)  0, 1, 2, 3
No. of subshells : 4
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 
If  = 0 (s subshell),
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 
If  = 1 (p subshell),
Possible range of m : 1 to +1  1 ,0,+1
No. of orbitals : 3
Names of orbitals : px, py, pz
146
Each subshell consists of one or more
orbitals depending on the value of 
If  = 2 (d subshell),
Possible range of m :
2 to +2  2, 1, 0, +1, +2
No. of orbitals : 5
Names of orbitals : dxy, dxz, dyz , dx2 y2 , dz2
147
Each subshell consists of one or more
orbitals depending on the value of 
If  = 3 (f subshell),
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
= 2 + 1
Energy of subshells : s<p<d<f
149
4. Spin quantum number (ms)
1
1
ms =  or 
2
2
They describe the spin property of
the electron, either clockwise, or
anti-clockwise
150
4. Spin quantum number (ms)
Each orbital can accommodate a maximum
of two electrons with opposite spins
151
Q.14(a)
Principal
quantum
shell
152
Subshells
Total no.
of orbitals
Total no.
of
electrons
n=1
1s
1(1s)
2
n=2
2s, 2p
1+3=4
8
n=3
3s, 3p, 3d
1+3+5=9
18
n=4
4s, 4p,
4d, 4f
1+3+5+7=16
32
14(b)
The total number of electrons in a
principal quantum shell = 2n2
153
Q.15(a)
The two electrons of helium are in the
1s orbital of the
1s subshell of the
first principal quantum shell.
1st
1
electron : n = 1, l = 0, m = 0, ms = 
2
2nd
1
electron : n = 1, l = 0, m = 0, ms = 
2
154
Q.15(b)
There are only 2 electrons in the 3rd
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.
1st
1
electron : n = 3, l = 0, m = 0, ms = 
2
2nd
1
electron : n = 3, l = 0, m = 0, ms = 
2
156
4.3
The Wavemechanical Model
of the Atom
157
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
Wave as particles : photons
E  h By Planck
2
(2) E  m c By Einstein
2
m c  h
  h
(3) m c  h  
 c   Wave property
(1)
Particle property :
momentum of a photon
160
Evidence :
photoelectric effect by
Albert Einstein (1905)
Nobel Prize Laureate in Physics,
1921
161
Particles as waves
L. De Broglie (1924)
Nobel Prize Laureate in Physics,
1929
 ν h
mc  h  
c 
h
(4)
mv 

162
(3)
h
mc 
(3)

h
(4)
mv 

Any particle (not only photon) in
motion (with a momentum, mv) is
associated with a wavelength
163
Q.16
h
mv 
λ
h

mv
Electron
34
6.63 10
10
λ

1
.
5

10
m
31
6
9.1  10  5.0  10
164
Q.16
h
mv 
λ
h
λ
mv
Helium atom
34
6.63 10
11
λ

7
.
2

10
m
27
3
6.6  10  1.4  10
165
Q.16
h
mv 
λ
h
λ
mv
100m world record holder
100 m
 9.52 s
1
10.5 m s
6.63 10 Js
37
λ
 7.3 10 m
-1
86 Kg 10.5 ms
34
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10m)
167
For very massive ‘particles’,
The wavelength associated with them
(1037m) are much smaller than the
dimensions of any physical system.
Wave properties cannot be observed.
168
2
= L
3
=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
The uncertainty principle
It is impossible to simultaneously
determine the exact position and
the exact momentum of an electron.
h
(x)( p ) 
4
uncertainty in position uncertainty in
measurement
momentum measurement
171
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
      8 m
 2  2  2 ( E  V )  0
2
x
y
z
h
2
2
2
2
 : wave function
m : mass of electron
h : Planck’s constant
E : Total energy of electron
V : Potential energy of electron
178
Schrodinger Equation
      8 m
 2  2  2 ( E  V )  0
2
x
y
z
h
2
2
2
2
The equation can only be solved for
certain i and Ei
d2sinx
 sinx
2
dx
179
Schrodinger Equation
The wave function of an 1s electron is
3
1 Z
 (1s)  12   e
  a0 
2
 Zr
a0
Radius of 1s orbit
of Bohr’s model
Z : nuclear charge (Z = 1 for Hydrogen)
a0 : Bohr radius = 0.529Å (1Å = 1010m)
r : distance of electron from the nucleus
180
Schrodinger Equation
The allowed energies of H atom are given by
 me Z
E
2 2 2
8 0 n h
4
2
n = 1, 2, 3,…
n is the principal quantum number,
All other terms in the expression are
constants
181
 me Z
E
2 2 2
8 0 n h
4
2
n = 1, 2, 3,…
E0
1
E 2
n
182
refer to p.6 of notes
E2
 1
1 
E
 hcR 2  2 

 n1 n2 
n = 2
hc
hcR hcR
E 2  2
n1
n2
E = E2 – E1
E1
 hcR   hcR 
   2     2   E2  E1
 n2   n1 
183
n = 1
4.4 Atomic orbitals (p.98)
2is the probability of finding an electron at a
particular point in space. (electron density)
2(1s)
Probability never becomes zero

184
There is no limit to the size
of an atom
relative probability of
finding the electron
  as r 
2
at the
nucleus
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
= 24r2dr
electron density
within the ‘shell’
total volume of
the ‘shell’
2 is the probability of finding the
electron per unit volume
190
 4r
2
2
2 as r  , 4r2  as r 
 a maximum at 0.529 Å
Orbital Model
191
1
probability
Bohr’s Orbit Model
0.529Å
192
r (Å)
Total probability of
finding the electron within
the ‘shell’ of thickness dr
= 24r2dr
The sum of the probabilities of finding
the electron within all ‘shells’ = 1
r
2
2

193
r0
ψ 4πr dr  1
 4r
2
2

r
r0
ψ 4πr dr  1
2
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
196
There is
no chance of
finding the
electron on the
nodal surface.
197
Probabilities
of finding the
electron at
A or B > 0
A
C
B
Probability of
finding the
electron at C = 0
How can the electron move between
A&B?
198
 can be considered as the amplitude
of the wave.
2 is always  0
199
2

200
=0
4.4 Atomic orbitals (p.100)
p Orbitals
Two lobes along an axis
201
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
202
xz plane
xy plane
4.4 Atomic orbitals (p.101)
Four lobes
between two axes
d Orbitals
Four lobes
along two axes
203
Two lobes &
one belt
Grand Orbital Table
http://www.orbitals.com/orb/orbtable.htm#table1
204
The END
205
4.1 The electromagnetic spectrum (SB p.82)
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
206
4.1 The electromagnetic spectrum (SB p.87)
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
207
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  1015 Hz
Planck constant = 6.626  10-34 J s
Avogadro constant = 6.02  1023 mol-1
208
Answer
4.1 The electromagnetic spectrum (SB p.88)
Back
For one hydrogen atom,
E = h
= 6.626  10-34 J s  3.29  1015 s-1
= 2.18  10-18 J
For one mole of hydrogen atoms,
E = 2.18  10-18 J  6.02  1023 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  1023 mol-1
Speed of light = 3  108 m s-1
1 nm = 10-9 m
210
Answer
4.1 The electromagnetic spectrum (SB p.88)
Back
For one mole of sodium atoms,
E = hL
hcL
=

34
8
-1
23
1
6.626

10
J
s

3

10
m
s

6.02

10
mol
=
-9
242

10
m
-1
= 494486 J mol
= 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)
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.
213
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)
215
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.
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 4s, 4p, 4d and 4f respectively.
217
Answer
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 1s and 2s 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 2p 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 2p orbitals are designated as 2px, 2py and 2pz.
Answer
219