1 the electronic structure of atoms 4.1the electromagnetic spectrum 4.2deduction of electronic...

1

The Electronic The Electronic Structure of AtomsStructure of Atoms

4.14.1 The Electromagnetic SpectrumThe Electromagnetic Spectrum

4.24.2 Deduction of Electronic Structure Deduction of Electronic Structure from Ionization Enthalpiesfrom Ionization Enthalpies

4.34.3 The Wave-mechanical Model of the The Wave-mechanical Model of the AtomAtom

4.4 Atomic Orbitals4.4 Atomic Orbitals

44

2

The electronic structure of atomsThe electronic structure of atoms Chapter 4 The electronic structure of atoms (SB p.80)

Two sources of evidence : -

(a) Study of atomic emission spectra of the elements

(b) Study of ionization enthalpies of the elements

3

Atomic Emission Spectra原子放射光譜

Spectra plural of Spectrum

Arises from light emitted from individual atoms

4

Frequency (Hz / s1)

Wavelength (m)

The electromagnetic spectrumThe electromagnetic spectrum

Speed of light (ms1) = Frequency Wavelength

c = 3108 ms1

5

Frequency (Hz / s1)

Wavelength (m)


red

violet

Visible light

6

Frequency (Hz / s1)

Wavelength (m)


red

violet

Ultraviolet light

Increasing energy

7

Frequency (Hz / s1)

Wavelength (m)


red

violet

X-rays Increasing energy

8

Frequency (Hz / s1)

Wavelength (m)


red

violet

Gamma rays Increasing energy

9

Frequency (Hz / s1)

Wavelength (m)

Decreasing energy


red

violet

Infra-red light

10

Frequency (Hz / s1)

Wavelength (m)


red

violet

Microwave & radio waves

Decreasing energy

11

Types of Emission SpectraTypes of Emission Spectra4.1 The electromagnetic spectrum (SB p.82)

1. Continuous spectra

E.g. Spectra from tungsten filament and sunlight

2. Line Spectra

E.g. Spectra from excited samples in discharge tubes

12

Continuous spectrum of white Continuous spectrum of white lightlight

Fig.4-5(a)

4.1 The electromagnetic spectrum (SB p.82)

13

Line spectrum of hydrogenLine spectrum of hydrogen

Fig.4-5(b)


14

How Do Atoms Emit Light ?

H2(g)Electric discharge

Or HeatingH(g)

Hydrogen atom in ground state

means

its electron has the lowest energy

15

Energy

Atom in ground state

H(g)Electric discharge

Or HeatingH*(g)

Excited Ground

Atoms in excited state

16

H*(g) H(g)Ground Excited

Energy



Not Stable

h

Atom returns to ground state

h

17

Energy




h

E = hPlanck’s equation

18

Energy




h

E = hPlanck : Nobel laureate Physics,

1918

19

Energy




h

E = hE = energy of the emitted light = E2 –

E1

E2

E1

20

Energy




h

E = h = Frequency of the emitted light

E2

E1

21

Energy




h

E = hh (Planck’s constant) = 6.63 1034 Js

E2

E1

22

Energy




h

E = h

h = 6.63 1034 Js

E2

E1

Energy cannot be absorbed or emitted by an atom in any arbitrary amount.

23

Energy




h

E = h

h = 6.63 1034 Js

E2

E1

Energy can only be absorbed or emitted by an atom in multiples of 6.631034 J.

24


Characteristic Features of the Characteristic Features of the Hydrogen Emission Line Hydrogen Emission Line Spectrum Spectrum 1. The visible region – The Balmer Series

green

red

violetultraviole

t

Convergence limit

25

Q.1

νhE ν

λc

λ

hcE

= 3.031019 J

m 10656.3

ms 103.00Js 106.639

1834

26

J103.03

E E19

2n3n

energy of one photon emitted

27

Q.2

(a) The spectral lines come closer at higher frequen

cy and eventually merge into a c

ontinuum(連續體 )

(b) n = n = 2

28

(c) The electron has been removed

from the atom. I.e. the atom

has been ionized.

Q.2e1

H(g) H+(g) + e

29

2nn corresponds to the last spectral line of the Balmer series

30

The Complete Hydrogen Emission The Complete Hydrogen Emission SpectrumSpectrum

UV

Visible IR


31

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.

32

Rydberg Equation

22H b

1a1

Rλ

1

Relates wavelength of the emitted light of hydrogen atom with the electron transition

33

22H b

1a1

R1

λ

1= wave number of the emitted light= number of waves in a unit length

e.g. 1m 100 1

100 waves in 1 meter

νλ

1

34

c

λ

1

22H b

1a1

R1

λ

1

35

22H b

1a1

Rλ

1

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

22H b

1a1

R1

Balmer series,

a = 2

b = 3, 4, 5,…

37

22H b

1a1

R1

Lyman series,

a = 1

b = 2, 3, 4,…

2 3 4 5 6

38

22H b

1a1

R1

Paschen series,

a = 3

b = 4, 5, 6,…

4 5 6 7

39

22H b

1a1

R1

RH is the Rydberg constant

40

Q.4

0.000

0.012

0.016

0.020

0.028

0.040

0.063

0.111

2.742.602.572.522.442.302.061.52(106

m1)

1

2b1

41

2b1

λ

1

106 m1

16m102.74

41

R1

21

R H22H

y-intercept

RH = 1.096107 m1

42

Interpretation of the atomic Interpretation of the atomic hydrogen spectrumhydrogen spectrum


Discrete spectral lines

energy possessed by electrons within hydrogen atoms cannot

be in any arbitrary quantities but only in specified amounts

called quanta.

43



Only certain energy levels are allowed for the electron in a hydrogen atom.

The energy of the electron in a hydrogen atom is quantized.

44



45

Niels Bohr (1885-1962)


Bohr’s Atomic Model of Hydrogen

Nobel Prize Laureate in Physics, 1922

46



for his services in the investigation of the structure of atoms and of the radiation emanating from them

47


Bohr’s model of H atom

48


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.

49


2. Different orbits have different energy levels.

An orbit with higher energy is further away from the nucleus.

50


3. Spectral lines arise from electron transitions from higher orbits to lower orbits.

E2

E1

n = 2

n = 1

51


For the electron transition E2 E1,

E = E2 – E1 = h

E2

E1

n = 2

n = 1

the energy emitted is related to the frequencyof light emitted by the Plank’s equation :

52


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.

53

4 3

5 2

4 2

3 2

4 1

3 1

1876

434.1

486.1

656.7

97.32

102.6

121.7

1876

434.3

486.5

656.6

97.28

102.6

121.62 1

Wavelength Determined by Experiment (nm)

Wavelength Predicted by Bohr (nm)

Electronic Transition

54


5. The theory failed when applied to elements other than hydrogen (multi-electron systems)

55

Illustrating Bohr’s Theory

First line of Lyman series, n = 2 n = 1

E = E2 – E1

E2

E1

n = 2

n = 1

By Planck’s equation,

hc

hE

56

E = E2 – E1

E2

E1

n = 2

n = 1

By Rydberg’s equation,

hc

hE

2

22

1

111

nnR

1

hcE

2

22

1

11

nnhcR

57

2

22

1

11

nnhcR

hcE

E = E2 – E1

E2

E1

n = 2

n = 122

21 n

hcR

n

hcRE

2

12

2 n

hcR

n

hcR12 EE

58

All electric potential energies have negative signs except E

E1 < E2 < E3 < E4 < E5 < …… < E = 0

All are negative

59

Balmer series,

Transitions from higher levels to n = 2

22

1

2

1

nhcRE n = 3, 4, 5, …

Visible region

60

Lyman series,


22

1

1

1

nhcRE n = 2, 3, 4,…

More energy released

Ultraviolet region

61

Paschen series,


22

1

3

1

nhcRE n = 4, 5, 6,…

Less energy released

Infra-red region

62

E = E2 – E1 = h


UVvisible

IR

63

Q.5 n = 100 n = 99

122

7

100

1

99

110096.1

1

m

m2105.4

Energy of the light emitted is extremely small.

64

Frequency (Hz / s1)

Wavelength (m)


Microwave

65

Q.5 n = 100 n = 99

122

7

100

1

99

110096.1

1

m

m2105.4

Microwave region

Energy of the light emitted is extremely small.

66

Q.6

67

121323 EEE

Energy of the first line n = 3

n = 2

n = 1

68

121323 EEE

Energy of the first line

121323 hhh

121323 Hz13106.2463.292

= 45.71013 Hz

69

121424 EEE

Energy of the second line n = 4

n = 2

n = 1

70

121424 EEE

121424 hhh

121424 Hz13106.2463.308

= 61.71013 Hz

Energy of the second line

71

121525 EEE

n = 5

n = 2

n = 1

Energy of the third line

72

121525 EEE

121525 hhh

121525 Hz13106.2467.315

= 69.11013 Hz

Energy of the third line

73

Convergence Limits and Ionization EnthalpiesIonization 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 mol1

74

Convergence Limits and Ionization Enthalpies

Convergence Limits

The frequency at which the spectral lines of a series merge.

75

Q.7

Ionization enthalpy

1E 11 hE

Convergence limit of Lyman series

H(g) H+(g) + e

n = 1

n =

76

Na

Li

He

n = 3 n =

n = 2 n =

n = 1 n =

n = 1 n =

n = 3

n = 2

n = 1

n = 1H

Electron transition for ionization of atom

Energy Level of e to be removed in ground state of

atomAtom

Lyman series ?

Balmer series ?

Paschen series ?

77

Q.9

Ionization enthalpy of helium

111 hEE

= (6.6261034 Js)(5.291015 s1)(6.021023 mol1)

= 2110 KJ mol1

> Ionization enthaply of hydrogen

78

E

E2

E1

E2’

E1’

HeH

Ionization enthalpy : He > H

Relative positions of energy levels depend on the nuclear charge of the atom.

79

The uniqueness of atomic emission The uniqueness of atomic emission spectraspectra

No two elements have identical atomic spectra


80

atomic spectra can be used to identify unknown elements.


The uniqueness of atomic emission The uniqueness of atomic emission spectraspectra

81

Flame Test

NaCl(s) Na(g) + Cl(g)heat

atomization

Na(g) + Cl(g) Na*(g) + Cl*(g)heat

Na*(g) + Cl*(g) Na(g) + Cl(g) hNa + hCl

82

Flame Test

The most intense yellow light is observed.

83

Q.10

hc

hE ν

m10450)ms 10Js)(3.0010(6.626

9

1834

= 4.421019 J

84

Bright lines against a dark background

Dark lines against a bright background

Emission vs AbsorptionEmission spectrum of hydrogen

(nm)

Absorption spectrum of hydrogen

(nm)

visible

visible

85

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

86

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.

87

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

88

Atomic Absorption Spectra

redshiftMoving at higher speed

Moving at lower speed

89

Redshift

Left : - Absorption spectrum from sunlight.

Right : - Absorption spectrum from a supercluster of distant galaxies

90

Atomic Absorption Spectra

Only

atomic emission spectrum of hydrogen

is required in A-level syllabus !!!

91

4.4.22 Deduction of Deduction of

Electronic Electronic Structure from Structure from

Ionization Ionization EnthalpiesEnthalpies

92

Evidence of the Existence of Evidence of the Existence of ShellsShells

For multi-electron systems,

Two questions need to be answered

93


1.How many electrons are allowed to occupy each electron shells ?

2.How are the electrons arranged in each electron shell ?

Existence of Shells

Existence of Subshells

94

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

Ionization enthalpyIonization enthalpy

95


Successive ionization enthalpies

Q. 11

96

Q.11(a)

Be(g) Be+(g) + e IE1

Be+(g) Be2+(g) + e IE2

Be2+(g) Be3+(g) + e IE3

Be3+(g) Be4+(g) + e IE4

97

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.

98

Q.11(c)

21060149051758900

IE4

(kJ mol1)

IE3

(kJ mol1)

IE2

(kJ mol1)

IE1

(kJ mol1)

99

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.

100

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.

101

Electron Diagram of Beryllium

Second Shell

First Shell

102

Energy Level Diagram of Beryllium

n =2

n= 1

103

E

IE1 = E - E2

E2

E1

Be

104

E

IE1

E2

E1

Be

E2’

E1’

Be+

IE2

IE1 < IE2

105

E

IE1

E2

E1

Be

E2’

E1’

Be+

E2’’

E1’’

Be2

+

IE2

IE1 < IE2 << IE3

IE3

106

E

IE1

E2

E1

Be

E2’

E1’

Be+

E2’’

E1’’

Be2

+

E2’’’

E1’’’

Be3

+

IE1 < IE2 << IE3 < IE4

IE3IE2 IE4

107

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

108

On the other hand,

orbital angular momentum of the Earth is associated with its annual motion around the Sun (公轉 )

109

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.

110

Paired electrons in an energy level should have opposite spins.

Electrons with opposite spins are represented by arrows in opposite directions.

Q.12

111

4 groups of electrons

Q.12

112

n = 4

n = 3 n = 2

n = 1 Which group of

electrons is in the first shell ?

Q.12

113

2, 8, 8, 2

2882

Q.12

114

n = 4

n = 2

n = 3

n = 1

Q.12

115


4.2 Deduction of electronic structure from ionization enthalpies (p.91)

2, 8, 1

116

n = 2

n = 3

n = 1

Na

117

Variation of IE1 with Atomic NumberEvidence for Subshell

118

Only patterns across periods are discussedRefer to pp.27-29 for further discussion

119

1.A general in IE1 with atomic number across Periods 2 and 3.13.(a)

120

2 – 3 - 3 13.(a)

121

2.IE1 value : Group 2 > Group 3

3. IE1 value : Group 5 > Group 6

13.(a)

122

2.Peaks appear at Groups 2 & 5

3. Troughs appear at Groups 3 & 6

13.(a)

123

2,1

2,2

2,3

2,4

2,5

2,6

2,7

2,8

2,8,1

2,8,2

2,8,32,8,4

2,8,5

2,8,6

2,8,7 2,8,8

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

es removed experience stronger attraction from the nucleus.

13.(b)

124

IE1 : B(2,3) < Be(2,2)

The electron removed from B occupies a subshell of higher energy within the n = 2 quantum shell

13.(b)

125

2s

2p

1s

n =

IE2s

IE2p

IE2s(Be) > IE2p(B)

IE1 : Be > B

13.(b)

126

IE1 : Al(2,8,3) < Mg(2,8,2)

The electron removed from Al occupies a subshell of higher energy within the n = 3 quantum shell

13.(b)

127

3s

3p

2p

n =

IE3s

IE3p

IE3s(Mg) > IE3p(Al)

IE1 : Mg > Al

13.(b)

128

3s

3p

2p

n =

Mg(12)

Al(13)

IE3s

IE3p

129

It is more difficult to remove an electron from a half-filled p subshell

13.(b)

IE1 : N(2,5) > O(2,6) ;

P(2,8,5) > S(2,8,6)

130

2s

2p

n =

IEN

O(8)N(7)

IEO

>First IE

131

2s

2p

n =

IEN

O(8)N(7)

IEO

>First IEThe three electrons in the half-filled 2p subshell occupy three different orbitals (2px , 2py , 2pz). repulsion between electrons is minimized.

132

3s

3p

n =

IEP

S(16)P(15)

IES

>First IEThe three electrons in the half-filled 3p subshell occupy three different orbitals (3px , 3py , 3pz).

repulsion between electrons is minimized.

133

The removal of an electron from O or S results in a half-filled p subshell with extra stability.(p.28, part 3)

Misleading !!!

134

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

135

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.

O(2,6) O+(2,5)

136

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).

137

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

138

Quantum Numbers

= 0, 1, 2,…,n-1

2.Subsidiary quantum number ( )

related to the shape of the subshells.

139

Quantum Numbers2. Subsidiary quantum number ( )

= 0, 1, 2,…,n-1

= 0 spherical s subshell= 1 dumb-bell p subshell

= 2 d subshell= 3 f subshell

complicated shapes

140

Each principal quantum shell can have one or more subshells depending on the value of n.

If n = 1,

possible range of :

name of subshell : 1s

0 to (1-1)

0No. of subshell : 1

141


0 to (2-1) 0, 1

If n = 2,

possible range of :

names of subshells : 2s, 2p No. of subshells : 2

142


names of subshells : 3s, 3p, 3d

0 to (3-1) 0, 1, 2

If n = 3,

possible range of :

No. of subshells : 3

143


names of subshells : 4s, 4p, 4d, 4f

0 to (4-1) 0, 1, 2, 3

If n = 4,

possible range of :

No. of subshells : 4

144

Quantum Numbers

3.Magnetic quantum number (m)

related to the spatial orientation of the orbitals in a magnetic field.

m = ,…0,…+

145

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

If = 0 (s subshell),

146

Possible range of m : 1 to +1 1 ,0,+1

No. of orbitals : 3

Names of orbitals : px, py, pz


If = 1 (p subshell),

147

Possible range of m : 2 to +2 2, 1, 0, +1, +2

No. of orbitals : 5

Names of orbitals : 222 zyxyzxzxy d,d,d,d,d


If = 2 (d subshell),

148

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


If = 3 (f subshell),

149

Total no. of orbitals in a subshell = 2 + 1

Energy of subshells : -

s < p < d < f

150

4.Spin quantum number (ms)

2

1 or

2

1ms =

They describe the spin property of the electron, either clockwise, or anti-clockwise

151

4.Spin quantum number (ms)

Each orbital can accommodate a maximum of two electrons with opposite spins

152

Q.14(a)

32n = 4

18n = 3

8

1+3+5+7=16

1+3+5=9

1+3=4

4s, 4p, 4d, 4f

3s, 3p, 3d

2s, 2pn = 2

21(1s)1sn = 1

Total no. of

electrons

Total no. of orbitalsSubshells

Principal quantum

shell

153

14(b)

The total number of electrons in a principal quantum shell = 2n2

154

Q.15(a)

The two electrons of helium are in the 1s orbital of the 1s subshell of the first principal quantum shell.

1st electron : n = 1, l = 0, m = 0, ms = 21

2nd electron : n = 1, l = 0, m = 0, ms = 21

155

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.

156

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 electron : n = 3, l = 0, m = 0, ms = 21

2nd electron : n = 3, l = 0, m = 0, ms = 21

157

4.4.33 The Wave-The Wave-

mechanical mechanical Model of the Model of the

AtomAtom

158

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)

159

Electrons display both particle nature and wave nature.

Particle nature : mass, momemtum,…

Wave nature : frequency, wavelength, diffraction,…

160

Wave as particles : photons

hmc 2

hE By Planck(1)2mcE By Einstein(2)

h

chmc

(3)

Wave propertyParticle

property :momentum of a photon

161

Evidence :

photoelectric effect by Albert Einstein (1905)


162

Particles as waves L. De Broglie (1924)


h

c

νhmc

(3)

h

mv (4)

163

h

mc

h

mv

Any particle (not only photon) in motion (with a momentum, mv) is associated with a wavelength

(3)

(4)

164

Q.16

λ

hmv

mv

h

Electron

m10631

34

105.1100.5101.9

1063.6

λ

165

Q.16

λ

hmv

mv

hλ

Helium atom

m11327

34

102.7104.1106.6

1063.6

λ

166

Q.16

λ

hmv

mv

hλ

m107.3ms 10.5Kg 86Js106.63 37

1-

34

λ

100m world record holder

s 9.52s m 10.5

m 1001

167

Wave nature of electrons (1927)Wave nature of electrons (1927)

4.3 The Wave-mechanical model of the atom (p.95)

Evidence

of electron inter-atomic spacing in metallic crystals (1010m)

168

For very massive ‘particles’,

The wavelength associated with them (1037m) are much smaller than the dimensions of any physical system.

Wave properties cannot be observed.

169

LStanding waves only have certain allowable modes of vibration

Similarly, electrons as waves only have certain allowable energies.

= 2L

= L

= L32

Standing waves in a

cavity

170

The uncertainty principle

Heinsenberg


171


It is impossible to simultaneously determine the exact position and the exact momentum of an electron.

4))((

hpx

uncertainty in position measurement

uncertainty in momentum measurement

172

high energy photon

Small and light electron

The momentum of electron would change greatly after collision


173

low energy photonSmall and

light electron

The position of electron cannot be located accurately


174

high energy photon

No problem in macroscopic world !


175

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.

176

Schrodinger Equation


de Broglie : electrons as waves

Use wave functions () to describe electrons

Pronounced as psi

177



Heisenberg : Uncertainty principle

Probability (2) of finding electron at a certain position < 1.

178

0)(8

2

2

2

2

2

2

2

2

VEh

m

zyx

: wave function

m : mass of electron

h : Planck’s constant

E : Total energy of electron

V : Potential energy of electron


179

0)(8

2

2

2

2

2

2

2

2

VEh

m

zyx


The equation can only be solved for certain i and Ei

sinxdxsinxd

2

2

180


The wave function of an 1s electron is

0

23

21

0

1)1( a

Zr

ea

Zs

Z : nuclear charge (Z = 1 for Hydrogen)

a0 : Bohr radius = 0.529Å (1Å = 1010m)

r : distance of electron from the nucleus

Radius of 1s orbit of Bohr’s model

181


The allowed energies of H atom are given by

n = 1, 2, 3,…222

0

24

8 hn

ZmeE

n is the principal quantum number,

All other terms in the expression are constants

182

n = 1, 2, 3,…222

0

24

8 hn

ZmeE

2

1

nE

0Erefer to p.6 of notes

183

2

22

1

11

nnhcR

hcE

E = E2 – E1

E2

E1

n = 2

n = 122

21 n

hcR

n

hcRE

2

12

2 n

hcR

n

hcR12 EE

184

22is the is the probability of finding an electron at of finding an electron at a particular point in space. (electron density)a particular point in space. (electron density)

4.4 Atomic orbitals (p.98)

22(1(1s)s)

Probability never becomes zero

There is no limit to the size of an atom

185

contour diagram

relative probability of finding the electron

at the nucleus

as r 2

186

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

187

The electron spends 90% of time within the boundary surface

188

A 3-dimensional time exposure diagram.

The density of the dots represents the probability of finding the electron at that position.

189

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.

190

Total probability of finding the electron within the ‘shell’ of thickness dr

= 24r2dr

electron density within the ‘shell’

total volume of the ‘shell’

2 is the probability of finding the electron per unit volume

dr is infinitesimally small

191

22 4 r 2 as r , 4r2 as r a maximum at 0.529 Å

Orbital Model

192

0.529Å r (Å)

1

probability

Bohr’s Orbit Model

193

Total probability of finding the electron within the ‘shell’ of thickness dr

= 24r2dr

The sum of the probabilities of finding the electron within all ‘shells’ = 1

1dr4πψr

0r

2

2r

194

22 4 r

The total area bounded by the curve and the x-axis = 1

Check Point 4-3Check Point 4-3

1dr4πψr

0r

2

2r

195

ss Orbitals Orbitals4.4 Atomic orbitals (p.98)

196

nodal surface

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

198

Probabilities of finding the electron at A or B > 0

A

B

Probability of finding the electron at C = 0

C

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

201

pp Orbitals Orbitals

Two lobes along an axis


202

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

203


dd Orbitals Orbitals

Four lobes along two

axes

Two lobes & one belt

Four lobes between two

axes

204

Grand Orbital Table

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

205

The END

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?

Back


Ultraviolet radiationAnswer

207

What does the convergence limit in the Balmer series correspond to?

Back


The convergence limit in the Balmer series corresponds to the electronic transition from

n = to n = 2.

Answer

208


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

Answer

209


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.

210


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 Answer

211


Back

For one mole of sodium atoms,

E = hL

=

=

= 494486 J mol-1

= 494 kJ mol-1

The ionization enthalpy of sodium of 494 kJ mol-1.

Lhc

m 10242mol10 6.02 s m 10 3 s J 106.626

9-

123-1834

212

(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


213


(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.

214

(b) Give a sketch of the logarithm of successive ionization enthalpies of potassium against no. of electrons removed. Explain your sketch. Answer


215


(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.

216

(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

217

(a)What are the limitations of Bohr’s atomic model?

(b)Explain the term “dual nature of electrons”.

(c) For principal quantum number 4, how many sub-shells are present? What are their symbols?

Back4.3 The Wave-mechanical model of the atom (p.97)

(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) Electrons can behave either as particles or a wave.

(c) When n = 4, l = 0, 1, 2 and 3, there are 4 sub-shells. The symbols are 4s, 4p, 4d and 4f respectively.

Answer

218

(a)Distinguish between the terms orbit and orbital.

(b)Sketch the pictorial representations of an s orbital and a p orbital. What shapes are they?


(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) s orbital is spherical in shape whereas p orbital is dumb-bell in shape.

Answer

219

(c)How do the 1s and 2s orbitals differ from each other?

(d) How do the 2p orbitals differ from each other?


Answer

(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) 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.

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1 the electronic structure of atoms 4.1the electromagnetic spectrum 4.2deduction of electronic...

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