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
New Way Chemistry for Hong Kong A-Level Book 1
Chapter 4 The electronic structure of atoms (SB p.80)
The electronic structure of atoms
Niels Bohr
Bohr’s model of H atom
2
New Way Chemistry for Hong Kong A-Level Book 1
Chapter 4 The electronic structure of atoms (SB p.80)
The electronic structure of atoms
Niels Bohr
Bohr’s model of H atom
3
New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.82)
The electromagnetic spectrum


4
c
New Way Chemistry for Hong Kong A-Level Book 1
c

4.1 The electromagnetic spectrum (SB p.82)
Continuous spectrum of white light
Fig.4-5(a)
5
New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.83)
Line spectrum of hydrogen
Fig.4-5(b)
6
New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.83)
The emission spectrum of atomic hydrogen
UV
7
Visible
New Way Chemistry for Hong Kong A-Level Book 1
IR
4.1 The electromagnetic spectrum (SB p.84)
Interpretation of the atomic hydrogen
spectrum
8
New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.84)
Interpretation of the atomic hydrogen
spectrum
9
New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.84)
Interpretation of the atomic hydrogen
spectrum
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New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.85)
Bohr proposed for a hydrogen atom:
1. An electron in an atom can only exist in
certain states characterized by definite energy
levels (called quantum).
2. Different orbits have different energy levels. An
orbit with higher energy is further away from the
nucleus.
3. When an electron jumps from a higher energy
level (of energy E1) to a lower energy level (of
energy E2), the energy emitted is related to
the frequency of light recorded in the emission
spectrum by:
E = E1 - E2 = h
11
New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.86)
How can we know
the energy levels
are getting closer
and closer
together?
12
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4.1 The electromagnetic spectrum (SB p.87)
E = E1 - E2 = h
Planck ’s
constant
13
Frequency of light emitted
New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.87)
Emission spectrum of hydrogen
Absorption spectrum of hydrogen
14
dark background
(photographic
plate)
bright background
(photographic
plate)
New Way Chemistry for Hong Kong A-Level Book 1
bright lines
dark lines
4.1 The electromagnetic spectrum (SB p.87)
Production of the absorption spectrum
Absorption spectrum of hydrogen
bright background
(photographic plate)
15
New Way Chemistry for Hong Kong A-Level Book 1
dark lines
4.1 The electromagnetic spectrum (SB p.87)
Convergence limits and ionization
What line in the H
spectrum corresponds to
this electron transition
(n= ∞  n=1)?
Last line in the Lyman Series
For n=∞  n=1:
H (g)
16
New Way Chemistry for Hong Kong A-Level Book 1
H+(g) + e-
4.1 The electromagnetic spectrum (SB p.87)
Example 4-1A
17
Example 4-1B
New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.89)
The uniqueness of atomic emission spectra
No two
elements have
identical
atomic spectra
atomic
spectra can be
used to identify
unknown
elements.
Check Point 4-1
18
New Way Chemistry for Hong Kong A-Level Book 1
4.2
Deduction of
Electronic Structure
from Ionization
Enthalpies
19
New Way Chemistry for Hong Kong A-Level Book 1
4.2 Deduction of electronic structure from ionization enthalpies (p.91)
Ionization enthalpy
Ionization enthalpy (ionization energy) of an
atom is the energy required to remove one mole
of electrons from one mole of its gaseous atoms
to form one mole of gaseous positive ions.
The first ionization enthalpy
M(g)  M+(g) + e-
H = 1st I.E.
The second ionization enthalpy
M+(g)  M2+(g) + e20
H = 2nd I.E.
New Way Chemistry for Hong Kong A-Level Book 1
4.2 Deduction of electronic structure from ionization enthalpies (p.91)
Evidence of shells
 shells
21
New Way Chemistry for Hong Kong A-Level Book 1
4.2 Deduction of electronic structure from ionization enthalpies (p.91)
Evidence of sub-shells
2,8
2,5
 subshells
2,7
2,2
2,4 2,6
2,3
2,1
Check Point 4-2
22
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4.3
The Wavemechanical Model
of the Atom
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New Way Chemistry for Hong Kong A-Level Book 1
4.3 The Wave-mechanical model of the atom (p.94)
Bohr’s atomic model and its limitations
Bohr considered the electron in the H atom (a
one-electron system) moves around the nucleus
in circular orbits.
Basing on classical mechanics, Bohr calculated
values of frequencies of light emitted for electron
transitions between such ‘orbits’.
The calculated values for the
frequencies of light matched with the
data in the emission spectrum of H.
24
New Way Chemistry for Hong Kong A-Level Book 1
4.3 The Wave-mechanical model of the atom (p.94)
Bohr tried to apply similar models to atoms of
other elements (many-electron system), e.g. Na
atom.
Basing on classical mechanics, Bohr calculated
values of frequencies of light emitted for electron
transitions between such ‘orbits’.
The calculated values for the frequencies of light
did NOT match with the data in the emission
spectra of the elements.
 The electron orbits in atoms may NOT be
simple circular path.
25
New Way Chemistry for Hong Kong A-Level Book 1
4.3 The Wave-mechanical model of the atom (p.95)
Wave nature of electrons
A beam of electrons shows diffraction phenomenon
 Electrons possess wave properties
(as well as particle properties).
26
New Way Chemistry for Hong Kong A-Level Book 1
4.3 The Wave-mechanical model of the atom (p.95)
Wave nature of electrons
Schrödinger used complex differential
equations/wave fucntions to describe
the wave nature of the electrons
inside atoms (wave mechanic model).
The solutions to the differential
equations describes the orbitals of the
electrons inside the concerned atom.
An orbital is a region of space having
a high probability of finding the
electron.
27
New Way Chemistry for Hong Kong A-Level Book 1
4.3 The Wave-mechanical model of the atom (p.95)
Quantum numbers
The solutions of
the wave
functions are the
orbitals -- which
are themselves
equations
describing the
electrons.
Electrons in orbitals are specified
with a set of numbers called
Quantum Numbers:
1. Principal quantum number (n)
n = 1, 2, 3, 4, …...
2. Subsidiary quantum number (l)
l = 0, 1, 2, 3…, n-1
s p d f
3. Magnetic quantum number (m)
m = -l, …, 0, …l
4. Spin quantum number (s)
s= +½, -½
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New Way Chemistry for Hong Kong A-Level Book 1
4.3 The Wave-mechanical model of the atom (p.96)
Principal
quantum
number
(n)
Subsidiary
quantum
number (l)
Number of
orbitals
(2l+1)
Symbol
of
orbitals
1
0
1
1s
Maximum
number
of
electrons
held
2
2
0
1
1
3
2s
2p
2
6
3
0
1
2
0
1
2
3
1
3
5
1
3
5
7
3s
3p
3d
4s
4p
4d
4f
5
6 18
10
2
6
32
10
14
4
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New Way Chemistry for Hong Kong A-Level Book 1
8
4.3 The Wave-mechanical model of the atom (p.97)
Check Point 4-3
3d
4s
3p
3s
2p
2s
1s
30
New Way Chemistry for Hong Kong A-Level Book 1
Each orbital
can
accommodate
2 electrons
with opposite
spin.
4.4
31
Atomic
Orbitals
New Way Chemistry for Hong Kong A-Level Book 1
4.4 Atomic orbitals (p.98)
s Orbitals
Graph of probability of finding an electron against
distance from nucleus
32
New Way Chemistry for Hong Kong A-Level Book 1
4.4 Atomic orbitals (p.98)
s Orbitals
33
New Way Chemistry for Hong Kong A-Level Book 1
4.4 Atomic orbitals (p.100)
p Orbitals
The shapes and orientations of 2px, 2py and 2pz
orbitals
34
New Way Chemistry for Hong Kong A-Level Book 1
4.4 Atomic orbitals (p.101)
d Orbitals
Check Point 4-4
The shapes and orientations of 3dxy, 3dyz, 3dx2-y2
and 3dz2 orbitals
35
New Way Chemistry for Hong Kong A-Level Book 1
The END
36
New Way Chemistry for Hong Kong A-Level Book 1
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
37
New Way Chemistry for Hong Kong A-Level Book 1
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 energy required for the
transition of an electron from n =2 to n = .
Back
38
New Way Chemistry for Hong Kong A-Level Book 1
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
39
New Way Chemistry for Hong Kong A-Level Book 1
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.
40
New Way Chemistry for Hong Kong A-Level Book 1
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
41
Answer
New Way Chemistry for Hong Kong A-Level Book 1
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.
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New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.90)
(a) The first line of the Balmer series of the emission
spectrum of atomic hydrogen corresponds to the energy
emitted in the transition of an electron from the third
energy level to the second energy level. It has a
wavelength of 656.3 nm. What is the energy difference
between the second and the third energy levels?
(Planck constant = 6.626  10-34 Js, Avogadro constant =
6.02  1023 mol-1)
Answer
43
New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.90)
c
λ
E = 6.626  10-34 J s 
(a) E = hv = h
3  10 8 m s 1
656.3  10  9 m
= 3.03  10-19 J (for one electron)
For 1 mole of electrons,
E = 3.03  10-19 J  6.02  1023 mol-1
= 182406 J mol-1
= 182 kJ mol-1
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New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.90)
(b) Given that the frequency of the convergence limit
corresponding to the ionization of helium is 5.29  1015 Hz,
calculate the ionization enthalpy of helium.
(Planck constant = 6.626  10-34 Js, Avogadro constant =
6.02  1023 mol-1)
Answer
(b) For 1 mole of helium atoms,
I.E. = hvL
= 6.626  10-34 J s  5.29  1015 s-1  6.02  1023 mol-1
= 2.11  106 J mol-1
= 2110 kJ mol-1
45
New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.90)
(c) The blue colour in fireworks is often achieved by heating
copper(I) chloride (CuCl) to about 1200 oC. The compound
then emits blue light with a wavelength of
450 nm. What is the energy released per copper(I) ion at
the specified condition?
Answer
hc
(c) E =
λ
34
8
-1
6.626

10
J
s

3

10
ms
=
450  10 9 m
= 4.42  10-19 J
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New Way Chemistry for Hong Kong A-Level Book 1
4.1 The electromagnetic spectrum (SB p.90)
(d) Name the element present in the sample when the
following flame colours are observed in flame tests.
(i) Golden yellow
(d) (i) Sodium
(ii) Lilac
(ii) Potassium
(iii) Brick-red
(iii) Calcium
(iv) Bluish green
(iv) Copper
Back
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New Way Chemistry for Hong Kong A-Level Book 1
Answer
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
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New Way Chemistry for Hong Kong A-Level Book 1
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.
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New Way Chemistry for Hong Kong A-Level Book 1
4.2 Deduction of electronic structure from ionization enthalpies (p.94)
(b) Give a rough a sketch of the logarithm of successive
ionization enthalpies of potassium. Explain your sketch.
Answer
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New Way Chemistry for Hong Kong A-Level Book 1
4.2 Deduction of electronic structure from ionization enthalpies (p.94)
(b)
51
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.
New Way Chemistry for Hong Kong A-Level Book 1
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
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New Way Chemistry for Hong Kong A-Level Book 1
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.
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New Way Chemistry for Hong Kong A-Level Book 1
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
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New Way Chemistry for Hong Kong A-Level Book 1
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 areas of high probability known as nodal surfaces.
(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
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New Way Chemistry for Hong Kong A-Level Book 1