CHM 131

CHAPTER 3

STRUCTURE OF ATOM
 Lesson Outcomes
• Understand the general regions of the electromagnetic
spectrum
• Understand how Bohr’s theory explained the line spectra of H
atom
• Find the energy change and wavelength of the photon
absorbed or emitted when H atom’s electron changes energy
level
• Determine quantum numbers and sublevel designations
• Draw the shapes of s, p and d orbitals
• Describe the arrangement of electrons in an atom based on
Aufbau Principle, Pauli Exclusion Principle and Hund’s Rule
• Write full and condensed electron configuration of an element
• Understand the periodic trends of elements
2
 Properties of wave

• Wavelength, λ (lambda)
- the distance between identical points on successive waves
• Amplitude
- the vertical distance from the midline of a wave to the peak &
trough
• Frequency, ν (nu)
- the number of waves that pass through a particular point in 1
second (Hz = 1 cycle/s)
The speed of the wave = λ x ν 3
 Electromagnetic Radiation
•Electromagnetic radiation is the emission and transmission of
energy in the form of electromagnetic wave

Speed of light (c) in vacuum = 3.00 x 108 m/s

All electromagnetic radiation

lxn=c
4
Example:
A photon has a frequency of 6.0 x 104 Hz. Convert this frequency
into wavelength (nm). Does this frequency fall in the visible
region?

l
Ans:
lxn=c n

l = c/n
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm

5
 Planck’s Quantum Theory
• Max Planck (1900) proposed the Quantum Theory
• Quantum theory states that energy in the form of
electromagnetic radiation can be emitted or absorbed in
discrete amount called quanta (singular: quantum)
• Quantum – the smallest quantity of energy that can be
emitted (or absorbed) in the form of electromagnetic radiation
• Planck’s equation:
E = hν
Where h = Planck’s constant = 6.63×10-34 J s

ν=c/λ

Therefore, E = hc / λ

6
7
 Example:
Calculate the energy (in joules) of (a) a photon with a wavelength of
5.00×104 nm (IR region)

Ans:
E = hc / λ
= (6.63×10-34 Js)(3.00×108 m/s)
(5.00×104 nm) 1×10-9 m
1 nm
= 3.98×10-21 J

8
 3.2 Introduction to
Hydrogen Emission
Spectrum, Bohr’s Theory

9
 The Line Spectrum of Hydrogen

• When electricity is passed through a discharge tube filled with

hydrogen gas at low pressure, a pink light is emitted
• When the light passes through a prism, a line spectrum is
obtained

10
• The energy supplied is absorbed by the electrons of hydrogen
atoms and causes them to be promoted from the ground
states (lowest energy level) to excited states (higher energy
levels).
• Transitions of electron between two energy levels produce
lines in the hydrogen spectrum
• The emission series obtained are classified according to
which level the electrons drop to
Level which Name of series Spectrum
electron drops region
n=1 Lymann Ultraviolet
n=2 Balmer Visible
n=3 Paschen Infrared
n=4 Brackett Infrared
n=5 Pfund Infrared

11
 Bohr’s Model of Atom (1913)
• e- can only have specific (quantized) energy values
• light is emitted as e- moves from one energy level to a lower
energy level
n (principal quantum number) = 1,2,3,…
1
En = -RH ( ) RH (Rydberg constant) = 2.18 x 10-18J
n2

12
 Ephoton = DE = Ef - Ei
ni = 3 ni = 3 1
Ef = -RH ( 2 )
nf
ni = 2 1
Ei = -RH ( 2 )
nf = 2 ni
1 1
DE = RH( 2 )
ni n2f

nnf f==11

13
Example:
Calculate;
a) The energy an electron has when it occupies a level
equivalent to the quantum number of n = 3 and n = 4
b) The energy of the photon emitted when one mole of electron
drops from the fourth energy level to the third energy level
c) The frequency and wavelength of this photon

Ans:
a) At n=3, E3= -RH = -2.18x10-18 J = -2.42x10-19 J
32 32
At n=4, E4= -RH = -2.18x10-18 J = -1.36x10-19 J
42 42

14
b) ΔE = E3 – E4,
= (-2.42x10-19 J) – (-1.36x10-19)
= -1.06x10-19 J
p/s: Note that the –ve sign indicates that energy is released
when an electron falls.
To calculate the amount of energy released by one
mole of electrons, multiply with NA

ΔE = -1.06x10-19 J x 6.022x1023
= -63.83 kJ/mol
c) Frequency and wavelength

Frequency, ν= E= -1.06x10-19J = 1.60x1014 s-1

h 6.63x10-34 Js-1
p/s: Ignore –ve sign, frequency value always +ve

Wavelength, λ = c = 3.00x108ms-1 = 1875 nm

15
ν 1.60x1014 s-1
 EXERCISE
1. What is the wavelength of the photon (in nm)
emitted during a transition from ni = 5 state to the nf
=2 state in the hydrogen atom (434 nm)
2. The first line of Balmer series occurs at wavelength
of 656.3 nm. What is the energy difference
between the two energy levels involved in the
emission that result in the spectral line? (3.027x10 -
19J)

3. A particular form of electromagnetic radiation has

a frequency of 8.11 x 1014 Hz. (a) What is the
wavelength in nanometers. (3.7 x 102 nm) (b) What
is the energy in joules of one quantum of this
radiation (5.38x10-19J) 16
 QUANTUM MECHANICS
Quantum Atomic Model
 Bohr atomic model

 Explained the spectra of atoms with one electron very well

 Not workable for atom with more than one electron

 Erwin Schrodinger (1926) proposed quantum atomic model which

derived a set wave functions for electrons

 The quantum atomic model is the more acceptable model today

 Schrodinger described the atomic orbital

Orbital = Region in space within which there is a high

probability of finding an electron in an atom
(atomic orbital) or molecule (molecular
orbital)

Electron cloud = Three dimensional space around the nucleus

which occupying by density of electron

 The wave function of each electron can be described as a set of four

quantum numbers which are n, l, m and s

 Quantum Number

Symbol Quantum number Description

n Principle Energy level

Azimuthal or angular
ℓ Shape of orbital
momentum

m Magnetic Orbital orientation in space

Direction that the electron spins;

s Electron spin
clockwise or anticlockwise
1. Principle quantum number

 Symbol – n

 Describes the shell or energy level

 Values – positive integer (1,2,3,4…..∞) or K, L, M, N…….

 E.g. :

▪ n= 1 or K (the shell closest to the nucleus, lowest energy)

▪ n= 4 or N (the 4th shell)

 The higher the value of n, the higher is the energy of the shell
 n.
.
4
3
n can be a positive
2
integer (1,2,3,…)
1

❖ n describes the coarse energy and size of the electron orbital.

❖ relates to the average distance of the electron from the nucleus.
2. Azimuthal quantum number

 Symbol - l

 Specifies a subshell in an atom

 Values: n-n…up to…n-1

 E.g.:

▪ When n = 1, l = (1-1) = 0

▪ When n = 2, l = (2-2), (2-1) = 0,1

▪ When n = 3, l = (3-3), (3-2), (3-1) = 0,1,2

 Note that the number of possible l values equals the value of n

 The values of ℓ are related to names or orbitals or letter codes for
subshells:

valueofl orbitalname
(subshelllettercode)

0 s
1 p
2 d
3 f
4 g

 To name a particular orbital (or subshell), we combine the value of n

followed by the letter code for the subshell

 E.g. The subshell with n = 3 and l = 1 is the 3p orbital

▪ l tell us the “shape” of the orbitals
▪ l can be zero or a positive integer
▪ l = ( 0 , 1 , 2 , 3 , 4 …, n-1)
3. Magnetic Quantum Number

 Symbol - m

 Describes the orientations (or components) of orbitals relative to

each other

 Values: - l ….0…..+ l

 For example:

▪ when l = 1, m= -1, 0, +1

▪ when l = 2, m= -2,-1, 0, +1, +2

 Degenerate orbitals (orbitals of similar energy). Each

components are represented by subscripts. (e.g. px, py, pz..)

for l = 1

ml = -1 ml = 0 ml = +1

• ml indicates the orientation of the electron’s orbital with

respect to the three axes in space.
4. Spin quantum number

 Symbol – s

 Describes the direction in which the electron spins on its axis

 Values:

▪ ms = +½ (e- spins in clockwise direction, ↑)

▪ ms = -½ (e- spins in anticlockwise direction, ↓)

 Only 2 electrons may occupy each one ‘orbital orientation’ and the

electrons must have opposite spins

Possibilities for electron spin:

N S

e-
e-

S N
Relationship among quantum number

 Each shell can be filled with 2n2 electrons and each subshell can be

filled by 2(2l+1) electrons

 For example:

▪ The 4th shell (n = 4) is filled by (2x42) = 32e

▪ each 4s subshell is filled by 2[(2x0)+1] = 2e

▪ each 4p subshell is filled by 2[(2x1)+1] = 6e

▪ each 4d subshell is filled by 2[(2x2)+1] =10e

▪ each 4f subshell is filled by 2[(2x3)+1] = 14e

n Max
l mℓ Subshell/orbital
electrons

1 0 0 1s(1) 2

0 0 2s(1) 2
2
1 -1, 0, +1 2p(3) 6

0 0 3s(1) 2

3 1 -1, 0, +1 3p(3) 6

2 -2, -1, 0, +1, +2 3d(5) 10

 s orbital

p orbital
d orbital
Example
What values of the angular momentum (l) and magnetic (m) quantum
numbers are allowed for a principal quantum number (n) of 3? How many
orbitals are allowed for n = 3?

SOLUTION:

For n = 3, l = 0, 1, 2

For l = 0 m= 0

For l = 1 m = -1, 0, or +1

For l = 2 m = -2, -1, 0, +1, or +2

There are 9 m values and therefore 9 orbitals with n = 3.

 EXERCISE 2
Give the name, magnetic quantum numbers, and number of orbitals for
each sublevel with the following quantum numbers:
(a) n = 3, l = 2 (b) n = 2, l = 0 (c) n = 5, l = 1 (d) n = 4, l = 3
SOLUTION:

n l sublevel name possible mvalues No. of orbital

3 2 3d -2, -1, 0, 1, 2 5

2 0 2s 0 1

5 1 5p -1, 0, 1 3

4 3 4f -3, -2, -1, 0, 1, 2, 3 7

 EXERCISE 3
State the orbitals and maximum number of electrons that can occupy
each orbital for each of quantum numbers given below
(a) n = 6, l = 0 (b) n = 4, l = 2 (c) n = 5, l = 1

SOLUTION:

Orbital Maximum no. of electron

6s 2

4d 10

5p 6
Three rules for assigning electrons

❑ Aufbau Principle

❑ Pauli Exclusion Principle

❑ Hund’s Rule
1. Aufbau Principle

Electrons in an atom will occupy the lowest-

energy orbitals available

(aufbau = “building up”)

Energy levels according to Aufbau

1s < 2s < 2p < 3s < 3p <

4s < 3d < 4p < 5s < 4d<
5p < 6s < 4f <5d < 6p < 7s
< 5f < 6d < 7p < 8s …
 There are some exceptions to the Aufbau Principle. This occurs mainly
with electrons in the d orbital where extra stability is obtained from a half
filled or fully filled d orbital. Therefore, if there are 4 electrons, or 9
electrons in the d orbital, it will move one electron from the s orbital below
it to fill the extra space.

• Cr's electron configuration, following the model

would be: 1s2 2s2 2p6 3s2 3p6 4s23d4, but instead it is
1s2 2s2 2p6 3s2 3p6 3d54s1, because there is extra
stability gained from the half-filled d orbital.
• Cu’ s electron configuration, following model would
be: 1s2 2s2 2p6 3s2 3p6 4s23d9, but instead it is 1s2 2s2
2p6 3s2 3p6 3d104s1 because there is extra stability
gained from the full-filled d orbital

39
 Electron configurations of Chromium and Copper

Cr: 1s2 2s2 2p6 3s2 3p6 3d5 4s1 or [Ar] 3d5 4s1

↑ ↑ ↑ ↑ ↑ ↑

3d 4s

Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 or [Ar] 3d10 4s1

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑

3d 4s
NOTE
d subshells that are half-filled or fully
filled are particularly stable compare
with partially filled orbitals
2. Pauli Exclusion Principle

A maximum of two electrons can occupy each

orbital. Each electron must have different spin
quantum numbers

↑↓

Electron with Electron with

clockwise spin anticlockwise spin
3. Hund’s Rule

The most stable arrangement of electrons is that

with the most unpaired electrons all with the
same spin

WRONG RIGHT
ELECTRONIC CONFIGURATIONS

“….describes how the electrons are

distributed among the various atomic
orbitals…”

Writing electronic configurations

1. Determine the number of electrons and atom has
2. Fill orbitals in order of increasing energy
 S
block P block

D block

F block

F block
 This letter tells This number tells you
you the subshell the number of electrons
in a sub shell

1s 2 2s2 2p63s1
This number tells you
the main energy level
or shell
 Ions & Isoelectronic Configurations

Aluminum (Al): 1s2 2s2 2p6 3s2 3p1

Aluminum ion (Al3+): 1s2 2s2 2p6

❖ Isoelectronic  different species having the same electron

configuration

O2-: 1s2 2s2 2p6

F- : 1s2 2s2 2p6
Ne : 1s2 2s2 2p6
Ground-state electron configurations

First three period

Electron configurations of atoms with proton number > 10
Example:
Write the electronic configurations of S2-, Ni(II)and Fe(III)

SOLUTION:

a) S = 16 electrons
S2-= (16 + 2 ) electrons = 18 electrons = 1s2 2s2 2p63s2 3p6

b) Ni = 28 electrons
Ni2+= (28 - 2) electrons = 26 electrons = 1s2 2s2 2p63s2 3p63d8

c) Fe = 26 electrons
Fe3+= (26 - 3) electrons = 23 electrons = 1s2 2s2 2p63s2 3p63d5
 TUTORIAL
• The electron configuration of the neutral atom is
1s22s22p63s2. Write a complete set of quantum
numbers for each of the electrons. Name the
element.
• Write the ground state electron configuration and
the orbital diagram of Cr and Cu. Explain why the
ground state electron configuration of Cr and Cu
are different from what we might expect.
• Indicate the number of unpaired electrons present
in each of the following atoms: B, Ne, P, Sc and Mn.
• Write the electronic configuration of the following
element. Al3+ , O2-, Ca2+ ,Se
50
3.5 Periodic Trends of
Elements

51
 Lesson Outcomes
• Indicate period, group and block s, p, d, f
• Specifies the position of metals, metalloids and non-
metals in the periodic table
• Describe the metallic behavior and acid-base
• Explain the variation in atomic , ionic and isoelectronic
radii
• Define the first and second ionization energies and
explain the variations in the first ionization energy across
period and down the group
• Define electron affinity and electronegativity
• Explain the variation in electronegativity of elements

52
Element

• Scientists have identified

90 naturally occurring
elements, and created
about 28 others
• The elements, alone or in
combinations, make up
our bodies, our world, our
sun, and in fact, the entire
universe.

53
The most abundant element in the earth’s
crust is oxygen

54
When the Elements Were Discovered

55
The Modern Periodic Table

56
 57

Noble Gas
Halogen
Classification of the Elements

Group
Period
Alkali Earth Metal
Alkali Metal
 Important Group Names

Group(s) Group Name

1 Alkali Metals
(Very Reactive)

Alkaline Earth Metals

2
(Very Reactive)

3-11 Transition Metals

17 Halogens
(Very Reactive)

18 Noble Gases
(Not Reactive)*Stable
58
 Electronic Configurations
Generally the periodic table can be divided into 4 major
regions which called blocks; s-block, p-block, d-block and f-
block, based on their electronic configurations

1.The s-block:
•The s-block is occupied by elements with a half-filled or a
completely-filled outermost s orbital. Having configuration of
ns1 at the Group 1. These elements are metals (except H), also
known as alkali metal.
•The next group in s-block is Group 2, comprised of
elements with a completely-filled outermost s orbital (ns2), also
known as alkaline earth metals

59
2. The p-block:
The p-block elements are elements that have electrons in the
outermost p orbital. The valence electronic configuration
varies from ns2 np1 (Group 13) to ns2 np6 (Group 18).
Consists of metals, metalloids and non-metals. Group 17
known as halogens and group 18 known as noble gas
3. The d-block:
The d-block elements are called the transition elements
which have partially-filled or completely filled d orbitals (d1 to
d10)
4. The f-block:
The f-block is known as the inner transition elements. The
first row elements are called lanthanides and second row is
called actinides. These elements also known as rare
earth elements

60
 ns1

4f
5f
ns2

d1

d5

d10

ns2np1
ns2np2
ns2np3
ns2np4
ns2np5
61

ns2np6
 Exercise 1

Determine the period, block and group for each element with
the following configuration:
1.A: 1s2 2s2 2p3 - Period 2, block p, group 15

2.B: 1s2 2s2 2p6 3s2 3p6 - Period 3, block p, group 18

3.C: 1s2 2s2 2p6 3s2 - Period 3, block s, group 2

62
 Atomic and Ionic Radii
• There are 2 major factors affecting the size of atoms:

1. The effective nuclear charge (Zeff)

- the residual nett charge felt by the velence electrons
2. The shielding effect
- also known as screening effect
- caused by mutual repulsion of electron for each other

63
Effective nuclear charge (Zeff) is is the "positive c harge" felt by
an electron

Zeff = Z - s 0 < s < Z (s = shielding constant)

Zeff  Z – number of inner or core

electrons
Z Core Zeff Radius (pm)

Na 11 10 1 186
Mg 12 10 2 160

Al 13 10 3 143

Si 14 10 4 132

64
Effective Nuclear Charge (Zeff)

65
 Size of Atom
• Across a period, from left to right of the periodic table, a
gradual decrease in size of atom is observed:

• The increase in proton number leads to the increase in zeff

• The valence electrons experience a greater attraction & being
pilled closer to the nucleus
• Results in decrease in size of atoms

66
• Down within a group, the increase in size of atoms is
observed

• Down the group, the additional electrons of each of element

go to bigger (larger value of n)
• The additional inner orbitals leads to an increase in the
shielding effect, hence the size of atom increases

67
68
 A look at one period

Li - 1s22s1 Be - 1s22s2
Be has a smaller
radius than Li
because there is one
1s 1s
3P+ 4P+ more electron and
the attraction
between them is
stronger. The
“shells” get pulled in
tighter around the
nucleus.

69
 A look at one group

• Moving down any group, 4s

another shell is added. 3s
2s
• Example: H, Li, Na, K
1s
• Each new shell is further
from the nucleus so an
atomic size increases
from top to bottom.

70
 The size of Ions
• The size of positive ions (cation)
- formed when atom loses electron
- always smaller than its neutral atom
- The atoms lose the valence electrons, leaving electrons of the
inner shells which tend to attract the nucleus more, thus
decreasing size
- E.g. Na+ ion < Na atom

71
• The size of negative ions (anion)
- formed when atom gains electrons
- always larger than its neutral atom
- electrons are added into the same shell and tend to
repel each other and so increase size
E.g. Cl- ion > Cl atom

72
The Radii (in pm) of Ions of Familiar
Elements

73
Comparison of Atomic Radii with Ionic
Radii

74
 Electron Configurations of Cations and
Anions of Representative Elements

Na: [Ne]3s1 Na+ [Ne] Atoms lose electrons so that

Ca: [Ar]4s2 Ca2+ [Ar] cation has a noble-gas outer
electron configuration.
Al: [Ne]3s23p1 Al3+ [Ne]

H: 1s1 H- 1s2 or [He]

Atoms gain electrons so
that anion has a noble- F: 1s22s22p5 F- 1s22s22p6 or [Ne]
gas outer electron
configuration. O: 1s22s22p4 O2- 1s22s22p6 or [Ne]

N: 1s22s22p3 N3- 1s22s22p6 or [Ne]

75
 Cations and Anions of Representative
+2 Elements

+3
+1

-3
-2
-1
76
 Isoelectronic: have the same number of electrons, and hence the
same ground-state electron configuration

Na+: [Ne] Al3+: [Ne] F-: 1s22s22p6 or [Ne]

O2-: 1s22s22p6 or [Ne] N3-: 1s22s22p6 or [Ne]

Na+, Al3+, F-, O2-, and N3- are all isoelectronic with Ne

What neutral atom is isoelectronic with H- ?

H- : 1s2 same electron configuration as He

77
 Electronegativity

•The ability of an atom to attract electrons (or electron density)

towards itself in a covalent bond
•The higher the associated electronegativity number, the more
an element or compound attracts electrons towards it.
•Across a period of elements, electronegativity increases.
As you go across a group, the nucleus contains one more proton
each element over. So, more protons means more power to
attract electrons.
•Down a group, electronegativity decreases. This is because
the atomic number increases down a group and thus there is an
increased distance between the valence electrons and nucleus,
or a greater atomic radius. Thus, less power to attract the
electron.

78
79
 Ionization Energy

• First Ionization Energy (energy required to remove the

outermost or the highest energy electron from a gaseous atom)
X(g) → X⁺(g) + е⁻

• Second Ionization Energy (energy required to remove any

subsequent high-energy electron from a gaseous cation)
X⁺(g)→ X²⁺(g) + e⁻

•This process is endothermic as energy must be supplied to

remove electron which is attracted to the positive charge of
the nucleus

80
• Ionization energy across a period:
- From left to right of periodic table, the atomic size decreases,
the outer e- are more closely attracted to the nucleus and thus
are more difficult to be removed
- The ionization energy increases across the period

• Ionization energy down a group:

- Down within a group, the ionization energy decreases
- The atomic radius increases as we go down the group
- The atomic size expands as more e- are added to the
successive energy level (shielding effect increases)
- The outer e- is far from nucleus and less attraction towards
nucleus. Thus easier to remove the electron.

81
General Trends in Ionization Energy

82
Successive inonzation energy value (KJ/mol)
Element First Second Third Fourth Fifth Sixth Seventh
Na 496 4,560
Mg 738 1,450 7,730
Al 577 1,816 2,881 11,600
Si 786 1,577 3,228 4,354 16,100
P 1,060 1,890 2,905 4,950 6,270 21,200
S 999.6 2,260 3,375 4,565 6,950 8,490 27,107
Cl 1,256 2,295 3,850 5,160 6,560 9,360 11,000
Ar 1,520 2,665 3,945 5,770 7,230 8,780 12,000

• Large jumps in the successive molar ionization energies

occur when passing noble gas configurations.

83
 Electron Affinity
• Electron affinity is the negative of the energy
change that occurs when an electron is accepted by an
atom in the gaseous state to form an anion.
• The more negative the electron affinity, the greater the
tendency of the atom to accept an electron
• Electron affinity generally increases across a period
in the periodic table and decreases down a group.
• Electron affinity follows the trend of electronegativity.

84
General Trends in Electron Affinity

85
86
 EXERCISE:
1. Arrange the following atoms in order of increasing
atomic radius,P: Si: N.
2. For each of the following pairs, indicate which one
of the two species is larger: a) N3- or F- (b) Mg2+ or
Ca2+ (c) Fe2+ or Fe3+
3. Which atom should have lower first ionization
energy, oxygen or sulphur. Give reason for your
answer.
4. The successive IE of the first four electrons of
representative element are 738 kJ/mol, 1450
kJ/mol, 7730kJ/mol and 10500 kJ/mol. Characterize
the element according to the periodic group.
87
 ANSWER
1. The order of increasing radius is N < P < Si
2. a) N3- and F- are isoelectronic anions, both containing 10
electrons. Because N3- has only seven protons and F- has
nine, the smaller attraction exerted by the nucleus on the
electrons result in a larger N3- ion. b) Both Mg and Ca belong
to group 2A. Thus, Ca2+ ion is larger than Mg2+ because Ca ‘s
valence electrons are in the larger shell (n=4) than are Mg ‘s
(n=3). C) Both ions have the same nuclear charge, but Fe2+
has one more electron and hence greater electron –electron
repulsion. Thus, radius Fe2+ is larger.

88
 ANSWER
3. Oxygen and sulphur are members of group 6A.
They have the same valence electron
configuration (ns2np4), but the 3p electron is farther
from the nucleus and experience less nuclear
attraction than the 2p electron in oxygen. Thus, we
predict that sulphur should have a smaller
ionization energy.
4. Group 2

89
Summary

90