Chapter 2

Electronic Structure of Atoms
2.1 INTRODUCTION
1. The way electrons are arranged around the nucleus of an atom (known as electronic configuration) is important as the electronic

configuration determines the chemical properties of an atom or element.
2. Atoms do not ordinarily emit any radiation, but when heated, electrons absorb energy and are thus excited to a higher energy

level. The colour seen is the portion of the visible light spectrum that is not absorbed.

2.2. A Continuous Spectrum (The Spectrum of Visible light)
(a) When sunlight passes through a prism onto a white screen, the colours of the rainbow are observed.
(b) This spectrum is composed of visible light of all frequencies or wave lengths. Such a spectrum is known as a continuous
spectrum. The visible spectrum is continuous. It ranges from blue to red light.

Separation of light by a prism according to wavelength to
give a continuous spectrum.

2. An Absorption Spectrum (Absorption Spectrum for Hydrogen gas)

3. An Emission Spectrum. (Eg. Line Spectrum for Hydrogen gas)

(a) On the other hand, light from a discharge tube containing a particular element would not contain all the frequencies of the visible
light, but is consist of a few frequencies (wavelengths) only,

(b) If this light is viewed through a spectroscope, a spectrum consisting of a number of lines, each corresponding to a particular
wavelength will be seen against a black background. Such a spectrum is known as a line spectrum.

(c) The relationship between wavelength (λ) and frequency ( ) is
λ =c

(c = velocity of light. 3.0 x 108 m s-1)

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EXAMPLE 2.1

Calculate the frequency of light with a wavelength of 590 nm.

ANSWER = 5.08 x 1014 Hz (or s-1)

Wavelength = 590 nm = 590 x 10-9 m

Using the formula: λ = c
λ  = 3 x 108
  = 3  108

590  10-9

EXAMPLE 2.2
Green light has a frequency of 5.75 x 1014 s-1. Calculate its wavelength

ANSWER λ =c
Using:

λ = 3  108 = 5.22 x 10-7 m or 522 nm
5.75  1014

2.1.2 LINE SPECTRUM OF HYDROGEN /HELIUM ION, He+ or Li2+ ion

1. When a vacuum tube filled with hydrogen (or helium ion, He+) gas, at low pressure, the electrons flowing from the negative
electrode to the positive electrode collide with the H2 molecules.
This collision caused the H2 molecules to dissociate into hydrogen atoms and the tube emits a pink light.

2. When this light is viewed through a spectroscope, it is seen to be composed of a few groups of lines, each with a particular
wavelength.
Some are in the visible region, while others are in the ultra-violet and infra-red region of the electromagnetic spectrum.

3. The series of lines in the visible region of the hydrogen spectrum are shown below

4. The groups of lines (known as series) are named after their discoverer.
The groups of lines in the ultra-violet region is called the Lyman series, those in the visible region is called the Balmer series,
while those in the infra-red region is called the Paschen series.

5. (a) (i) Each series consists of sharp discrete lines with fixed frequencies.
(ii) The lines are formed because the energy of the electron is quantised, that is having a fixed amount of energy.
(iii) This is because the electron orbits the nucleus in a fixed energy level.
(iv) These energy levels are characterised by a whole number called the principal quantum number, n.
(v) The greater the energy of the electronic transition, the greater the frequency of the light emitted.

(b) (i) In each series, the intervals between the wavelengths of the lines become smaller and smaller towards the low
wavelength (high frequency) end of the spectrum, and finally merge to form a continuous spectrum.

(ii) The further the energy levels are from the nucleus, the closer the electronic energy levels to each other.
At continum, the electronic energy levels eventually come together and the lines converge /becomes closer.

6. (a) Each series ends with a minimum wavelength (or maximum frequency).
(b) This type of spectrum is called a converging spectrum.

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Explaination of formation of line spectrum of H atom/He+ / Li2+ ion in the Balmer series using Bohr's model.
He+ ion has only one electron in its valence shell.
The electron of the He+ ion at the ground state can absorb energy and get excited to a higher energy level. [1 +1]
The electron is said to be in the excited state. However this excited state is unstable.
The excited electron will eventually drop to a lower energy level, the excess energy will be released in the form of
electromagnetic radiation/light/wave with a specific frequency or wavelength.

Frequency of each line can be determined by the formula ∆E = h

Discrete lines are formed because energy of the electrons is quantised.
Balmer series is caused by electronic transitions from n > 2 to n = 2.
The energy difference between the two energy levels corresponds to the wavelength or frequency of the line in the spectrum
As difference in energy between energy levels become smaller, the lines become convergent/closer.

7. In 1885, J. J. Balmer discovered that the wavelengths of lines in the visible region the Ryberg's equation.

where RH = 1.097 x 107 m-1 (The Ryberg’s constant)
n = integers greater than 2 (i.e. 3,4,5,6…....)

8. The wavelengths of the Lyman series can also be calculated from the Ryberg’s equation which takes the following form.

or :
where C = 3 x 108 m s-1 (velocity of light) n = 2,3,4,5,6…...............

9. The wavelength or frequency of each and every line in the hydrogen spectrum can be calculated using the general form of the
Ryberg’s equation:

3

or :
n1 and n2 are integers known as the principle quantum numbers.
10. The values for n land n 2 for each series is given in the table below.

Series Region nl n2

Lyman Ultra-violet 1 2,3,4,5…...

Balmer Visible 2 3,4,5,6…...

Paschen Infra-red 3 4,5,6,7…...

Bracket Infra-red 4 5,6,7,8…...

2.1.3 Differences between Lyman series and Balmer series

Lyman series Balmer series
1. Lines emitted lies in the ultraviolet region Lines emitted lies in the visible light range
2. Can be used to calculate the ionisation energy of Can be used to determine the frequency of light
(only red, blue, indigo, violet)
hydrogen
In the emission spectrum, electrons from a higher energy level
3. In the emission spectrum, electrons from a higher n > 2 de-excites to n = 2.
energy level of n > 1 de-excites to n = 1
Set of lines with lower frequencies (longer wavelengths).
4. Set of lines with higher frequencies.

EXAMPLE 2.3
Calculate the wavelength of (a) the first line and, (b) the last line, in the Balmer series

ANSWER

(a) Using nl = 2 and n2 = 3

1 = 1.097 x 107 ( )



1 = 1.534 x 106 m-1



 λ = 6.52 x 10-7 m or = 652 nm

b) Using nl = 2 and n2 = 

1 = 1.097 x 107 ( 1 1 )
 22 - 2

1 = 2.74 x 106 m-1



 λ = 3.65 x 10-7 m or = 365 nm

EXAMPLE 2.4

Calculate the frequency of the first line in the Lyman series.

ANSWER

Using n1 = 1 and n2 = 2

v = (3 x 108)(1.097 x 107)( 1 - 1 )= 2.47 x 10 15 Hz
12 22

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Example 2.5

The frequencies of the first four lines in the Lyman Series are given below :
2.56 ; 2.92; 3.08 ; 3.16 (X 1015 Hz)

(a) Plot a suitable graph to determine the frequency of the last line in the Lyman Series.
(b) use your answer in question (a) to calculate ionisation energy of hydrogen.

Answer

Using the equation :

The Rydberg's equation for the Lyman series is :

1 = RH( 1 - 1 )
 12 n2

Or f = CRH( 1 - 1 )
12 n2

Rearranging : f = CRH - CRH ( 1 )
n2

Graph of frequency against 1 will give a straight line where the maximum frequency can be obtained by the intercept at the
n2

frequency axis where n =  and 1 = 0.
n2

f(x 1015) / s-1 2.46 2.92 3.08 3.16

n2345

1 0.25 0.11 0.063 0.040

n2

The maximum frequency is 3.29 X 1015 Hz.

(b) Using the equation : ΔE = h NA

.'. ΔE = (6.63 x 10-34) (3.29 x 1015) X 10-3 x (6.02 x 1023) = 1312.7 kJ mol-1

Alternative method:

The convergence frequency of the Lyman series occurs when the difference in frequency of successive lines (Δf is zero).

Thus, if we plot a graph of frequency (f) against Δf, the intersection on the f axis (where Δf = 0) would give the value of the

convergence frequency.

f(x 1015) / Hz 2.46 2.92 3.08 3.16

Δf(x 1015) / Hz - 0.36 0.16 0.08

Using the equation : ΔE = h NA

.'. ΔE = (6.63 x 10-34) (3.29 x 1015) x 10-3 x (6.02 x 1023) = 1312.7 kJ mol-1

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2.1.4 THE BOHR MODEL OF HYDROGEN ATOM [Or He+(g) or Li2+(g)]

1. In order to explain the origin of the spectra lines of the hydrogen atom, N. Bohr put forward his model of the hydrogen atom in 1913.
2. He referred to M. Planck's quantum theory which states that energy can be absorbed or emitted in fixed amounts called quanta.
3. He postulates that:

(a) The lone electron in the hydrogen atom revolves round the nucleus in fixed circular paths known as orbits with different radii,
just as the planets revolve around the sun.

(b) Each orbit is associated with a fixed amount of energy. The electron can reside in the orbit but cannot exist anywhere between
the orbits.

(c) The energy of the orbits increases as the orbits distance from the nucleus increase.
The orbits are sometimes called energy levels or electronic shells or principle shells.

(d) (i) Each orbit is represented by a integer known as the principle quantum number, n.
(ii) The orbit closest to the nucleus (i.e. the shell or level with the lowest energy) is given the principle quantum number n = 1.
(iii) In a hydrogen atom, the single electron is in the lowest energy level, it is stable and is said to be in the ground state.

(e) The electron of the hydrogen atom at the ground state can be excited to a higher energy level if it absorbs the necessary (light)

energy. The electron is said to be in the excited state.

However this excited state is unstable and the electron will lose energy and will eventually return to the ground state.

(f) In the transition of the electron between a higher energy level and a lower energy level, the excess energy (i.e. the difference

of energy between the two levels concerned, ∆ E) will be released in the form of light.

(g) Since an electron can only be in a specific energy level, therefore the energy absorbed or radiated in each electronic transition

is in a discrete amount. The light emitted with a discrete amount of energy is called a photon.

(h) Each photon of light has a fixed frequency,  , and frequency of each line can be calculated from the equation :

∆ E = h hc
=

where ∆ E = energy difference between the levels
h = Plank's constant (6.63 x 10-34 Js or 3.99 x 10-10 Js mol-1)

 = frequency of light

c = speed of light = 3.00 x 108 m s-1

 = wavelength of light

(i) The energy of the light emitted or absorbed is the difference in the energy between the two electron shells.

(j) Each electronic transition from a higher energy level to a lower energy level corresponds to radiation with a specific frequency

or wavelength.

Energy level are quantised and hence discrete lines are formed.

(k) The lines in the Lyman series are formed when the excited electrons ‘drop back’ from a higher energy level (n = 2,3…) to

the orbit n = 1.

The Balmer series is produced when the excited electrons drop back from higher energy levels (n =3,4, 5..) to the orbit n = 2.

(l) The convergence lines of the hydrogen spectra shows that the difference in energy between successive energy levels.

The energy level become closer and finally converge to form a continuous spectrum as the distance of the orbit from the

nucleus increases.

4. The diagram below shows the formation of the Lyman series, the Balmer series and the Paschen series.

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EXAMPLE 2.6

The difference in energy between the second and third orbits of the hydrogen atom is 3.03 x 10 -19 J. Calculate the frequency of
the emitted light when electron transition takes place between these two levels.

ANSWER

Using the formula: of = 3.031019 = 4.57 x 1014 S - 1
6.63  103 4


lonisation Energy of an Atom

1. All electrons are bounded within the atom by the nucleus.

2. Ionisation energy of an element is defined as the minimum energy required to remove one mole of electrons from one mole

of gaseous atom of the element to form one mole of unipositive gaseous ions.

The equation best describes first ionisation energy is given below:
X(g) X+(g) + e- ; ΔH = + x kJ mol-1

The removal of electrons must begin from the outer shell (valence electrons) inwards.

3. In the event there are more than one electron that is removed from the atom, then the term first ionisation energy, second

ionisation energy, third ionisation energy and so on will be used.

The equation for the second ionisation energy is given below;
X+(g) X2+(g) + e- ; ΔH = + y kJ mol-1

2.2 Sub-shells

1. An energy level with a collection of orbitals is called a shell. A study of the line spectrum shows that the lines
representing transitions between the principle shell are in fact split into finer lines. This indicates the presence of
sub-shells with different energy values in each principle shell.

2. The number of sub-shells in a principle shell is the same as the principle quantum number of the principle shell.

Principle shell n No. of sub-shell Type of subenergy levels No. of orbitals Max. number of electrons

First 1 1 1s 1 2
Second 2 2 2s, 2p 4 8
Third 3 3 3s, 3p, 3d 9 18
Fourth 4 4 4s, 4p, 4d, 4f 16 32
5 5 5s, 5p, 5d, 5f, 5g 25 50
Fifth

3. In any sub-shell, the shell with the lowest energy is represented by the letter s (which stands for sharp), followed
by p (principal), d (diffuse), f(fundamental), g and so on.

4. Further experiments on the line spectra (in the presence of a magnetic field) shows that each sub-shell is further
made up of several orbitals where the electrons are placed.

5. The total number of electrons that can occupy any principle shell (with quantum number n) is 2n2.

6. The number of orbitals depends on the type of sub-shell.

Sub-shell No. of orbitals Orbitals
(Subenergy level)

s1 s

p 3 pX, p y , pZ
d 5 dXy, dyZ, dxz, dX2 - Y2, dZ2

f7 -

7. The arrangement of the orbitals in order of increasing energy is as follows:

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[ 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s and so on.] Note : Refer to your Periodic Table

2.2.1 A T O M I C O R B I T A L S

1. (a) An orbital is defined as a region (or volume) in space around the nucleus where the probability of finding a particular electron is
maximum (high).

(b) An orbit on the other hand, is the circular path of an electron around the nucleus.

2. Using complex mathematics, scientists were able to deduce the shape of the orbitals.
3. (a) The s orbital is spherical (like a round ball) with its nucleus in the centre. The radii of the s orbitals increases with the principle

quantum numbers.
(b) The shaded area represents the region in where the chances of finding the s electron is more than 95 %.

Shapes of the 1s and 2s orbital individually

Shapes of the 1s and 2s orbital collectively
4. The p orbital has a ‘dumb bell’ shape. The three p orbitals are arranged along the x, y and z axes in space.

Shape of the Px orbital

The three p orbitals

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5. The order of the increasing size and energy is 2p < 3p < 4p.
(i) Individually

(ii) Collectively

6. The diagram below shows the 1s, 2s and 2p orbitals in a single atom.

7. There are four sets of quantum numbers to describe an electron orbit.

Quantum number Symbol Description Values
Principal n Size of orbital n = 1,2,3………..
Azimuthal l Shape of orbital 0 = s, 1 = p, 2= d, 3 = f
Magnet m Orientation of space s subshell : m = 0
p subshell : m = -1, 0, +1
Spin s Spin of electron in orbital d subshell : m = -2, -1, 0, +1, +2
ELECTRONIC CONFIGURATIONS
+1 ; -1

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• electrons exist in different energy levels ( n = 1,2,3,4 etc. )
• energy levels can be split into sub levels
• each level contains orbitals (s, p, d, or f)
• each orbital can hold up to 2 electrons

2.3 FILLING OF ELECTRONS
The electronic configuration of an atom can be determined by using the following rules.

2.3.1 The Aufbau Principle
(a) The Aufbau principle states that electrons must occupy orbitals of the lowest energy first before occupying orbitals
of higher energy.
(b) The diagram below shows the arrangement of the orbitals in order of increasing energy.
(c) The three boxes representing the three p orbitals are joined indicating that they have the same energy,
that is they are degenerated.
The same is true for the five d orbitals.

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2.3.2 Pauli Exclusion Principle

(a) The Pauli exclusion principle states that each orbital can be occupied by a maximum of two electrons of opposite spins
only.
The spins of the electrons are represented by and .

(b) Hence, the total number of electrons in an orbital or set of orbitals is as follows:

Type of orbital Maximum no. of electrons

S 2x1=2
p 2x3=6
d 2x5= 10

f 2x7=14
The total number of electrons th at can occu py a prin ci ple shell is 2 n 2.

First 1 2
Second 2 8
Third 3 18
Fourth 4 32
5 50
Fifth

2.3.3 Hund's Rule
(a) In a given set of degenerate orbitals (orbitals of equivalent energy) (e.g. the three p orbitals or the five d orbitals),
electrons tend to occupy the orbitals singly first and are in parallel spins before pairing up.

(b) For example, the two electrons in a p orbital is and not

(c) This is because, with such an arrangement, the electrons are at maximum distance of separation and the repulsive

forces between the electrons are minimum.

Filling energy levels

Rules 1. Electrons enter the lowest energy orbital available.
2. Orbitals can hold a max. of 2 electrons provided they have opposite spin.
3. Orbitals of the same energy remain singly occupied before pairing up.
4. This is due to the repulsion between electron pairs.

Order Orbitals are not filled in numerical order (e.g. 4s is filled before 3d)

2.3.4 Working out electronic configuration of the elements in the Periodic Table

1. Hydrogen (Z = 1) has only one electron, Therefore, it will occupy the lowest energy level/orbital available, that is, the 1s orbital.
The electronic configuration of hydrogen is written as 1s1.

1s
2. The electronic configuration of helium (Z = 2) is written as 1s2 (or 2).

1s
According to Pauli’s Exclusion principle, each orbital can occupy a maximum of two electrons with opposite spins, therefore
both the electrons will occupy the 1s orbital, but with opposite spins.
3. The lithium atom (Z = 3) has three electrons. The first two go into the 1s orbital.
The third electron will occupy the next available empty orbital of 2s based on Aufbau’s principle.

The electronic configuration is written as 1s2 2s1 (or 2.1).
1s 2s
4. Beryllium (Z =4) has the electronic configuration

This is written as 1s2 2s2 (or 2.2).
5. Boron (Z = 5) has five electrons and are filled into orbitals with the lowest energy first based on Aufbau’s Principle.

Energy level of increasing energy is 1s < 2s < 2p.
The first four go into the 1s and 2s orbitals while the fifth electron occupies one of the empty 2p orbitals.

10

The electronic configuration is written as 1s2 2s2 2p1 (or 2.3).
6. Carbon (Z = 6) has six electrons. The 5th and the 6th electrons will occupy the 2p orbital based on Hund’s rule.

Since the three 2p orbitals are degenerate, therefore the 5th electron can occupy the 2px, 2py or 2pz orbital.
The 6th electron cannot occupy the same orbital as the fifth electron, but instead occupies one of the two remaining empty 2p
orbitals with parallel spin (Hund's Rule)

and not

1s 2s 2p 1s 2s 2p
The electronic configuration written as ls2 2s2 2p2 (or 2.4).

7. Nitrogen (Z = 7) has the electronic configuration

The electronic configuration written as 1s2 2s2 2p3 (or 2.5). The p orbitals are half-filled.
According to Hund’s rule, in the filling of the 2p orbitals, three electrons are filled singly into each of the p-orbitals as this is a
stable electronic configuration as electrons will have the maximum number of parallel spins.
8. Oxygen (Z = 8) has eight electrons.
The electrons are filled into orbitals with the lowest energy first based on Aufbau Principle that is in increasing energy
of 1s < 2s < 2p.
The filling of electrons in the degenerate 2p orbitals are based on Hund’s rule, that is, the electrons are filled singly and in
parallel spins before the 8th electron is paired up in any of the half-filled 2p orbitals.

The electronic configuration written as 1s2 2s2 2p4 .
9. The electronic configuration of argon (Z = 18) is written as 1s2 2s2 2p6 3s2 3p6 (or 2.8.8).

1 s 2s 2p 3s 3p

The p orbitals are completely filled. This is a very stable electronic configuration.

10. The next element potassium (Z = 19) has nineteen electrons . The electrons are filled according to Aufbau Principle, where

electrons are filled into orbital of lowest energy level first that is 1s followed by 2s, 2p,3s, 3p and 4s.

The nineteenth electron will occupy the 4s orbital and not the 3d orbital because the 4s orbital is of lower energy than that

of the 3d orbital.

The electrons are also filled according to Pauli’s Exclusion principle where each orbital can occupy a maximum of two

electrons with opposite spins

The electrons that are filled in the degenerate 2p orbitals and degenerate 3p orbitals follows Hund’s rule where they are filled

singly and in parallel spins before pairing occurs.

The electronic configuration written as 1s2 2s2 2p6 3s2 3p6 4s1 (or 2.8.8.1).
11. The electronic configuration of the first 36 elements is shown in the following table.

This is known as the ground state electronic configuration that is, the configuration with the minimum energy possible.

2.3.5 Electron configuration

The electron configuration for an element is a representation of the positions of the atom's electrons in the various atomic
orbitals, s-, p-, d-, etc., see below,

Atom Electron configuration Atom Electron configuration
H 1s1 K 1s22s22p63s23p64s1
He 1s2 Ca 1s22s22p63s23p64s2
Li 1s22s1 or [He]2s1 Sc 1s22s22p63s23p63d14s2
Be 1s22s2 or [He]2s2 Ti 1s22s22p63S23p63d24s2
B 1s22s22p1 or [He]2s22p1 V 1s22s22p63s23p63d34s2
C 1s22s22p2 or [He]2s22p2 Cr 1s22s22p63S23p63d54s1
N 1s22s22p3 or [He]2s22p3 Mn 1s22s22p63s23p63d54s2
O 1s22s22p4 or [He]2s22p4 Fe 1s22s22p63s23p63d64s2
F 1s22s22p5 or [He]2s22p5 Co 1s22s22p63S23p63d74s2
Ne 1s22s22p6 or [He]2s22p6 Ni 1s22s22p63s23p63d84s2

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Na 1s22s22p63s1 or [Ne]3s1 Cu 1s22s22p63s23p63d104s1
Mg 1s22s22p63s2 or [Ne]3s2 Zn 1s22s22p63s23p63d1°4s2
Al 1s22s22p63s23p1 or [Ne]3s23p1 Ga 1s22s22p63s23p63d104s24p1

Si 1s22s22p63s23p2 or [Ne]3s23p2 Ge 1S22s22p63s23p63d1°4s24p2

P 1s22s22p63s23p3 or [Ne]3s23p3 As 1S22s22p63s23p63d1°4s24p3

S 1s22s22p63s23p4 or [Ne]3s23p4 Se 1s22s22p63s23p63d1°4s24p4

Cl 1s22s22p63s23p5 or [Ne]3s23p5 Br 1s22s22p63s23p63d104s24p5

Ar 1s22s22p63s23p6 or [Ne]3s23p6 Kr 1s22s22p63s23p63d1°4s24p6

N.B.: As can be seen above it is sometimes acceptable to write a shorthand version of the electron configuration, by using
"[noble gas]" to represent the electron configuration of the nearest previous noble gas.

2.3.6 Electronic configuration of d-block elements.

1. For the electronic configuration of the transition elements
(Z = 21 to Z = 30) the energy level of the 4s orbital is lower
than that of the 3d orbitals. However, once the electron/s
is/are filled into the 3d orbital, the order is reversed.
The 3d orbitals now have lower energy than the 4s orbital.

Therefore the electronic configuration of scandium (Z = 21) is
1s2 2s2 2p6 3s2 3p6 3d1 4s2 (or 2.8.9.2) and not 1s2 2s2 2p6 3s2 3p6 4s2 3d1

The first electron ionised off from the scandium atom is from the 4s orbital and not the 3d orbital.
2. Orbitals that are fully filled (eg. s2, p6 or d10) or are half-filled (eg. p3 or d5) have extra stability due to their

symmetrical charge (or electron) distribution. This is evident in the electronic configuration of 24Cr or 29Cu.
3. The electronic arrangement of chromium (Z = 24) is [Ar] 3d5 4s1 and not [Ar] 3d4 4s2

This is because an electronic configuration of a half-filled set of 3d orbitals (d5) is more stable energetically than a
partially-filled set of 3d orbitals (d4).

energetically more stable energetically less stable as electrons are not symmetrically distributed
4. The electronic configuration of copper (Z = 29) is [Ar] 3d104s1 and not [Ar]3d9 4s2
This is because a completely-filled 3d orbitals (d10) is more stable energetically than a partially-filled set of 3d orbitals (d9).

energetically more stable energetically less stable

2.3.7 Electronic Configuration of Ions

1. In the formation of cations (positive ions), electrons are removed in reverse order from of electron filling.
That is, the last electron is removed first.

2. The electronic configuration of oxygen (Z = 8) is

The first electron to be removed is from the paired electron in the 2p orbital.
Thus, the electronic arrangement for the O+ ion is

This is written as 1s2 2s2 2p3 (or 2.5)

3. In the formation of anions (negative ions), electrons are added in the same manner as the filling of electrons in the neutral atoms.

4. The electronic configuration of fluorine is

The fluoride ion, F- will have the following configuration :

EXAMPLE 2.7
Write the electronic configuration of the following ions.
(a) Mg2+ (b) Cu2+ (c) Fe2+ (d) P 3 - ( e ) Cl-

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ANSWER (4s electrons are furthest from the nucleus. Thus, they will be removed first)
(a) 1s2 2s2 2p6 (4s electrons are furthest from the nucleus. Thus, they will be removed first)
(b) 1s2 2s2 2p6 3s2 3p6 3d9
(c) 1s2 2s2 2p6 3s2 3p6 3d6
(d) 1s2 2s2 2p6 3s2 3p6
(e) 1s2 2s2 2p6 3s2 3p6

2.4 ELECTRONIC CONFIGURATION AND THE PERIODIC TABLE

1. At present, there are 92 naturally-occurring elements. Elements with proton numbers greater than 92 are all artificially

synthesised and are radioactive.

2. All elements in the Periodic Table are grouped under 7 Periods (the horizontal rows) and 18 Groups (the vertical columns).

3. Elements in Group 1 (or I) and Group 2 (or II) are known as the s-block elements because their valence electrons occupy the s

orbitals.

Group 1 3Li 11Na 19K 37Rb 55Cs 87Fr

Valence shell configuration 2s1 3s1 4s1 5s1 6s1 7s1

Group 2 4Be 12Mg 20Ca 38Sr 56Ba 88Ra
Valence shell configuration 2s2 3s2 4s2 5s2 6s2 7s2

4. Elements in Group 13 (or III) to Group 18 (or VIII) are known as the p block elements as their valence electrons occupy the p

orbitals.

Group 13 14 15 16 17 18
(III) (IV) (V) (VI) (VII) (VIII)

Element 13Al 14Si 15P 16S 170 18Ar
3s2 3p1 3s2 3p2 3s2 3p3 3s2 3p4 3s2 3p5 3s2 3p6
Valence shell
configuration

5. Elements in Group 3 through Group 12 are known as the d block elements where filling of electrons involves the d orbitals.

Group 3 4 5 6 7 8 9 10 11 12

Element 21Sc 22Ti 23V 24Cr 25Mn 26Fe 27Co 28Ni 29Cu 30Zn
3 d 1 3 d 2 3 d 3 3 d 5 3 d 5 3 d 6 3 d 7 3 d 8 3 d 10 3 d 10
Valence shell 4s2 4s2 4s2 4s1 4s2 4s2 4s2 4s2 4s1 4s2
configuration

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

H He
1 2

1

Li Be B C N O F Ne
2 4 5 6 7 8 9 10

3

Na Mg Al Si P S Cl Ar
3 13 14 15 16 17 18

11 12

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
4

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
5

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

6 Cs Ba * Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

55 56 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

7 Fr Ra ** Rf Db Sg Bh Hs Mt Uun Uuu Uub

87 88 104 105 106 107 108 109 110 111 112

* La Ce Tb Dy Ho Er Tm Yb Lu
57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

** Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
89 90 91 92 93 94 95 96 97 98 99 100 101 102 103

Element Groups (Families)
Alkali Earth Alkaline Earth Transition Metals
Rare Earth Other Metals Metalloids
Non-Metals Halogens Noble Gases

6. Elements with proton numbers 58 → 71 and from 90 → 103 are known as the f-block elements. It consists of two series – the
Lanthanide and the Actinide series. Most of the elements in the Actinide series are synthesized by chemists.

Proton number 5 8 59 60 61 62 63 64 65

Element Ce Pr Nd Pm Sm Eu Gd Tb
Electronic 4f 15d1 4f 15d2 4f 45d0 4f 55d0 4f 65d0 4f 75d0 4f 75d1 4f 95d0
configuration
6s2 6s2 6s2 6s2 6s2 6s2 6s2 6s2
Proton number 70 71
66 67 68 69 Yb Lu
Element 4f145d0 4f145d1
Dy Ho Er Tm 6s2 6s2
Electronic
configuration 4f 105d0 4f 115d0 4f 125d0 4f135d0
6s2 6s2 6s2 6s2

7. The Group number of an element indicates the number of valence electrons while the Period number indicates the outermost
principle shell that is filled with electrons.

8. For example, Bromine (proton number = 35) is found in the 4th Period and Group 17 of the Periodic Table.
Hence, its valence shell configuration is 4s2 4p5

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