Ramallah Friends School

Chemistry – 11 IB HL
Unit Two HL: Atomic Structure – HL

Ionization Energy (I.E)

The minimum energy required to remove one electron from outermost shell of an isolated gaseous atom.
Or
The minimum energy required to remove one mol of electrons from the outermost shells of one mol of
isolated gaseous atoms.
X(g)  X+(g) + e- First I.E

The unit of ionization energy is KJ mol-1

Trend of ionization energy across period and down group

Across period:
As we proceed from left to right within a period, the ionization energy increases because electrons are added
to the same energy level, the number of protons or effective nuclear charge increases. So, the radius
decreases, the electron removed is closer to the nucleus and there will be more nuclear attraction. Thus
needs more energy to remove electron.

Down group:
As we proceed down within a group the ionization energy decreases because the number of energy levels
increases, the electron to be removed becomes further from the nucleus and is repelled by more complete
electron shells or more inner electrons. As we proceed down, the shielding effect increases. So it will
experience less effective nuclear charge, less nuclear attraction and easier to remove.

Removal of Successive Electrons from an Atom:

1- If the next removed electron belongs to the outer energy level then this electron experiences more
attraction from the nucleus because the atom becomes more positively charged. The ionization energy for
removing this electron will be higher.

Al (g)  Al+1 (g) + e- First I.E

Al+1 (g)  Al+2 (g) + e- Second I.E
Al+2 (g)  Al+3 (g) + e- Third I.E

2- If the next removed electron is from an inner energy level that is closer to the nucleus then it will be more
strongly attracted and more difficult to remove. The ionization energy for removing this electron will be
higher. So, there will be a sudden large rise in ionization energy.

Question: Define the second ionization energy of Mg and represent it in the form of an equation.

Second ionization energy of Mg:

The amount of energy needed to remove an electron from the gaseous ion (Mg+1) to form the gaseous ion
(Mg+2).
Mg+1 (g)  Mg+2 (g) + e- 2nd I.E

◘ The more electrons that have been removed from an atom, the greater the energy required to remove the
next electron because the next removed electron experiences more attraction from the nucleus because the
atom becomes more positively charged. The ionization energy for removing this electron will be higher.

1ST I.E < 2nd I.E < 3rd I.E ……..

◘ Sometimes the next electron must be removed from a filled inner (lower) energy level, so that this electron
will be closer to the nucleus and more strongly attracted by the nucleus and more difficult to remove and

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Trend of Ionization Energy for the first 20 elements:

-A graph of first ionization energies plotted against atomic number shows periodicity or a repeating pattern.
Periodicity means a repeating pattern of physical and chemical properties. It can be seen that the highest
value is for helium, an atom that contains two protons and two electrons. The two electrons are in the lowest
level and are held tightly by the two protons.

-For lithium it is relatively easy to remove an electron, which suggests that the third electron in lithium is in a
higher energy level than the first two. The value then generally increases until element 10, neon, is reached
before it drops sharply for sodium which suggests that the 11 th electron is removed from a higher energy
level. The value then generally increases until element 18, argon, is reached before it drops sharply for
potassium which suggests that the 19th electron is removed from a higher energy level.

-The general increase in ionization energy as we proceed from left to right within a period and having the
highest values of ionization energies for the noble gases and having the lowest values of ionization energies
for alkali metals, provides evidence for the existence of the main energy levels and also provides
evidence that the levels can contain different numbers of electrons before they become full.

- Exceptions of ionization energies: the above graph dips slightly for boron, aluminum, oxygen and sulfur.

1-The ionization energy of B is slightly less than the ionization energy of Be.
2 2 2 2 1
4Be: 1s 2s 5B:1s 2s 2P

Valence shell of Be is 2s2 while in B is 2s2 2p1. The outer electron of B is in 2p while the outer electron of Be
is in 2s. The outer electron of B is higher in energy because the 2p sub–level has a higher energy than the
2s sub - level and also the outer electron of B is further from nucleus, shielded by 2s electrons and weakly
attracted to the nucleus. Hence the outer electron of B is easier to remove. This provides evidence for the
existence of sub - levels

2-The ionization energy of Al is slightly less than the ionization energy of Mg.

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 12 Mg: 1s2 2s2 2p6 3s2 13 Al:1s2 2s2 2P6 3s2 3p1

Valence shell of Mg is 3s2 while Al is 3s2 3p1. The outer electron of Al is in 3p while the outer electron of Mg
is in 3s. The outer electron of Al is higher in energy because the 3p sub – level has a higher energy than the
3s sub - level and also the outer electron of Al is further from nucleus, shielded by 3s electrons and weakly
attracted to the nucleus. Hence the outer electron of Al is easier to remove. This provides evidence for the
existence of sub – levels.

3-The ionization energy of O is slightly less than the ionization energy of N.

7 N: 1s2 2s2 2p3 8 O:1s2 2s2 2P4

Valence shell of N is 2s2 2p3 while O is 2s2 2p4. In 2p3 of N the electrons are arranged singly while in 2p 4 of O
the electron removed is paired. There is a greater repulsion between paired electrons in O due to two
electrons in the same p orbital. This reduces the nuclear attraction and therefore it is easier to remove the
electron. The ionization energy decreases. This provides evidence for the existence of orbitals in sub-levels.

4-The ionization energy of S is slightly less than the ionization energy of P.

15 P: 1s2 2s2 2p6 3s2 3P3 16 S:1s2 2s2 2P6 3s2 3p4

Valence shell of P is 3s2 3p3 while S is 3s2 3p4. In 3p3 of P the electrons are arranged singly while in 3p 4 of S
the electron removed is paired. There is a greater repulsion between paired electrons in S due to two
electrons in the same p orbital. This reduces the nuclear attraction and therefore it is easier to remove the
electron. The ionization energy decreases. This provides evidence for the existence of orbitals in sub-levels.

Graphs of Successive ionization energies for the same element:

Question: Draw a graph of successive ionization energies for sodium against number of electrons
lost. Give your reasoning.

By counting back to the first large jump you can determine the number of electrons in the outer energy level
and hence the group in the Periodic Table to which the element belongs.

Explanation of the graph:

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1-The electronic configuration of Na = 1s22s22p63s1. The first value on the graph is due to removal of the 3s 1
electron. The next eight values are due to the removal of the eight electrons of the second energy level,
2s22p6. The last two values are due to the removal of the two electrons of the first energy level, 1s2.

2-The more electrons that have been removed, the more positive the nucleus becomes. The remaining
electrons will experience more electrostatic attraction from the nucleus and each main energy level becomes
closer to the nucleus. For that, there is a continuous increase in the ionization energy as more electrons are
removed from the atom.

3-The first electron is the easiest to remove as it is furthest from nucleus. There is a large increase between
first and second ionization energies as electron now removed from n = 2. The eight electrons of 2s 22p6 are
more difficult to remove than 3s1 electron because they are in a lower energy level and hence closer to the
nucleus and exposed to greater effective nuclear charge. These eight electrons show relatively small
increase in ionization energy as these electrons are in the same energy level (n = 2).

4-There is a large increase between 9 th and 10th ionization energy as electron now removed from n = 1. The
10th electron is closer to the nucleus and not shielded by inner electrons and hence exposed to a greater
effective nuclear charge and hence more difficult to remove. Electron 11 also comes from 1s, so shows a
small increase in its ionization energy.

Evidence for the presence of three main energy levels in sodium:

There is a large increase in ionization energy when the second and tenth electrons are removed. Thus, the
first electron is further from nucleus than the second electron and the ninth electron is further from nucleus
than the tenth electron. The large increase in ionization energy indicates changes in main energy levels. This
proves the existence of three main energy levels.

Evidence for sub-levels:

As more electrons are removed the pull of the protons holds the remaining electrons more tightly so
increasingly more energy is required to remove them, hence a logarithmic scale is usually used. If the graph
is examined more closely then it can be seen that graph does not increase regularly. This provides evidence
that the main energy levels are split into sub-levels.

Value of Ionization energy:

When sufficient energy is supplied to the atom or to the electron in order to move the electron to the n = ∞,
the electron will be lost completely, the electron is no longer under the influence of nucleus and the hydrogen
atom will be ionized to produce H+1. by knowing the exact frequency at which convergence takes place, the
I.E could be measured and determined. The ionization energy is equal to the energy difference between the
first level, n = 1, and the infinite level, n = ∞.

H(g) → H+1(g) + e-1

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Ionization energies from emission spectra:

It can be seen from the emission spectrum of hydrogen that the energy levels converge. Hydrogen contains
just one electron, which will be in the lowest energy level in its ground state. If sufficient energy is supplied, it
can be promoted to the infinite level- that it is has been removed from the atom and the atom has become
ionized to form H+1 ion. This amount of energy corresponds to the energy it would emit if it fell back
from n= infinity to n= 1, which produces a convergence line in the ultraviolet region of the spectrum
at a wavelength of 91.2 nm (9.12 x 10-8 m). We can use this value to calculate the energy involved.

The electromagnetic spectrum:

Electromagnetic waves can travel through space or matter. The velocity of travel, c, is related to its
wavelength, λ, and its frequency, v. Velocity (c) is measured in m s-1, wavelength in m and frequency in s-1.

c=λ×v

Electromagnetic radiation is a form of energy. The energy of electromagnetic radiation is related to its
frequency, v, and to a constant known as Planck’s constant, h:

E=hXv

Units:
v :s-1 h: J.s E: J

v=c/λ

E=hc/λ

∆E = E∞ - E1 = I.E = hc/λ

Ionization energy per electron, in Joules = h X v = hc/λ

Ionization energy per mol of electrons, in J mol-1 = I.E per electron in Joules x 6.02 x 1023
Ionization energy per mol of electrons, in KJ mol-1 = I.E in J mol-1/1000

I.E in J x 6.02 x 1023 → J mol-1 / 1000 → KJ mol-1

KJ mol-1 x 1000 → J mol-1 /6.02 x 1023 → J

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The smaller the wavelength, the higher the frequency, and the more energy the wave possesses.

1 m = 1 x 109 nm
1 nm = 1 x 10-9 m
1 KJ = 1000 J

Example (1)
Determine the energy, in J, of a photon of red light given that the wavelength = 650.0 nm. (Use the
information given in sections 1 and 2 of IB data booklet).

E = hv = hc/λ = [(6.63 x 10-34 J s) x 3.00 x 108 m s-1)] = 3.06 X 10-19 J

[650.0 x 10-9 m]

Example (2)
Calculate the first ionization energy, in KJ mol-1, for hydrogen given that its shortest wavelength in the
Lyman series is 91.2 nm. (Use the information given in sections 1 and 2 of IB data booklet).

IE1 = hv = hc/λ = [(6.63 x 10-34 J s) x 3.00 x 108 m s-1)] = 2.179 X 10-18 J

[91.2 x 10-9 m]

IE1 in KJ mol-1:

IE1 = (2.179 X 10-18 J) x (6.02 X 1023 mol-1) = 1312 KJ mol-1

1000

Example (3)
The first ionization energy of hydrogen is 1312 KJ mol-1. Determine the frequency and wavelength of the
convergence line in the ultraviolet emission spectrum of hydrogen. (Use the information given in sections 1
and 2 of IB data booklet).

IE for one electron

IE1 = (1312 KJ mol-1) x (1000) = 2.179 X 10-18 J

(6.02 X 1023 mol-1)

E = hv
Frequency = v = E/h = (2.179 X 10-18 J) = 3.29 x 1015 s-1
(6.63 x 10-34 J s)

c= vλ

Wavelength = λ= c/v = (3.00 X 108 m s-1) = 9.12 x 10-8 m = 91.2 nm

(3.29 x 1015 s-1)

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IB Questions:
Q(1) The graph below represents the successive ionization energies of sodium. The vertical axis plots log
(ionization energy) instead of ionization energy to allow the data to be represented without using an
unreasonably long vertical axis.

a- State the full electron configuration of sodium and explain how the successive ionization energy
data for sodium are related to its electron configuration.

b- Explain how successive ionization energies account for the existence of three main energy levels
in the sodium atom.

Answer:
a- Na = 1s22s22p63s1
The first electron is the easiest to remove as it is furthest from nucleus. There is a large increase between
first and second ionization energies as electron now removed from n = 2. The eight electrons of 2s 22p6 are
more difficult to remove than 3s1 electron because they are in a lower energy level and hence closer to the
nucleus and exposed to greater effective nuclear charge. These eight electrons show relatively small
increase in ionization energy as these electrons are in the same energy level (n = 2). There is a large
increase between 9th and 10th ionization energy as electron now removed from n = 1. The 10 th electron is
closer to the nucleus and not shielded by inner electrons and hence exposed to a greater effective nuclear
charge and hence more difficult to remove. Electron 11 also comes from 1s, so shows a small increase in its
ionization energy.

b- There is a large increase in ionization energy when the second and tenth electrons are removed. Thus,
the first electron is further from nucleus than the second electron and the ninth electron is further from
nucleus than the tenth electron. The large increase in ionization energy indicates changes in main energy
levels. This proves the existence of three main energy levels in sodium.

Q(2)

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a- Explain why the first ionization energy of aluminum is less than that of magnesium.
b- Explain why the first ionization energy of sulfur is less than that of phosphorus.

Answer:

a-
12 Mg: 1s2 2s2 2p6 3s2 13 Al:1s2 2s2 2P6 3s2 3p1

Valence shell of Mg is 3s2 while Al is 3s2 3p1. The outer electron of Al is in 3p while the outer electron of Mg
is in 3s. The outer electron of Al is higher in energy because the 3p sub – level has a higher energy than the
3s sub - level and also the outer electron of Al is further from nucleus, shielded by 3s electrons and weakly
attracted to the nucleus. Hence the outer electron of Al is easier to remove. This provides evidence for the
existence of sub – levels.
b-
2 2 6 2 3 2 2 6 2 4
15P: 1s 2s 2p 3s 3P 16S:1s 2s 2P 3s 3p

Valence shell of P is 3s2 3p3 while S is 3s2 3p4. In 3p3 of P the electrons are arranged singly while in 3p 4 of S
the electron removed is paired. There is a greater repulsion between paired electrons in S due to two
electrons in the same p orbital. This reduces the nuclear attraction and therefore it is easier to remove the
electron. The ionization energy decreases. This provides evidence for the existence of orbitals in sub-levels.

Q(3) Magnesium is the eighth most abundant element in the earth’s crust. The successive ionization
energies of the element are shown below.

a- Explain the general increase in successive ionization energies of the element

b- Explain the large increase between the tenth and eleventh ionization energies.

Answer:

a- The more electrons that have been removed, the more positive the nucleus becomes. The remaining
electrons will experience more attraction from the nucleus and each main energy level becomes closer to the
nucleus. For that, there is a continuous increase in the ionization energy as more electrons are removed
from the atom.

b- The 11th electron is removed from the first energy level while the 10 th electron is removed from the second
energy level. The 11th electron is closer to the nucleus and not shielded by inner electrons and exposed to
greater effective nuclear charge.

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Q(4)The graph of the first ionization energy plotted against atomic number for the first twenty elements
shows periodicity.

a- Explain how information from this graph provides evidence for the existence of main energy levels
and sub-levels within atoms.

b- State what is meant by the term second ionization energy.

c- Sketch and explain the shape of the graph obtained for the successive ionization energies of
potassium using a logarithmic scale for ionization energy on the y-axis against number of electrons
removed on the x-axis.

Answer:

a- The general increase in ionization energy as we proceed from left to right within a period and having the
highest values of ionization energies for the noble gases and having the lowest values of ionization energies
for alkali metals, provides evidence for the existence of the main energy levels.
The drop in ionization energy from Be to B and from Mg to Al, and the drop in ionization energy from N to O
and from P to S, provides evidence for the existence of sub – levels.

b- The minimum energy needed to remove an electron from X+1 gaseous ion to form X+2 gaseous ion.

X+1 (g) → X+2 (g) + e Second I.E

3-

The electronic configuration of K = 1s22s22p63s23p64s1. The first value on the graph is due to removal of the
4s1 electron. The next eight values are due to the removal of the eight electrons of the third energy level,

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3s23p6. The next eight values are due to the removal of the eight electrons of the second energy level,
2s22p6, and last two values are due to the removal of the two electrons of the first energy level, 1s2.

The more electrons that have been removed, the more positive the nucleus becomes. The remaining
electrons will experience more attraction from the nucleus and each main energy level becomes closer to the
nucleus. For that, there is a continuous increase in the ionization energy as more electrons are removed
from the atom. If the next removed electron is removed from a lower energy level, then this electron will be
much closer to the nucleus and more strongly attracted by the nucleus and more difficult to remove and there
will be a sudden large rise in ionization energy.

It is relatively easy to remove the first electron. After that there is a jump in value when the second electron is
removed. After removing the eight electrons of the third energy level, a second jump is encountered when
the tenth electron is removed. After removing the eight electrons of the second energy level, a third jump is
encountered when the eighteenth electron is removed. This proves the existence of four main energy levels
in potassium.

Q(5)
a- Describe how the first four ionization energies of aluminum vary. (You may wish to sketch a graph
to illustrate your answer.)

b- State the electronic configurations of aluminum and boron. Explain how the first ionization energy
of aluminum compares with the first ionization energy of boron.

Answer:
a-

b-

5 B: 1s22s22p1 13 Al: 1s22s22p63s23p1

The removed electron in Al is in a higher energy level than that of B and further from the nucleus of Al and is
repelled by more complete electron shells or more inner electrons. The shielding effect is higher in Al. The
removed electron in Al experiences less effective nuclear charge and less nuclear attraction and easier to
remove. The ionization energy of Al is lower than that of B.

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Q(6) The successive ionization energies of germanium are shown in the following table:

1st 2nd 3rd 4th 5th

Ionization 760 1540 3300 4390 8950
energy / KJ mol-1

a- Identify the sub-level from which the electron is removed when the first ionization energy of germanium is
measured.

b- Write an equation, including state symbols, for the process occurring when measuring the second
ionization energy of germanium.

c- Explain why the difference between the 4th and 5th ionization energies is much greater than the difference
between any two other successive values.

Answer:

a- The electronic structure of germanium (Z = 32) is 1s22s22p63s23p64s23d104p2

The sub-level is 4p.

b- Ge+1 (g) → Ge+2 (g) + e

c- The 4th ionization energy corresponds to the removal of the 4 th electron from the 4th energy level (4s sub-
level) while the 5th ionization energy corresponds to the removal of the 5 th electron from the 3rd energy level
(3d sub-level). The 5th electron is removed from an energy level that is lower and closer to the nucleus and
as a result it will be more strongly attracted by the nucleus and more difficult to remove.

Q(7) The first four ionization energies (kJ mol-1) for a particular element are 550, 1064, 4210 and 5500
respectively. This element should be placed in the same Group as

a- Li
b- Be (the correct answer)
c- B
d- C

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