Ionisation Energy

AS · 14 min

Ionisation energy is the most direct evidence we have for the shell and sub-shell structure of atoms. By measuring how much energy it takes to pull electrons off an atom one at a time, chemists can see where shells begin and end, which group an element is in, and even the small effects of sub-shells and electron pairing. It is a reliable source of explanation marks in Paper 2: almost every question asks you to explain a trend using the same four factors, so this note shows you exactly how to phrase them.

What ionisation energy measures

An electron in an atom is held by its attraction to the positive nucleus. Removing it means doing work against that attraction, so ionisation always requires energy: ionisation energies are always positive (endothermic).

Definition

The first ionisation energy, $IE_1$ , of an element is the energy required to remove one electron from each atom in one mole of gaseous atoms of the element to form one mole of gaseous $1+$ ions.

\ce{Na(g) -> Na+(g) + e-} \qquad IE_1 = +494\ \text{kJ mol}^{-1}

Successive ionisation energies remove further electrons, one at a time, from the gaseous ion:

Key result

First: $\ce{X(g) -> X+(g) + e-}$
Second: $\ce{X+(g) -> X^2+(g) + e-}$
Third: $\ce{X^2+(g) -> X^3+(g) + e-}$
$n$ th: $\ce{X}^{(n-1)+}\text{(g)} \longrightarrow \ce{X}^{n+}\text{(g)} + \ce{e-}$

Watch out

Three mistakes cost the equation mark most often:

Missing state symbols. Every species must be (g): ionisation energies are defined for gaseous atoms and ions.
Writing the second ionisation as $\ce{X(g) -> X^2+(g) + 2e-}$ . That is the sum of the first and second ionisation energies, not the second. The second IE starts from $\ce{X+(g)}$ and removes one electron.
Writing the electron as $\ce{e}$ without its charge. Write $\ce{e-}$ .

The four factors

Every ionisation energy explanation is built from these factors. The ionisation energy is large when the outer electron is strongly attracted to the nucleus.

Key result

Nuclear charge. More protons means a stronger attraction for the electron being removed. IE increases.
Atomic (or ionic) radius. The further the electron is from the nucleus, the weaker the attraction. IE decreases.
Shielding. Electrons in inner shells (and, to a smaller extent, inner sub-shells) repel the outer electron and reduce the attraction it feels. More shielding, lower IE.
Spin-pair repulsion. An electron that shares an orbital with another electron is repelled by its partner, so it is easier to remove. IE decreases.

Exam answers name the factor and say what it does to the attraction: "the outer electron is further from the nucleus and more shielded, so it is less strongly attracted by the nucleus, so less energy is needed to remove it."

Trend down a group

First ionisation energy decreases down a group.

Group 1	Li	Na	K	Rb	Cs
$IE_1$ / kJ mol $^{-1}$	519	494	418	403	376

Group 2	Be	Mg	Ca	Sr	Ba
$IE_1$ / kJ mol $^{-1}$	900	736	590	548	502

Down a group, each element has an extra shell. The outer electron is further from the nucleus and is shielded by more inner shells. The nuclear charge increases too, but its effect is outweighed by the greater distance and shielding. The outer electron is attracted less strongly, so less energy is needed to remove it.

Trend across a period

Across a period, first ionisation energy shows a general increase, with two small dips.

First ionisation energies from hydrogen to calcium (data from the 9701 data booklet). Peaks at the noble gases, sharp drops to the Group 1 metals, and small dips at boron and aluminium (Group 13) and at oxygen and sulfur (Group 16).

The general increase

Across a period the nuclear charge increases by one each step. The electrons are added to the same shell, so the distance from the nucleus decreases slightly and the shielding by inner shells stays about the same. The outer electrons are attracted more strongly, so the ionisation energy increases. The noble gas at the end of each period has the highest value.

The big drop from a noble gas to the next alkali metal

From neon (2080) to sodium (494), or argon (1520) to potassium (418), the ionisation energy falls sharply. The electron removed from sodium is in a new shell (3s), further from the nucleus and shielded by two full inner shells.

First dip: Group 2 to Group 13 (Be to B, Mg to Al)

\text{Mg}: [\text{Ne}]\,3s^2 \qquad \text{Al}: [\text{Ne}]\,3s^2 3p^1

Aluminium's outer electron is in a 3p sub-shell, which is higher in energy than 3s and slightly further from the nucleus. It is also partly shielded by the 3s electrons. So, despite the extra proton, it needs less energy to remove: $IE_1(\text{Al}) = 577 < IE_1(\text{Mg}) = 736\ \text{kJ mol}^{-1}$ .

Second dip: Group 15 to Group 16 (N to O, P to S)

atom	3s	3p	3p	3p
P	↑↓	↑	↑	↑
S	↑↓	↑↓	↑	↑

In phosphorus the three 3p electrons are each in a separate orbital. In sulfur the fourth 3p electron must go into an orbital that already contains one electron. The two electrons in that orbital repel each other (spin-pair repulsion), so one of them is easier to remove. That is why $IE_1(\text{S}) = 1000 < IE_1(\text{P}) = 1060\ \text{kJ mol}^{-1}$ .

Watch out

For the Group 15 to 16 dip, do not write "half-filled sub-shells are extra stable" as your whole answer. The examiner wants the mechanism: in S (or O) the electron removed is paired in a 3p (or 2p) orbital, and repulsion between the paired electrons makes it easier to remove.

Successive ionisation energies

For any element, each successive ionisation energy is larger than the one before. After each electron is removed:

the remaining electrons are held by the same number of protons, so each feels a greater share of the nuclear charge;
there is less repulsion between the remaining electrons;
the ion is smaller, so the next electron is closer to the nucleus.

The increase is gradual within a shell and very large when the next electron has to come from a shell closer to the nucleus. Those big jumps reveal the shell structure.

(1, 2.70) (2, 3.66) (3, 3.84) (4, 3.98) (5, 4.13) (6, 4.22) (7, 4.30) (8, 4.41) (9, 4.46) (10, 5.15) (11, 5.20)

The graph plots $\log_{10}$ of the successive ionisation energies of sodium (in kJ mol $^{-1}$ ) against the number of the electron removed. A log scale is used because the values range from about 500 to 159 000 kJ mol $^{-1}$ . The pattern is 1, then 8, then 2: one electron is easy to remove (the 3s electron), then there is a big jump to the eight electrons of the second shell, then another big jump to the two 1s electrons. So sodium is $2,8,1$ , in Group 1.

Method

Deducing the group from successive ionisation energies

Look for the first big jump (often a factor of three or more, much bigger than the steps before it).
Count how many electrons are removed before that jump. That number is the number of outer-shell electrons.
That gives the group: 1 or 2 for the s-block; for the p-block, add 10 to get the modern group number (3 outer electrons means Group 13).
If all the data are given, the positions of the other big jumps show the number of electrons in each inner shell.

Within a shell you can sometimes see smaller jumps where the sub-shell changes. For aluminium ( $[\text{Ne}]\,3s^2 3p^1$ ), the gap between $IE_1$ (577, a 3p electron) and $IE_2$ (1820, a 3s electron) is proportionally larger than that between $IE_2$ and $IE_3$ (2740), because the first electron comes from the higher-energy 3p sub-shell.

Writing ionisation equations

Write equations, with state symbols, for (a) the first ionisation energy of chlorine and (b) the third ionisation energy of aluminium.

Solution

(a) $\ce{Cl(g) -> Cl+(g) + e-}$

(b) $\ce{Al^2+(g) -> Al^3+(g) + e-}$

The third ionisation starts from the $2+$ ion and removes one electron.

Explaining a dip across a period

Explain why the first ionisation energy of boron (799 kJ mol $^{-1}$ ) is lower than that of beryllium (900 kJ mol $^{-1}$ ).

Solution

Beryllium is $1s^2 2s^2$ ; boron is $1s^2 2s^2 2p^1$ . The electron removed from boron is in the 2p sub-shell, which is of higher energy than the 2s sub-shell from which beryllium's electron is removed. The 2p electron is also shielded by the 2s electrons and is slightly further from the nucleus. These effects outweigh boron's greater nuclear charge, so the 2p electron is less strongly attracted and less energy is needed to remove it.

Identifying the group from data

The first seven ionisation energies of element E, in kJ mol $^{-1}$ , are:

$IE_1$	$IE_2$	$IE_3$	$IE_4$	$IE_5$	$IE_6$	$IE_7$
1000	2252	3357	4556	7004	8496	27107

E is in Period 3. Identify E and explain your answer.

Solution

Calculate the ratio of each value to the previous one: $2.25, 1.49, 1.36, 1.54, 1.21, 3.19$ . The only large jump is between $IE_6$ and $IE_7$ (factor of about 3.2, an increase of over 18 000 kJ mol $^{-1}$ ).

So six electrons are removed from the outer shell before the seventh must come from an inner shell closer to the nucleus with much less shielding. E has six outer electrons and is in Group 16. In Period 3, E is sulfur.

Exam-style: successive ionisation energies across a period

The table shows the first four ionisation energies of three consecutive elements in Period 3, P, Q and R (not their symbols), in kJ mol $^{-1}$ .

element	$IE_1$	$IE_2$	$IE_3$	$IE_4$
P	494	4560	6940	9540
Q	736	1450	7740	10500
R	577	1820	2740	11600

(a) Identify P, Q and R. (b) Explain why $IE_1$ of R is lower than $IE_1$ of Q. (c) Explain why $IE_2$ of P is much larger than $IE_2$ of Q.

Solution

(a) P has a big jump after one electron: Group 1, sodium. Q has a big jump after two: Group 2, magnesium. R has a big jump after three: Group 13, aluminium.

(b) R (Al) loses a 3p electron; Q (Mg) loses a 3s electron. The 3p electron is in a higher-energy sub-shell, slightly further from the nucleus and shielded by the 3s electrons, so it is less strongly attracted despite the greater nuclear charge of Al.

(c) The second electron removed from P (Na) comes from the second shell ( $2p$ ), which is much closer to the nucleus and shielded only by the 1s electrons. The second electron removed from Q (Mg) is still in the third shell (3s). The 2p electron in $\ce{Na+}$ is far more strongly attracted, so much more energy is needed.

Exam-hard: interpreting a full set of successive ionisation energies

A graph of $\log_{10}(IE)$ against electron number for element Z shows: a gradual rise for electrons 1 to 4, a big jump between electrons 4 and 5, a gradual rise for 5 to 12, a big jump between 12 and 13, and then a rise for 13 to 14. Deduce the electronic configuration of Z and identify Z. The values for the first four electrons are 786, 1577, 3232 and 4356 kJ mol $^{-1}$ . Explain why the rise from electron 2 to electron 3 is proportionally larger than the rise from electron 3 to electron 4.

Solution

Count electrons in each group of the pattern: 4, then 8 (electrons 5 to 12), then 2 (13 to 14). Fourteen electrons in total, arranged $2,8,4$ . Z is silicon, $1s^2 2s^2 2p^6 3s^2 3p^2$ , in Group 14.

Ratios: $IE_3 / IE_2 = 3232 / 1577 = 2.05$ but $IE_4 / IE_3 = 4356 / 3232 = 1.35$ .

The first two electrons removed are 3p electrons and the next two are 3s electrons. Electron 3 is the first one taken from the 3s sub-shell, which is lower in energy and not shielded by any 3p electrons, so there is an extra increase on top of the usual rise in charge on the ion. Electrons 3 and 4 both come from 3s, so the step between them reflects only the increasing charge on the ion and is smaller. Both steps are far smaller than the jump between shells (after the fourth electron), because all four electrons are in the third shell.

Exam tip

"Explain" a trend: state the factor, link it to attraction between the nucleus and the outer electron, then link attraction to energy needed. Missing the attraction link is the most common reason for losing marks.
Use "outer electron" or "electron being removed", not "outer shell" alone.
In data questions, show the comparison you used (for example a calculated jump or ratio). One sentence of evidence earns the mark that a bare answer does not.
The data booklet lists first to fourth ionisation energies for many elements. You can quote values to support your explanation.

Summary

$IE_1$ : energy to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous $1+$ ions. Equations always use (g).
Factors: nuclear charge, distance (radius), shielding by inner shells and sub-shells, spin-pair repulsion.
Down a group: IE decreases (more shells, more shielding, outweighs extra protons).
Across a period: general increase (more protons, same shell, similar shielding).
Dips at Group 13 (new, higher-energy p sub-shell, shielded by s) and Group 16 (electron removed is paired, spin-pair repulsion).
Successive IEs always increase; big jumps mark a change to an inner shell. Count electrons before the first big jump to find the group.

Practice

Question

Define first ionisation energy and write the equation for the first ionisation energy of magnesium.
Write the equation for the second ionisation energy of calcium.
Explain why the first ionisation energy of potassium is lower than that of sodium.
Explain why the first ionisation energy of oxygen is lower than that of nitrogen.
Explain why the first ionisation energy of argon is higher than that of chlorine.
Calcium has successive ionisation energies 590, 1150, 4940, 6480 kJ mol $^{-1}$ . Explain how these values show that calcium is in Group 2.
Explain why the second ionisation energy of any element is always greater than its first.
An element in Period 3 has $IE_1$ to $IE_5$ of 1012, 1907, 2914, 4964, 6274 and $IE_6 = 21267$ kJ mol $^{-1}$ . Identify the element and explain.
Sketch the shape of a graph of $\log_{10}(IE)$ against electron number for all 13 electrons of aluminium, and explain each big jump.
The first ionisation energy of gallium (577 kJ mol $^{-1}$ ) is almost identical to that of aluminium (577 kJ mol $^{-1}$ ), even though gallium is below aluminium in Group 13. Suggest an explanation. (Hint: gallium is $[\text{Ar}]\,3d^{10} 4s^2 4p^1$ .)

Answers

The energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous $1+$ ions. $\ce{Mg(g) -> Mg+(g) + e-}$ .
$\ce{Ca+(g) -> Ca^2+(g) + e-}$ .
Potassium's outer electron is in the fourth shell (4s), sodium's in the third (3s). The 4s electron is further from the nucleus and more shielded by inner shells. This outweighs the greater nuclear charge of potassium, so the electron is less strongly attracted and less energy is needed to remove it.
N is $1s^2 2s^2 2p^3$ , one electron in each 2p orbital. O is $1s^2 2s^2 2p^4$ , so one 2p orbital contains a pair of electrons. The paired electrons repel each other (spin-pair repulsion), so one of them is easier to remove, despite oxygen's greater nuclear charge.
Argon has one more proton than chlorine. The electron removed is in the same sub-shell (3p) at a similar distance with the same shielding, so the greater nuclear charge attracts it more strongly; more energy is needed. (Both remove a paired 3p electron, so spin-pair repulsion does not change between them.)
$IE_1$ and $IE_2$ are relatively low and close together; the jump from $IE_2$ (1150) to $IE_3$ (4940) is a factor of about 4.3. So two electrons are removed from the outer shell, and the third must come from an inner shell closer to the nucleus and less shielded. Two outer electrons means Group 2.
After the first electron is removed, the remaining electrons are attracted by the same number of protons, there is less repulsion between electrons, and the ion is smaller, so the next electron is closer to the nucleus and more strongly attracted. More energy is needed to remove it.
Ratios: $1.88, 1.53, 1.70, 1.26, 3.39$ . The big jump is between $IE_5$ and $IE_6$ , so 5 outer electrons: Group 15, so phosphorus.
The graph rises steadily for electrons 1 to 3, with a slightly larger step from 1 to 2 (3p then 3s). A big jump between 3 and 4: the fourth electron comes from the second shell, much closer to the nucleus, less shielded. Gradual rise for 4 to 11 (eight electrons of shell 2, with a slightly larger step between the 2p and 2s electrons, i.e. between electrons 9 and 10). A second big jump between 11 and 12: the last two electrons are in the first shell, closest to the nucleus with no shielding. Overall pattern 3, 8, 2.
Gallium has an extra shell, which on its own would lower the IE. But it also has 18 more protons than aluminium, and ten of the extra electrons are in 3d orbitals, which shield the outer electron poorly. So the outer 4p electron feels a much larger effective attraction than the extra shell alone would suggest, and the two effects roughly cancel.

What ionisation energy measures

The four factors

Trend down a group

Trend across a period

The general increase

The big drop from a noble gas to the next alkali metal

First dip: Group 2 to Group 13 (Be to B, Mg to Al)

Second dip: Group 15 to Group 16 (N to O, P to S)

Successive ionisation energies

Practice

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