Ionisation energy

1. Overview

Ionisation energy is the energy needed to pull electrons away from atoms and ions. Its values are some of the strongest evidence that electrons are arranged in shells and sub-shells. This section covers how to define ionisation energies and write equations for them, why they change across a period, down a group and from one electron to the next, and how to use successive ionisation energies to work out an element's electronic configuration and its place in the Periodic Table.

Key Definitions

• First ionisation energy (IE₁): the energy needed to remove one electron from each atom in one mole of gaseous atoms, to form one mole of gaseous 1+ ions.
• Second ionisation energy (IE₂): the energy needed to remove one electron from each ion in one mole of gaseous 1+ ions, to form one mole of gaseous 2+ ions.
• Successive ionisation energies: the first, second, third and later ionisation energies of one element; the nth ionisation energy is the energy needed to remove one electron from each ion in one mole of gaseous (n−1)+ ions to form one mole of gaseous n+ ions.
• Nuclear charge: the positive charge on the nucleus, set by the number of protons; it goes up by one from each element to the next in the Periodic Table.
• Shielding: the effect of electrons in inner shells and inner sub-shells, which repel an outer electron and so reduce the attraction it feels from the nucleus.
• Spin-pair repulsion: the repulsion between two electrons of opposite spin that share one orbital; it makes one of them easier to remove than an electron alone in an orbital of the same sub-shell.
• Atomic radius: a measure of the size of an atom, usually taken as half the distance between the nuclei of two neighbouring atoms of the same element bonded together; the larger it is, the further the outer electrons are from the nucleus.

Content

What ionisation energy measures

An atom holds its electrons by electrostatic attraction between the positive nucleus and the negative electrons. Removing an electron means supplying energy to overcome that attraction. The first electron to go is an outer electron, the one held least strongly.

• The stronger the attraction between the nucleus and the electron being removed, the larger the ionisation energy.
• Every ionisation energy is positive: energy must always be put in, so ionisation is endothermic. Losing an electron is also oxidation.
• Values are in kJ mol⁻¹. The Data Booklet lists up to the first four ionisation energies of selected elements.
• The atoms are gaseous, so each one is separate and the value measures only the removal of the electron.
• One electron is removed from each atom in a mole of atoms. "An electron from one mole of atoms" would mean a single electron in total.

Writing equations for ionisation energies

Each ionisation energy removes exactly one electron from each gaseous particle. For nitrogen:

Ionisation energy	Equation	Value / kJ mol⁻¹
First	N(g) → N⁺(g) + e⁻	1400
Second	N⁺(g) → N²⁺(g) + e⁻	2860
Third	N²⁺(g) → N³⁺(g) + e⁻	4590

In general, for the nth ionisation energy of element X:

$$\text{X}^{(n-1)+}(\text{g}) \rightarrow \text{X}^{n+}(\text{g}) + \text{e}^-$$

1. Put (g) on every atom and ion. The electron is written e⁻.
2. Start from single atoms, even for an element made of molecules: nitrogen's first ionisation starts from N(g), not from N₂(g).
3. Remove one electron per equation. N(g) → N²⁺(g) + 2e⁻ is not the second ionisation energy: it needs IE₁ + IE₂.

Worked example 1: turning gaseous chromium atoms into Cr³⁺ ions

Question: For chromium ($A_{\text{r}}$ = 52.0) the first three ionisation energies are 653, 1590 and 2990 kJ mol⁻¹. (a) Write the three equations. (b) Calculate the energy needed to turn 0.250 g of gaseous chromium atoms into gaseous Cr³⁺ ions.

Step 1: equations.

• Cr(g) → Cr⁺(g) + e⁻ (653 kJ mol⁻¹)
• Cr⁺(g) → Cr²⁺(g) + e⁻ (1590 kJ mol⁻¹)
• Cr²⁺(g) → Cr³⁺(g) + e⁻ (2990 kJ mol⁻¹)

Adding them gives Cr(g) → Cr³⁺(g) + 3e⁻, so the energy per mole is the sum: $$653 + 1590 + 2990 = 5233\ \text{kJ mol}^{-1}$$

Step 2: amount of chromium. $$n = \frac{0.250}{52.0} = 4.808 \times 10^{-3}\ \text{mol}$$

Step 3: energy for the sample. $$E = 4.808 \times 10^{-3} \times 5233 = 25.16\ \text{kJ}$$

Answer: (a) the three equations in Step 1. (b) 25.2 kJ (3 s.f.). Using 3 × 653 kJ mol⁻¹ instead of the sum would give only 9.42 kJ: each electron is harder to remove than the one before.

What controls the size of an ionisation energy

Four factors decide how strongly the nucleus attracts the electron being removed.

Factor	What it is	Effect on ionisation energy
Nuclear charge	the number of protons	more protons: stronger attraction, larger value
Atomic or ionic radius	how far the outer electron is from the nucleus	further away: weaker attraction, smaller value
Shielding by inner shells and sub-shells	repulsion from electrons between the nucleus and the outer electron	more shielding: weaker attraction, smaller value
Spin-pair repulsion	repulsion between two electrons sharing one orbital	a paired electron is easier to remove: smaller value

Electrons in the same shell shield one another only a little; most shielding comes from the inner shells. When two species differ in more than one factor, name each factor and say which one wins.

The trend across a period

Group	Period 2	Period 3
1	Li 519	Na 494
2	Be 900	Mg 736
13	B 799	Al 577
14	C 1090	Si 786
15	N 1400	P 1060
16	O 1310	S 1000
17	F 1680	Cl 1260
18	Ne 2080	Ar 1520

First ionisation energies / kJ mol⁻¹.

General increase. Across a period:

1. the nuclear charge goes up by one at each step;
2. the new electron goes into the same outer shell, so shielding by the inner shells stays similar;
3. so the outer electrons are attracted more strongly and the atomic radius decreases;
4. so more energy is needed to remove an outer electron.

The noble gases sit at the peaks. Helium has the largest first ionisation energy of all the elements: its two electrons are in the first shell, close to a 2+ nucleus, with no inner shell to shield them.

Dip from Group 2 to Group 13 (Be to B, Mg to Al). Boron is 1s² 2s² 2p¹. Its outer electron is in the 2p sub-shell, which is higher in energy than 2s, and it is partly shielded by the two 2s electrons. So it is easier to remove than a 2s electron of beryllium, even though boron has one more proton. Aluminium's 3p electron, shielded by the 3s pair, behaves the same way.

Dip from Group 15 to Group 16 (N to O, P to S).

Atom	2s	2p (three orbitals)
N, 1s² 2s² 2p³	[↑↓]	[↑] [↑] [↑]
O, 1s² 2s² 2p⁴	[↑↓]	[↑↓] [↑] [↑]

In nitrogen each 2p orbital holds one electron. In oxygen, one 2p orbital holds a pair. The two electrons in that orbital repel each other (spin-pair repulsion), so one of them is easier to remove, and IE₁ of oxygen is lower than that of nitrogen despite the extra proton. Sulfur (3p⁴) and phosphorus (3p³) follow the same pattern. After each dip the rise starts again: C is above B and F is above O.

Big drop from Group 18 to Group 1. From Ne (2080) to Na (494), the outer electron is in a new shell, n = 3. It is further from the nucleus and shielded by two full inner shells, and these effects far outweigh the one extra proton.

The trend down a group

Group	Period 2	Period 3	Period 4
1	Li 519	Na 494	K 418
17	F 1680	Cl 1260	Br 1140

First ionisation energies / kJ mol⁻¹.

Down a group, first ionisation energy decreases:

1. the nuclear charge increases, which on its own would raise the value;
2. but each step down adds a shell, so the outer electron is further from the nucleus (larger atomic radius);
3. and it is shielded by more inner shells;
4. the extra distance and shielding outweigh the extra protons, so the outer electron is attracted less strongly and less energy is needed to remove it.

This is a general trend. It holds at every step in the Data Booklet for Groups 1, 2 and 17, but in Group 13 aluminium and gallium share the value 577 kJ mol⁻¹. Between them come the d-block elements scandium to zinc, whose ten extra protons are shielded only poorly by the ten 3d electrons added with them, so gallium's 4p electron is held more strongly than the extra shell alone would suggest.

Worked example 2: explaining a difference in first ionisation energy

Question: Zinc (proton number 30) and gallium (proton number 31) are next to each other in Period 4. Their first ionisation energies are 908 and 577 kJ mol⁻¹. Explain why gallium's value is lower, even though gallium has one more proton.

Step 1: write both configurations.

• Zn: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²
• Ga: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p¹

Step 2: identify the electron removed. Zinc loses a 4s electron. Gallium loses its single 4p electron.

Step 3: compare the factors.

• Nuclear charge: one higher for gallium, which on its own would raise the value.
• Sub-shell: 4p is higher in energy than 4s.
• Shielding: gallium's 4p electron is partly shielded by the filled 4s sub-shell as well as by the inner shells.

Step 4: decide which effect wins. The value falls by 331 kJ mol⁻¹, so the sub-shell and shielding effects outweigh the extra proton.

Answer: Gallium's outer electron is in the 4p sub-shell, which is higher in energy than 4s and partly shielded by the 4s electrons. This outweighs the increase in nuclear charge, so the electron is attracted less strongly and less energy is needed to remove it: the same pattern as Be to B and Mg to Al.

Successive ionisation energies of one element

The ionisation energies of any element always increase in order: IE₁ < IE₂ < IE₃ and so on. After each electron is removed:

• the same number of protons attracts fewer electrons;
• there is less repulsion between the remaining electrons, so the ion is smaller (its ionic radius decreases) and they are held closer to the nucleus;
• so the next electron needs more energy.

This happens even inside one sub-shell: IE₂ of calcium (1150 kJ mol⁻¹) is almost twice IE₁ (590 kJ mol⁻¹), although both electrons come from 4s and the nucleus is the same.

Big jumps. A much larger increase appears when the next electron has to come from a shell closer to the nucleus, where it is less shielded and held far more strongly. For sodium, IE₁ is 494 kJ mol⁻¹ but IE₂ is 4560 kJ mol⁻¹, more than nine times as much: the first electron is the single 3s electron, the second comes from 2p in shell 2.

Counting the electrons between the big jumps gives 1, then 8, then 2: sodium is 2, 8, 1. The line never dips. On this log graph it rises gently inside each shell and steeply where the shell changes.

Using successive ionisation energies: configuration and position

1. Find the big jumps. A change of shell shows up as a rise far larger than the rises either side of it. In a short list this is usually clear at a glance. In a long list, divide each value by the one before it (or read a log graph like the one above) and look for the ratio that stands out: raw rises in kJ mol⁻¹ can mislead, because values deep inside a shell are large and rise by large amounts. For sodium, the rise from the 7th to the 8th value (about 5400 kJ mol⁻¹) is bigger than the rise from the 1st to the 2nd (about 4100 kJ mol⁻¹), yet only the 1st-to-2nd rise marks a change of shell: its ratio is 9.2, against 1.3 for the 7th to 8th.
2. Group. For s- and p-block elements, the number of electrons removed before the first big jump is the number of outer-shell electrons. One or two give Group 1 or 2; three to eight give Groups 13 to 18. A jump between the 4th and 5th values means four outer electrons, Group 14. Hydrogen and helium have only one shell, so their values show no big jump.
3. Shells and period. If all the values are given, the electrons between jumps give the number in each shell, the number of values is the proton number, and the number of shells is the period.
4. Configuration. Fill the electrons into sub-shells in order of energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p.

The method is clearest for elements up to calcium. For scandium to zinc, the 3d electrons are close in energy to the 4s ones, so there is no clear jump when 4s empties and the group rule does not apply. For gallium to krypton, the next electrons after the outer shell come from the full 3d sub-shell, so the jump is much smaller. With gallium's Data Booklet values, the 1st-to-2nd ratio (3.43) is even larger than the 3rd-to-4th (2.09), so compare the rises instead: the 3rd-to-4th rise (3230 kJ mol⁻¹) is more than twice either rise before it, and it marks the change of shell after the 3rd electron.

Worked example 3: identifying an element from all its ionisation energies

Question: Element D has six ionisation energies, given in the first two columns of the table below. Deduce (a) the group of D, (b) its full electronic configuration, and (c) its period and identity.

Step 1: divide each value by the one before it.

Ionisation	Energy / kJ mol⁻¹	Ratio to the value before
1st	1090
2nd	2350	2.16
3rd	4610	1.96
4th	6220	1.35
5th	37 800	6.08
6th	47 300	1.25

Step 2: find the big jump. One ratio stands out: 6.08, from the 4th to the 5th value. The others lie between 1.25 and 2.16. So the 5th electron is the first to come from a shell closer to the nucleus.

Step 3: electrons in each shell. Electrons 1 to 4 are in the outer shell and electrons 5 and 6 are in the inner shell: 2, 4.

Step 4: configuration. Six electrons, filled in order: 1s² 2s² 2p².

Answer: (a) Group 14, because four electrons are removed before the big jump. (b) 1s² 2s² 2p². (c) Two occupied shells, so Period 2; six electrons, so proton number 6: D is carbon. The first four values match the Data Booklet entries for carbon.

Key Equations

First ionisation energy (always positive, in kJ mol⁻¹): $$\text{X}(\text{g}) \rightarrow \text{X}^{+}(\text{g}) + \text{e}^-$$

nth ionisation energy (one electron removed from each gaseous (n−1)+ ion): $$\text{X}^{(n-1)+}(\text{g}) \rightarrow \text{X}^{n+}(\text{g}) + \text{e}^-$$

Energy to form an n+ ion from the gaseous atom (per mole, in kJ mol⁻¹): $$E_{\text{total}} = \text{IE}_1 + \text{IE}_2 + \dots + \text{IE}_n$$

Energy for a sample ($m$ = mass of gaseous atoms in g; $A_{\text{r}}$ = relative atomic mass; energy in kJ): $$E = \frac{m}{A_{\text{r}}} \times E_{\text{total}}$$

Group from successive data (s- and p-block only): number of electrons removed before the first big jump = number of outer-shell electrons (1 → Group 1, 2 → Group 2, 3 to 8 → Groups 13 to 18).

Common Mistakes to Avoid

• Mistake: leaving out state symbols, or starting from the element as it is at room temperature. Fix: put (g) on every atom and ion and start from single atoms: Br(g) → Br⁺(g) + e⁻, not Br₂(l) → Br⁺(g) + e⁻.
• Mistake: writing the second ionisation energy as the loss of two electrons. Fix: IE₂ of chromium is Cr⁺(g) → Cr²⁺(g) + e⁻, 1590 kJ mol⁻¹. The change Cr(g) → Cr²⁺(g) + 2e⁻ needs IE₁ + IE₂ = 653 + 1590 = 2243 kJ mol⁻¹.
• Mistake: a vague definition, such as "the energy to remove an electron from one mole of atoms". Fix: say one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions.
• Mistake: saying the nuclear charge increases from one successive ionisation to the next. Fix: an element's ions all have the same nucleus. Calcium has 20 protons for both IE₁ (590) and IE₂ (1150); the second value is larger because Ca⁺ has fewer electrons, less repulsion and a smaller radius.
• Mistake: across a period, saying shielding "stays constant", or stating "more protons" without linking it to the outer electron. Fix: from Na to Cl, every outer electron is in shell 3 with the same ten inner electrons, so shielding is similar while the nuclear charge rises, and the outer electrons are attracted more strongly.
• Mistake: explaining the Group 15 to 16 dip with "sulfur has paired electrons" or "a half-filled sub-shell is extra stable". Fix: phosphorus has paired 3s electrons too. Say that sulfur (3p⁴) has two electrons sharing one 3p orbital, and their repulsion makes one easier to remove, so IE₁ of S (1000) is below P (1060).
• Mistake: explaining a big jump without saying where the electron comes from. Fix: name the shell. For sodium, the 2nd electron comes from shell 2, which is closer to the nucleus and less shielded than the 3s electron, so IE₂ (4560) is far above IE₁ (494).
• Mistake: sketching successive ionisation energies with a dip, or counting the electrons after the jump. Fix: every value is larger than the one before. For a jump between the 4th and 5th values, the group comes from the four electrons before it: Group 14, not 15.

Exam Tips

• Construct an equation with the symbol of the element named, (g) on every atom and ion, and one e⁻.
• Explain a trend or difference as a chain: compare nuclear charge, distance and shielding (and spin-pair repulsion when an orbital holds a pair), say which effect outweighs the others, link it to the attraction between the nucleus and the outer electron, and finish with "so more (or less) energy is needed to remove it".
• Deduce from successive data by stating where the big jump is ("between the 2nd and 3rd"), what it means (the 3rd electron comes from a shell closer to the nucleus), and only then the group or configuration.
• Sketch first ionisation energy across a period as a general rise with small dips at Groups 13 and 16 and a big drop to the next Group 1 metal. Sketch successive values as always rising, with steep steps where the shell changes.
• Suggest a missing value by placing it against its neighbours, including the dips. For example, selenium (Group 16) should be expected slightly below arsenic, not between arsenic and bromine.