Electron Configurations of Atoms and Ions

Lessons

Senior 4 · Chemistry

Electron Configurations of Atoms and Ions

5 lessons

Lesson 5 of 5

Ionisation Energy and the Factors That Affect It

25 min

Evidence you can measure

Everything in this unit so far, shells, subshells, orbitals, configurations, has been a model. How do we know it is right? One answer is the energy needed to pull electrons off atoms. Sodium gives up its outer electron for 496 kJ per mole; its neighbour neon needs 2081 kJ per mole, more than four times as much. Sodium reacts violently with water and neon reacts with nothing. Ionisation energies rise and fall across the periodic table in a pattern that matches the configurations you wrote in Lesson 4, almost electron by electron.

Definition

First ionisation energy: the energy needed to remove one mole of electrons from one mole of gaseous atoms, forming one mole of gaseous 1+ ions. For sodium:

\mathrm{Na(g)} \longrightarrow \mathrm{Na^{+}(g)} + e^- \qquad \Delta H = +496\,\mathrm{kJ\,mol^{-1}}

Successive ionisation energies: the energies for removing a second, third, … electron from the ion. The second ionisation energy of sodium is for $\mathrm{Na^{+}(g)} \rightarrow \mathrm{Na^{2+}(g)} + e^-$ .

Ionisation energies are always positive (endothermic): energy must be supplied to pull a negative electron away from a positive nucleus.

Five factors

How strongly an outer electron is held depends on the balance between the pull of the nucleus and everything that weakens it:

Nuclear charge: more protons pull harder, so ionisation energy rises.
Distance from the nucleus (atomic radius): the attraction falls off with distance, so an electron in a higher shell is easier to remove.
Shielding: inner electrons repel the outer electron and screen it from the nucleus. More inner shells mean more shielding. Electrons in the same shell shield each other only a little.
Type of subshell: within one shell, a p electron is slightly higher in energy than an s electron, because the s electron penetrates closer to the nucleus. A p electron is easier to remove.
Spin-pair repulsion: two electrons sharing one orbital repel each other, so one of a pair is easier to remove than an unpaired electron in the same subshell.

The first three are often combined as the effective nuclear charge: the nuclear charge minus the shielding, the net pull the outer electron actually feels.

First ionisation energies from hydrogen to calcium

Down a group the first ionisation energy falls (He 2372 → Ne 2081 → Ar 1521; Li 520 → Na 496 → K 419). Each element has one more shell, so the outer electron is further away and more shielded. This outweighs the greater nuclear charge.
Across a period it rises overall, from the Group 1 metal to the noble gas. The nuclear charge increases by one each step, the electrons are added to the same shell, so shielding hardly changes and the atom gets smaller.
A big fall from each noble gas to the next alkali metal: the new electron starts a new shell.

Two dips that prove the subshells

Table 5.1: First ionisation energies across Period 3, in $\mathrm{kJ\,mol^{-1}}$

Na	Mg	Al	Si	P	S	Cl	Ar
496	738	578	787	1012	1000	1251	1521

Mg → Al falls (738 → 578). Magnesium's outer electron is in 3s; aluminium's is the first electron in 3p, which is higher in energy and partly shielded by the 3s electrons. It is easier to remove. The same happens from Be to B.
P → S falls slightly (1012 → 1000). Phosphorus has three unpaired 3p electrons (↑ ↑ ↑). In sulfur the fourth 3p electron must pair up (↑↓ ↑ ↑); the repulsion between the paired electrons makes one of them easier to remove. The same happens from N to O.

If electrons were simply arranged in shells, the values across a period would rise smoothly. The dips appear exactly where the s/p boundary and the first p pairing occur, which is direct evidence for subshells.

Worked example 1: explaining a comparison

Explain why the first ionisation energy of nitrogen (1402) is higher than that of carbon (1086) but also higher than that of oxygen (1314).

Solution. C is $1s^2\,2s^2\,2p^2$ , N is $1s^2\,2s^2\,2p^3$ , O is $1s^2\,2s^2\,2p^4$ .

N > C: nitrogen has one more proton; the extra electron is in the same 2p subshell with similar shielding, so the effective nuclear charge is greater and the electron is held more tightly.
N > O: in oxygen the electron removed is one of a pair in a 2p orbital. Repulsion between the paired electrons outweighs oxygen's extra proton, so less energy is needed. Nitrogen's three 2p electrons are all unpaired.

Successive ionisation energies reveal the shells

Each successive ionisation energy of an element is larger than the one before, because each electron is removed from an ion that is more positive and smaller, with fewer electrons left to repel it. But the increase is not steady. When all the electrons of one shell have gone, the next electron must come from a shell closer to the nucleus and much less shielded, and the energy jumps. The values span such a range that they are plotted as logarithms (Figure 5.2).

For sodium, the jumps divide the 11 electrons into 1, then 8, then 2: exactly 2,8,1, or $1 s^{2}$ , $2s^2\,2p^6$ , $3 s^{1}$ . The number of electrons removed before the first big jump is the number of outer-shell electrons, which gives the group in the periodic table.

Worked example 2: identifying an element

The first five ionisation energies of element X, in $\mathrm{kJ\,mol^{-1}}$ , are 578, 1817, 2745, 11 577, 14 842. X is in Period 3. Identify X and write an equation for its third ionisation energy.

Solution. Compare each value with the one before: $\frac{1817}{578} = 3.1$ , $\frac{2745}{1817} = 1.5$ , $\frac{11\,577}{2745} = 4.2$ , $\frac{14\,842}{11\,577} = 1.3$ . The big jump is between the 3rd and 4th electrons, so X has 3 outer electrons: Group 13. In Period 3 that is aluminium.

Third ionisation energy: $\mathrm{Al^{2+}(g)} \rightarrow \mathrm{Al^{3+}(g)} + e^-$ .

Answer: aluminium, $1s^2\,2s^2\,2p^6\,3s^2\,3p^1$ . Its three outer electrons are lost fairly easily, which is why it forms $\mathrm{Al^{3+}}$ ions.

Warning

Common mistakes. (1) Leaving out the state symbols: the atoms and ions must be gaseous, (g). (2) Writing the second ionisation energy as $\mathrm{Mg(g)} \rightarrow \mathrm{Mg^{2+}(g)} + 2e^-$ : that is the sum of the first two. (3) Saying "a full or half-full subshell is stable" to explain the N/O dip; examiners want the reason: spin-pair repulsion in oxygen. (4) Saying shielding increases across a period: electrons are added to the same shell, so it stays almost constant.

Tip

For any ionisation-energy explanation, use the three main factors in a fixed order: nuclear charge, distance (or shell), shielding. Then add one of the two special cases only if it applies: s versus p (Groups 2 → 13) or pairing (Groups 15 → 16).

Mike Sugiyama Jones (MSJ Chem): Trends in ionisation energy (3 min). Watch for the explanations of the general rise across a period and the two dips, and compare them with Table 5.1.

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Practice

Write equations, with state symbols, for (a) the first ionisation energy of chlorine, (b) 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 beryllium (900) is greater than that of boron (801).
Explain why the second ionisation energy of sodium is about nine times its first.
Element Y has successive ionisation energies (kJ mol⁻¹) 590, 1145, 4912, 6491, 8153. Which group is Y in? Explain.
Sketch the graph of $\log_{10}$ (successive ionisation energy) against electron number for the 13 electrons of aluminium. Mark where the big jumps occur.
Predict, with a reason, whether the first ionisation energy of selenium (Group 16, Period 4) is greater or less than that of sulfur.

Show answersHide answers

(a) $\mathrm{Cl(g)} \rightarrow \mathrm{Cl^{+}(g)} + e^-$ ; (b) $\mathrm{Ca^{+}(g)} \rightarrow \mathrm{Ca^{2+}(g)} + e^-$ .
Potassium's outer electron is in the fourth shell (4s), further from the nucleus and shielded by three full inner shells; the extra nuclear charge is outweighed, so it is held less tightly than sodium's 3s electron.
Boron's outer electron is in 2p, which is higher in energy than 2s and partly shielded by the 2s electrons, so it is easier to remove than beryllium's 2s electron.
The second electron comes from the full 2p subshell of $\mathrm{Na^{+}}$ : a shell closer to the nucleus with much less shielding, and from a positive ion. A jump of this size marks the change of shell.
The big jump is between the 2nd (1145) and 3rd (4912) values: 2 outer electrons, so Group 2 (Y is calcium).
Jumps between the 3rd and 4th electrons ( $3s^2\,3p^1$ to 2p) and between the 11th and 12th (2s to 1s); electrons 4 to 11 rise gradually.
Less: selenium's outer electron is in the fourth shell, further from the nucleus and more shielded, so the extra nuclear charge is outweighed (the real values are S 1000, Se 941).

Quick check

Why is the first ionisation energy of aluminium lower than that of magnesium?

Key points

First ionisation energy: $\mathrm{X(g)} \rightarrow \mathrm{X^{+}(g)} + e^-$ , per mole; always endothermic.
It depends on nuclear charge, distance, shielding, the type of subshell (s or p) and spin-pair repulsion.
It falls down a group and rises across a period, with dips at Groups 2 → 13 (new p subshell) and 15 → 16 (first paired p electron).
Successive ionisation energies rise, with big jumps at each change of shell; the electrons before the first big jump give the group number.

1Evidence you can measure

2Five factors

3First ionisation energies from hydrogen to calcium

4Two dips that prove the subshells

5Successive ionisation energies reveal the shells

Evidence you can measure

Five factors

First ionisation energies from hydrogen to calcium

Two dips that prove the subshells

Successive ionisation energies reveal the shells