Atomic Structure: Question 5

Worked solution

Part (a): Defining first ionisation energy

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

$X(g) \rightarrow X^+(g) + e^-$

Both the atoms and the ions must be in the gaseous state, and the quantity refers to one mole of atoms.

Part (b): Deducing the outer-shell electron count

Look at the ratio between each pair of consecutive ionisation energies:

$\frac{IE_2}{IE_1} = \frac{1060}{580} \approx 1.8 \qquad \frac{IE_3}{IE_2} = \frac{4900}{1060} \approx 4.6 \qquad \frac{IE_4}{IE_3} = \frac{6300}{4900} \approx 1.3 \qquad \frac{IE_5}{IE_4} = \frac{8000}{6300} \approx 1.3$

The increase from $IE_1$ to $IE_2$ is modest, but the increase from $IE_2$ to $IE_3$ is far larger than any of the others, a sharp jump. Removing the third electron requires disproportionately more energy because it is being pulled from a shell that sits much closer to the nucleus (with much less shielding) than the first two electrons removed.

This means the first two electrons removed came from the outermost occupied shell, and the third electron came from a completely different, inner shell. Element T therefore has 2 electrons in its outermost occupied shell.

Part (c): Identifying the group

An element with 2 electrons in its outermost shell belongs to Group 2 of the Periodic Table.

Part (d): Why $IE_2 > IE_1$

Removing the first electron leaves the ion $T^+$ with one fewer electron than the neutral atom, but the same number of protons (same nuclear charge). This means:

• there is less electron-electron repulsion among the remaining electrons, and
• each remaining electron effectively experiences a greater attraction per electron to the nucleus (the same nuclear charge is now shared among fewer electrons).

Both effects make the second electron more strongly held, so more energy ($IE_2$) is needed to remove it than was needed to remove the first ($IE_1$).

Part (e): The magnesium–aluminium anomaly

Magnesium ($1s^2\,2s^2\,2p^6\,3s^2$) and aluminium ($1s^2\,2s^2\,2p^6\,3s^2\,3p^1$) are next to each other in Period 3, and aluminium has one more proton than magnesium. Based on nuclear charge alone, aluminium’s first ionisation energy would be expected to be higher than magnesium’s, but it is actually lower.

The explanation lies in which sub-shell the outermost electron occupies:

• Magnesium’s outermost electron is removed from a full $3s^2$ sub-shell.
• Aluminium’s outermost electron is removed from the $3p$ sub-shell, which is at a slightly higher energy than $3s$, and this single $3p$ electron is also shielded to some extent by the underlying, complete $3s^2$ sub-shell.

Because the $3p$ electron in aluminium is, on balance, further from the nucleus in energy terms and better shielded than a $3s$ electron, it is easier to remove despite aluminium’s larger nuclear charge. This sub-shell effect outweighs the extra proton, giving aluminium the lower first ionisation energy.

Final answers

• (a) Energy to remove 1 electron from each atom in 1 mole of gaseous atoms, forming 1 mole of gaseous $1+$ ions.
• (b) $\boxed{2}$ electrons in the outermost shell (sharp jump between $IE_2$ and $IE_3$).
• (c) Group 2.
• (d) Same nuclear charge, one fewer electron $\Rightarrow$ less repulsion / greater effective attraction per electron.
• (e) Al’s outer electron is in the higher-energy, shielded $3p$ sub-shell rather than the full $3s^2$ sub-shell of Mg, outweighing Al’s extra proton.