2017 H2 Chem P1 Q8 — Atomic structure Solution

The answer is B (\(Q\text{Cl}_3\)).
A very high ionisation energy means the electron is removed from an inner principal quantum shell, close to the nucleus.
The fifth IE drops sharply from R (37 832 kJ \(\text{mol}^{-1}\)) to S (9445 kJ \(\text{mol}^{-1}\)). So for S the fifth electron still comes from the valence shell, but for R it comes from an inner shell. R therefore has exactly 4 valence electrons: \(R^{4+}\) has a noble gas configuration and its fifth electron is taken from the inner shell.
So R is in Group 14, S in Group 15 and T in Group 16, and Q (the element before R) is in Group 13 with 3 valence electrons. Q loses its 3 valence electrons (or forms 3 covalent bonds) to form the chloride \(Q\text{Cl}_3\). The values match boron, carbon, nitrogen and oxygen, so Q is boron and the chloride is \(\text{BCl}_3\).
• A is wrong: \(Q\text{Cl}_2\) is the chloride of a Group 2 element, but Q has 3 valence electrons (Group 13), one fewer than R.
• C is wrong: \(Q\text{Cl}_4\) is the chloride of a Group 14 element. That is R, not Q.
• D is wrong: \(Q\text{Cl}_5\) is the chloride of a Group 15 element. That is S, whose fifth IE is low, not Q.