2024 H2 Chem P1 Q3 — Chemical bonding Solution

The answer is C (2 and 3 only).
• Statement 1 is incorrect: S is more electronegative than Se, so the H-S bond is more polar than the H-Se bond and \(\text{H}_2\text{S}\) has stronger pd-pd attractions than \(\text{H}_2\text{Se}\). \(\text{H}_2\text{S}\) boils lower because it has fewer electrons and a smaller, less polarisable electron cloud than \(\text{H}_2\text{Se}\), so its id-id attractions are weaker. The reason given is wrong.
• Statement 2 is correct: \(\text{H}_2\text{O}\) has far fewer electrons than \(\text{H}_2\text{Se}\) (10 against 36), so its electron cloud is smaller and less polarisable and its id-id attractions are weaker. (Water still boils highest because of its hydrogen bonding.)
• Statement 3 is correct: O is more electronegative than S, so the H-O bond is more polar than the H-S bond and \(\text{H}_2\text{O}\) has stronger permanent dipole attractions than \(\text{H}_2\text{S}\).
• A and B are wrong: both include statement 1, which is incorrect.
• D is wrong: it leaves out statement 2, which is correct.