2018 H2 Chem P1 Q18 — Transition elements Solution

The answer is A.
When excess aqueous ammonia is added to \(\text{Cu}^{2+}\)(aq), four \(\text{NH}_3\) ligands replace four of the six water ligands:
\([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}\)
This complex is dark blue and its sulfate is soluble in water, so \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\text{SO}_4^{2-}\) is a dark blue solution.
• B is wrong: the dark blue copper(II) ammine complex sulfate is water-soluble, so it gives a dark blue solution, not a dark blue precipitate. In aqueous ammonia only four \(\text{NH}_3\) replace water, giving \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).
• C is wrong: \(\text{Cu}(\text{OH})_2(\text{H}_2\text{O})_4\) is a pale blue precipitate, not a solution.
• D is wrong: \([\text{Cu}(\text{H}_2\text{O})_6]^+\) would be a copper(I) complex. Cu(I) is \(3\text{d}^{10}\), so there are no d-d transitions and it would not be blue; copper(I) chloride is in fact an insoluble white solid.