2019 H2 Chem P1 Q27 — Transition elements Solution

The answer is C (\([\text{Cr(OH)}_6]^{3-}\)).

In aqueous solution, chromium(III) is present as \([\text{Cr(H}_2\text{O)}_6]^{3+}\).
• A little \(\text{OH}^-\) gives the grey-green precipitate \([\text{Cr(OH)}_3(\text{H}_2\text{O})_3]\), i.e. \(\text{Cr(OH)}_3\):
\([\text{Cr(H}_2\text{O)}_6]^{3+} + 3\text{OH}^- \rightarrow [\text{Cr(OH)}_3(\text{H}_2\text{O})_3] + 3\text{H}_2\text{O}\)
• \(\text{Cr(OH)}_3\) is amphoteric, so it dissolves in excess \(\text{OH}^-\) to give the deep green complex anion \([\text{Cr(OH)}_6]^{3-}\):
\([\text{Cr(OH)}_3(\text{H}_2\text{O})_3] + 3\text{OH}^- \rightarrow [\text{Cr(OH)}_6]^{3-} + 3\text{H}_2\text{O}\)

• A is wrong: excess NaOH(aq) supplies \(\text{OH}^-\) ligands, not \(\text{Cl}^-\); also Cr with four \(\text{Cl}^-\) ligands and a 2- charge would be Cr(II), not Cr(III).
• B is wrong: \([\text{Cr(H}_2\text{O)}_6]^{3+}\) is the ion in the chromium(III) chloride solution before NaOH is added; \(\text{OH}^-\) removes \(\text{H}^+\) from its water ligands.
• D is wrong: \([\text{Cr(OH)}_3(\text{H}_2\text{O})_3]\) is the neutral initial precipitate, which dissolves in excess NaOH(aq); it is not the species in the green solution.