Coordination Compounds Part 4

Concept:Copper (Cu) is the only first-row transition metal that does not liberate $\mathrm{H}_2$ from dilute HCl because it lies below hydrogen in the reactivity series.Chemical Equation / Formula:$\mathrm{CuSO}_4 + \text{excess KCN} \rightarrow [\mathrm{Cu}(\mathrm{CN})_4]^{3-}$$[\mathrm{Cu}(\mathrm{CN})_4]^{3-} + \mathrm{H}_2\mathrm{S} \rightarrow$ no reactionExplanation:Step 1: $\mathrm{Cu}$ does not react with dilute HCl, so metal M is $\mathrm{Cu}$.Thus $\mathrm{MSO}_4$ is $\mathrm{CuSO}_4$.Step 2: When 1 mol of $\mathrm{CuSO}_4$ is treated with excess aqueous KCN, all $\mathrm{Cu}^{2+}$ ions form the extremely stable complex $[\mathrm{Cu}(\mathrm{CN})_4]^{3-}$.This complex is a perfect cyanide complex, meaning no free $\mathrm{Cu}^{2+}$ remains in solution.Step 3: When $\mathrm{H}_2\mathrm{S}$ gas is passed, it cannot precipitate $\mathrm{CuS}$ because $\mathrm{Cu}^{2+}$ ions are not available.Therefore, no metal sulphide ($\mathrm{CuS}$) is formed.The amount of MS produced is zero moles.Answer:0 mol (Option C)