Question

A first row transition metal (M) does not liberate \( \mathrm{H_2} \) gas from dilute HCl.

1 mol of aqueous solution of \( \mathrm{MSO_4} \) is treated with excess aqueous KCN and then \( \mathrm{H_2S(g)} \) is passed through the solution.

The amount of \( \mathrm{MS} \) (metal sulphide) formed from the above reaction is ________ mol.

• A. \(1\)
• B. \(0\)
• C. \(2\)
• D. \(3\)

Hint:

Copper does not displace hydrogen from dilute acids and forms insoluble sulphides even in presence of complexing agents.

Answer 1

The Correct Option is A

The problem involves a transition metal (M) that does not liberate hydrogen gas from dilute HCl. Therefore, M is likely copper (Cu), since copper does not displace hydrogen from acids like HCl due to its positive reduction potential.

Now, consider the steps given in the problem:

1. One mole of aqueous solution of \(\mathrm{MSO_4}\) is treated with excess aqueous KCN.
2. \(\mathrm{H_2S(g)}\) is passed through the solution.

In the first step, excess potassium cyanide (KCN) forms a complex with copper sulfate (CuSO₄). The stable complex formed is potassium tetracyanocuprate (K₃[Cu(CN)₄]), which can be represented as follows:

\(\mathrm{Cu^{2+} + 4CN^- \rightarrow [Cu(CN)_4]^{3-}}\)

This complex is quite stable and the formation of such complexes prevents the release of free metal ions.

In the next step, hydrogen sulfide gas (H₂S) is introduced. The complexation with cyanide means that no free Cu²⁺ ions are available to react with H₂S. As a result, copper sulfide (CuS) will not precipitate since the free copper ions are unavailable.

Therefore, the amount of \(\mathrm{MS}\) (metal sulfide) formed in this reaction is 0 mol. So, the correct answer is \(0\).