A first row transition metal (M) does not liberate H2 gas from dilute HCl. 1 mol of aqueous solution of MSO4 is treated with excess of aqueous KCN and then H2S(g) is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is _____

Amount of metal sulphide formed after KCN and H2S treatment

Q. A first row transition metal (M) does not liberate H₂ gas from dilute HCl. 1 mol of aqueous solution of MSO₄ is treated with excess of aqueous KCN and then H₂S (g) is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is ______ mol.

A. 1

B. 2

C. 0

D. 3

Correct Answer: 0

Explanation

The metal M is a first row transition metal which does not liberate hydrogen gas from dilute HCl. This indicates that the metal is less reactive and lies below hydrogen in the electrochemical series.

When MSO₄ is treated with excess aqueous KCN, the metal ion forms a stable cyanide complex.

General reaction:

M²⁺ + excess CN⁻ → [M(CN)_n]⁽²⁻ⁿ⁾

Cyanide ion is a strong ligand and forms very stable complexes with transition metal ions. As a result, the free M²⁺ ions are no longer available in solution.

Now, when H₂S gas is passed through the solution, metal sulphide (MS) can precipitate only if free metal ions are present.

Since all metal ions are locked inside the stable cyanide complex, no free M²⁺ ions remain to react with sulphide ions.

Therefore, no metal sulphide precipitate is formed.

Amount of MS formed = 0 mol

This question is based on coordination chemistry and qualitative analysis, and is very important for JEE Main 2026, JEE Advanced and IIT JEE.

Explanation

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