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.

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Copper does not displace hydrogen from dilute acids and forms insoluble sulphides even in presence of complexing agents.
Updated On: Feb 4, 2026
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The Correct Option is A

Solution and Explanation

Step 1: Identify the metal.
A first-row transition metal that does not liberate hydrogen gas from dilute HCl is Cu.
Step 2: Reaction with excess KCN.
Copper forms a stable cyanide complex: \[ \text{Cu}^{2+} + 4\text{CN}^- \rightarrow [\text{Cu(CN)}_4]^{2-}. \] However, on passing \( \mathrm{H_2S} \), copper sulphide precipitates.
Step 3: Sulphide formation.
Each mole of \( \mathrm{Cu^{2+}} \) gives one mole of \( \mathrm{CuS} \).
Thus, from 1 mol of \( \mathrm{MSO_4} \), \[ \text{Amount of MS formed} = 1 \text{ mol}. \]
Final Answer: \[ \boxed{1} \]
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