JEE Main 2026 — Jan 22, Shift 1 — Chemistry solution

JEE Main2026Jan 22, Shift 1Chemistry

A first row transition metla (M) does not liberate $H_{2}$ gas from dilute HCl. 1 mol of aqueous solution of $M S O_{4}$ is treated with excess of aqueous KCN and then $H_{2} S (g)$ is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is _______ mol.

Solution

A first-row TM that does not release $H_{2}$ from dilute HCl sits below hydrogen in the electrochemical series. Among +2 sulphate-forming first-row metals, this points to Cu (so M = Cu, $M S O_{4} = C u S O_{4}$ ).

With excess KCN, $C u^{2 +}$ is first reduced to $C u^{+}$ (cyanide reduces it) and then trapped as the very stable tetracyanocuprate(I) complex: $[C u (C N)_{4}]^{3 -}$ . This complex is so stable that the free $C u^{+}$ concentration becomes negligible.

When $H_{2} S$ is passed, the ionic product $[C u^{+}]^{2} [S^{2 -}]$ stays below $K_{s p}$ of $C u_{2} S$ , so no metal sulphide precipitates.

Amount of MS formed $= 0$ mol.

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