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

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Coordination Compounds

A first row transition metla (M) does not liberate

H_{2}

gas from dilute HCl .1 mol of aqueous solution of

MSO_{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.

(A)2
(B)1
(C)3
(D)0✓ Correct

Answer: (D)

Solution

A first-row transition metal that cannot displace hydrogen from dilute acid lies below hydrogen in the electrochemical series, which marks it as copper, so

MS O_{4}

CuS O_{4}

. When excess potassium cyanide is added to

C u^{2 +}

, the cupric ion is first reduced to the cuprous state and ends up held in the very stable tetracyanidocuprate(I) ion

[Cu (CN)_{4}]^{3 -}

, whose dissociation is almost negligible. Passing

H_{2} S

then provides sulphide ions, but the concentration of free

C u^{+}

set loose by so stable a complex is far too small for the ionic product to reach the solubility product of copper sulphide, so none separates. The amount of metal sulphide formed is

0

mol.

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Coordination Compounds

Solution

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