Question 64

A first row transition metal (M) does not liberate $$H_{2}$$ gas from dilute HCI. 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.

Solution

Metal Identification:

Reaction with KCN (Masking): When excess $$KCN$$ added to $$CuSO_4$$, a two-step reaction occurs. First, $$Cu^{2+}$$ is reduced to $$Cu^{+}$$, and then it forms an extremely stable complex called Potassium tetracyanocuprate(I), represented as $$K_3[Cu(CN)_4]$$.

Reaction with $$H_2S$$: In Group II of basic radical analysis, $$Cu^{2+}$$ usually precipitates as black $$CuS$$. However, in this specific experiment, the Cyanide ligands ($$CN^-$$) bind to the copper so tightly that they "lock" the metal ions away. The concentration of free copper ions left in the solution is so low that it cannot react with $$H_2S$$ to form a precipitate.

Final Result: Since the copper ions are effectively hidden, 0 mol of Metal Sulphide ($$MS$$) is formed.

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A first row transition metal (M) does not liberate $$H_{2}$$ gas from dilute HCI. 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.

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