A first row transition metal ( 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.

Step-by-step solution

The first row transition metal (M) that does not liberate H₂ gas from dilute HCl is Copper ( Cu ), as it has a positive reduction potential ( E^ _ Cu²⁺/Cu = +0.34 V ). When 1 mol of CuSO₄ reacts with excess aqueous KCN , it first forms Cu(CN)₂ , which is unstable and decomposes to CuCN and cyanogen gas (CN)₂ . 2CuSO₄ + 4KCN 2Cu(CN)₂ + 2K₂SO₄ 2Cu(CN)₂ 2CuCN + (CN)₂ The CuCN then dissolves in excess KCN to form a highly stable soluble complex, potassium tetracyanocuprate(I): CuCN + 3KCN K₃[Cu(CN)₄] The complex [Cu(CN