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.
A. 0
The first row transition metal (M) that does not liberate H_2 gas from dilute HCl is Copper (Cu), as it has a positive reduction potential (E^ _ Cu^ 2+ /Cu = +0.34 V ). When 1 mol of CuSO_4 reacts with excess aqueous KCN, it first forms Cu(CN)_2, which is unstable and decomposes to CuCN and cyanogen gas (CN)_2. 2CuSO_4 + 4KCN 2Cu(CN)_2 + 2K_2SO_4 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 K_3[Cu(CN)_4] The complex [Cu(CN)_4]^ 3- is so stable that its dissociation constant is extremely low. When H_2S gas is passed through this solution, the concentration of Cu^+ ions is insufficient to exceed the solubility product of Cu_2S. Therefore, no precipitate of metal sulphide is formed. The amount of MS (or M_2S) formed is 0 mol .
Related: Chemistry — d and f Block Elements · All PYQ Banks