A first row transition metal (M) does not liberate H2 gas from dilute HCl. 1 mol of aqueous solution of MSO4 is treated with excess of aqueous KCN and then H2S(g) is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is mol.
- 1
- 2
- 0
- 3
Answer
(3) 0
Solution
Among the 3d metals only Cu has a positive E M2+/M (+0.34 V) it does not liberate H2 from dilute HCl M = Cu, MSO4 = CuSO4.
With excess KCN, Cu2+ is reduced and forms the very stable complex:
2CuSO4 + 10KCN 2K3[Cu(CN)4] + 2K2SO4 + (CN)2
[Cu(CN)4]3- is so stable (very high stability constant) that it gives practically no free copper ions in solution.
On passing H2S, no CuS is precipitated (this is used to separate Cu2+ from Cd2+ in qualitative analysis).
Amount of MS formed = 0 mol
Answer: (3)