A first row transition metal (M) does not liberate H_2$H_{2}$ gas from dilute HCl. 1 mol of aqueous solution of MSO_4$MSO_{4}$ is treated with excess of aqueous KCN and then H_2S(g)$H_{2}S(g)$ is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is ____ mol.

### Related Formula Cu^2+ + 4CN^- rightarrow [Cu(CN)_4]^3- quad (textafter redox with CN^-)$$Cu^{2+} + 4CN^{-} \rightarrow [Cu(CN)_{4}]^{3-} \quad (\text{after redox with } CN^-)$$ ### Core Logic The first-row transition metal that does not liberate H_2$H_{2}$ gas from dilute HCl is Copper (Cu), because its standard reduction potential is positive (E^circ_Cu^2+/Cu = +0.34text V$E^{\circ}_{Cu^{2+}/Cu} = +0.34\text{ V}$). When CuSO_4$CuSO_{4}$ is treated with excess KCN, it forms a very stable soluble cyano complex: CuSO_4 + 2KCN rightarrow Cu(CN)_2 + K_2SO_4$$CuSO_{4} + 2KCN \rightarrow Cu(CN)_{2} + K_{2}SO_{4}$$ 2Cu(CN)_2 rightarrow 2CuCN + (CN)_2$2Cu(CN)_{2} \rightarrow 2CuCN + (CN)_{2}$ CuCN + 3KCN rightarrow K_3[Cu(CN)_4]$CuCN + 3KCN \rightarrow K_{3}[Cu(CN)_{4}]$ The complex ion [Cu(CN)_4]^3-$[Cu(CN)_{4}]^{3-}$ is highly stable (a perfect complex). When H_2S$H_{2}S$ is passed through this solution, it does not yield sufficient Cu^+$Cu^{+}$ ions to exceed the solubility product (K_sp$K_{sp}$) of Cu_2S$Cu_{2}S$. ### Step 1: Final Conclusion Since no copper sulphide precipitates, the amount of MS formed is 0 moles. ### Pattern Recognition Cu and Cd separation: Cu^2+$Cu^{2+}$ forms a very stable cyanide complex that does not precipitate with H_2S$H_2S$, whereas Cd^2+$Cd^{2+}$ forms a less stable complex that does precipitate as CdS$CdS$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements Class 12 Chemistry: Coordination Compounds