Consider the reaction CuSO_4 + KCN to CuCN + | vedclass

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(D) The reaction proceeds as follows: $1$. $2CuSO_4 + 4KCN 	o 2CuCN + (CN)_2 + 2K_2SO_4$. Here,$Y$ is $(CN)_2$ (cyanogen). $2$. Cyanogen $(CN)_2$ has

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$Z$ is a diamagnetic complex with $sp^3$ hybridization.

Vedclass · Answer

(D) The reaction proceeds as follows: $1$. $2CuSO_4 + 4KCN 	o 2CuCN + (CN)_2 + 2K_2SO_4$. Here,$Y$ is $(CN)_2$ (cyanogen). $2$. Cyanogen $(CN)_2$ has a linear structure,not an open book-like structure. $3$. When $CuCN$ reacts with excess $KCN$,it forms the complex $K_3[Cu(CN)_4]$. $4$. In $K_3[Cu(CN)_4]$,the copper is in the $+1$ oxidation state $(Cu^+)$,which has a $d^{10}$ configuration. $5$. The complex $[Cu(CN)_4]^{3-}$ is tetrahedral with $sp^3$ hybridization. $6$. Since all $d$-orbitals are filled $(d^{10})$,there are no unpaired electrons,making it a diamagnetic complex. $7$. Therefore,statement $D$ is correct.

Consider the reaction $CuSO_4 + KCN \to CuCN + Y \xrightarrow{\text{excess } K^{+}CN^{-}} Z$. Identify the correct statement regarding the products $Y$ and $Z$.

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