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.

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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 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.

List-$I$ (Molecules/ion)	List-$II$ (Hybridisation of central atom)
$A. PF_5$	$I. dsp^2$
$B. SF_6$	$II. sp^3d$
$C. Ni(CO)_4$	$III. sp^3d^2$
$D. [PtCl_4]^{2-}$	$IV. sp^3$

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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Match the List-$I$ with List-$II$
List-$I$ (Molecules/ion) List-$II$ (Hybridisation of central atom)
$A. PF_5$ $I. dsp^2$
$B. SF_6$ $II. sp^3d$
$C. Ni(CO)_4$ $III. sp^3d^2$
$D. [PtCl_4]^{2-}$ $IV. sp^3$
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The geometry possessed by $[Ni(CO)_4]$ is

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