The coordination number of $Ni^{2+}$ is $4$.
$NiCl_2 + KCN$ (excess) $\rightarrow A$ (cyano complex)
$NiCl_2 + \text{Conc. } HCl$ (excess) $\rightarrow B$ (chloro complex)
$1.$ The $IUPAC$ names of $A$ and $B$ are:
$(A)$ Potassium tetracyanonickelate $(II)$,potassium tetrachloronickelate $(II)$
$(B)$ Tetracyanopotassiumnickelate $(II)$,tetrachloropotassiumnickelate $(II)$
$(C)$ Tetracyanonickel $(II)$,tetrachloronickel $(II)$
$(D)$ Potassium tetracyanonickel $(II)$,potassium tetrachloronickel $(II)$
$2.$ Predict the magnetic nature of $A$ and $B$:
$(A)$ Both are diamagnetic.
$(B)$ $A$ is diamagnetic and $B$ is paramagnetic with one unpaired electron.
$(C)$ $A$ is diamagnetic and $B$ is paramagnetic with two unpaired electrons.
$(D)$ Both are paramagnetic.
$3.$ The hybridization of $A$ and $B$ are:
$(A)$ $dsp^2, sp^3$
$(B)$ $sp^3, sp^3$
$(C)$ $dsp^2, dsp^2$
$(D)$ $sp^3 d^2, d^2 sp^3$
Give the answers for questions $1, 2$ and $3$.

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(C) is $K_2[Ni(CN)_4]$ and $B$ is $K_2[NiCl_4]$.
$1.$ The $IUPAC$ name of $K_2[Ni(CN)_4]$ is Potassium tetracyanonickelate $(II)$ and $K_2[NiCl_4]$ is Potassium tetrachloronickelate $(II)$. Thus,option $(A)$ is correct.
$2.$ In $A$,$CN^-$ is a strong field ligand,so $Ni^{2+}$ $(3d^8)$ undergoes $dsp^2$ hybridization,making it diamagnetic. In $B$,$Cl^-$ is a weak field ligand,so $Ni^{2+}$ $(3d^8)$ undergoes $sp^3$ hybridization,leaving two unpaired electrons,making it paramagnetic. Thus,option $(C)$ is correct.
$3.$ As determined,$A$ has $dsp^2$ hybridization and $B$ has $sp^3$ hybridization. Thus,option $(A)$ is correct.
The sequence of answers is $A, C, A$.

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