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(C) To determine the hybridization of the central metal atom,we analyze the coordination number and the nature of the ligands:$1$. In $Na_{2}[NiCl_{4}

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$K_{2}[Ni(CN)_{4}]$

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(C) To determine the hybridization of the central metal atom,we analyze the coordination number and the nature of the ligands:$1$. In $Na_{2}[NiCl_{4}]$,$Ni$ is in the $+2$ oxidation state ($3d^{8}$ configuration). $Cl^-$ is a weak field ligand,leading to $sp^{3}$ hybridization (tetrahedral geometry).$2$. In $NiCl_{2} \cdot 6 H_{2}O$,the complex is $[Ni(H_{2}O)_{6}]^{2+}$. $Ni^{2+}$ is in an octahedral environment with $sp^{3}d^{2}$ hybridization.$3$. In $K_{2}[Ni(CN)_{4}]$,$Ni$ is in the $+2$ oxidation state ($3d^{8}$ configuration). $CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals. This leaves one $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals available for hybridization,resulting in $dsp^{2}$ hybridization (square planar geometry).$4$. In $[Ni(CO)_{4}]$,$Ni$ is in the $0$ oxidation state ($3d^{8} 4s^{2}$ configuration). $CO$ is a strong field ligand,causing the $4s$ electrons to pair into the $3d$ orbitals,resulting in $sp^{3}$ hybridization (tetrahedral geometry).Therefore,the correct compound is $K_{2}[Ni(CN)_{4}]$.

According to the valence bond theory,the hybridization of the central metal atom is $dsp^{2}$ for which one of the following compounds?

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