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(B) $1$. In $[Ni(CO)_{4}]$, $Ni$ is in $0$ oxidation state $(3d^{8} 4s^{2})$. $CO$ is a strong field ligand, causing pairing of electrons. Hybridisati

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$2$

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(B) $1$. In $[Ni(CO)_{4}]$, $Ni$ is in $0$ oxidation state $(3d^{8} 4s^{2})$. $CO$ is a strong field ligand, causing pairing of electrons. Hybridisation is $sp^{3}$, unpaired electrons $= 0$.$2$. In $[NiCl_{4}]^{2-}$, $Ni^{2+}$ is $3d^{8}$. $Cl^{-}$ is a weak field ligand, no pairing occurs. Hybridisation is $sp^{3}$, unpaired electrons $= 2$.$3$. In $[PtCl_{2}(NH_{3})_{2}]$, $Pt^{2+}$ is $5d^{8}$. $Pt$ is a $5d$ series metal, so pairing occurs even with weak ligands. Hybridisation is $dsp^{2}$, unpaired electrons $= 0$.$4$. In $[Ni(CN)_{4}]^{2-}$, $Ni^{2+}$ is $3d^{8}$. $CN^{-}$ is a strong field ligand, causing pairing. Hybridisation is $dsp^{2}$, unpaired electrons $= 0$.$5$. In $[Pt(CN)_{4}]^{2-}$, $Pt^{2+}$ is $5d^{8}$. $CN^{-}$ is a strong field ligand. Hybridisation is $dsp^{2}$, unpaired electrons $= 0$.Total unpaired electrons $= 0 + 2 + 0 + 0 + 0 = 2$.

Total number of unpaired electrons present in the central metal atoms/ions of $[Ni(CO)_{4}]$, $[NiCl_{4}]^{2-}$, $[PtCl_{2}(NH_{3})_{2}]$, $[Ni(CN)_{4}]^{2-}$ and $[Pt(CN)_{4}]^{2-}$ is . . . . . . .

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