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The hybridization involved in complex $[Ni(CN)_4]^{2-}$ is (At. No. Ni = 28)
A
$d^2sp^2$
B
$d^2sp^3$
C
$dsp^2$
D
$sp^3$
Detailed Solution
Oxidation state of Ni: $x - 4 = -2 \Rightarrow x = +2$; $Ni^{2+}$: $[Ar]3d^84s^0$
$CN^-$ is a strong field ligand, so the two unpaired 3d electrons pair up, leaving one 3d orbital empty.
One 3d, one 4s and two 4p orbitals hybridise: $dsp^2$ (square planar).