The hybridization involved in the complex [Ni | vedclass

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(D) $1$. The oxidation state of $Ni$ in $[Ni(CN)_4]^{2-}$ is calculated as: $x + 4(-1) = -2$,so $x = +2$.$2$. The electronic configuration of $Ni^{2+}

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$dsp^2$

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(D) $1$. The oxidation state of $Ni$ in $[Ni(CN)_4]^{2-}$ is calculated as: $x + 4(-1) = -2$,so $x = +2$.$2$. The electronic configuration of $Ni^{2+}$ is $[Ar]^{18} 3d^8 4s^0$.$3$. $CN^-$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.$4$. After pairing,one $3d$,one $4s$,and two $4p$ orbitals are available for hybridization.$5$. Thus,the hybridization is $dsp^2$.

The hybridization involved in the complex $[Ni(CN)_4]^{2-}$ is (Atomic number of $Ni = 28$):

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