The hybridisations of Ni(CO)_4 and [Cr(H_2O)_ | vedclass

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(C) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$. Its electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,the $4s$ e

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$sp^3$ and $sp^3d^2$

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(C) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$. Its electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,the $4s$ electrons pair up in the $3d$ orbitals,resulting in $sp^3$ hybridisation.The oxidation state of $Cr$ in $[Cr(H_2O)_6]^{2+}$ is $+2$. Its electronic configuration is $[Ar] 3d^4$. Since $H_2O$ is a weak field ligand,it does not cause pairing of electrons. However,the complex $[Cr(H_2O)_6]^{2+}$ involves $sp^3d^2$ hybridisation (outer orbital complex).

List-$I$ (Compound)	List-$II$ (Hybridization of Central atom)
$A. [Ni(NH_3)_6]^{2+}$	$1. sp^3$
$B. [PtCl_4]^{2-}$	$2. sp^3d^2$
$C. [Ni(CO)_4]$	$3. dsp^2$
$D. [Co(ox)_3]^{3-}$	$4. d^2sp^3$

The hybridisations of $Ni(CO)_4$ and $[Cr(H_2O)_6]^{2+}$ respectively,are

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