The hybridisation of Ni centre in [Ni(PPh_3)_ | vedclass

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Question

(A) The oxidation number of $Ni$ in both $[Ni(PPh_3)_2Cl_2]$ and $[NiCl_4]^{2-}$ is $+2$. The electronic configuration of $Ni(II)$ is $[Ar] 3d^8 4s^0$

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$dsp^2$ and $sp^3$

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(A) The oxidation number of $Ni$ in both $[Ni(PPh_3)_2Cl_2]$ and $[NiCl_4]^{2-}$ is $+2$. The electronic configuration of $Ni(II)$ is $[Ar] 3d^8 4s^0$.In $[Ni(PPh_3)_2Cl_2]$,$PPh_3$ is a strong field ligand,which forces the electrons in the $3d$ orbital to pair up,leaving one $3d$ orbital vacant. This allows for $dsp^2$ hybridisation,resulting in a square planar geometry.In $[NiCl_4]^{2-}$,$Cl^-$ is a weak field ligand,which does not cause pairing of electrons in the $3d$ orbital. Thus,the $4s$ and $4p$ orbitals are used for $sp^3$ hybridisation,resulting in a tetrahedral geometry.Therefore,the hybridisation of $Ni$ in $[Ni(PPh_3)_2Cl_2]$ and $[NiCl_4]^{2-}$ are $dsp^2$ and $sp^3$ respectively.

List - $I$ (Complex)	List – $II$ (Geometry)
$(a)$ $[Co(NH_3)_6]^{3+}$	$(i)$ Trigonal bipyramidal
$(b)$ $[NiCl_4]^{2-}$	(ii) Octahedral
$(c)$ $[Ni(CN)_4]^{2-}$	(iii) Tetrahedral
$(d)$ $[Fe(CO)_5]$	(iv) Square planar

The hybridisation of $Ni$ centre in $[Ni(PPh_3)_2Cl_2]$ and $[NiCl_4]^{2-}$,respectively are

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