Geometrical shapes of the complexes formed by | vedclass

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Question

(B) $Ni^{2+}$ has a $3d^8$ electronic configuration.$1$. For $[NiCl_4]^{2-}$,$Cl^{-}$ is a weak field ligand,so it does not cause pairing of electrons

Vedclass · Accepted Answer

tetrahedral,square planar and octahedral

Vedclass · Answer

(B) $Ni^{2+}$ has a $3d^8$ electronic configuration.$1$. For $[NiCl_4]^{2-}$,$Cl^{-}$ is a weak field ligand,so it does not cause pairing of electrons. The hybridization is $sp^3$,resulting in a tetrahedral geometry.$2$. For $[Ni(CN)_4]^{2-}$,$CN^{-}$ is a strong field ligand,which causes pairing of $3d$ electrons. The hybridization is $dsp^2$,resulting in a square planar geometry.$3$. For $[Ni(H_2O)_6]^{2+}$,$H_2O$ is a ligand that forms an octahedral complex with $Ni^{2+}$ involving $sp^3d^2$ hybridization.Thus,the shapes are tetrahedral,square planar,and octahedral,respectively. Therefore,option $B$ is correct.

Geometrical shapes of the complexes formed by the reaction of $Ni^{2+}$ with $Cl^{-}$,$CN^{-}$,and $H_2O$,respectively,are

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