Which of the following compounds has a differ | vedclass

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(A) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing. Hybridization is $dsp^2$,geometry is square planar.$

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$[Ni(CN)_4]^{2-}$

Vedclass · Answer

(A) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing. Hybridization is $dsp^2$,geometry is square planar.$2$. $[Cu(CN)_4]^{3-}$: $Cu^+$ is $3d^{10}$. Hybridization is $sp^3$,geometry is tetrahedral.$3$. $[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing. Hybridization is $sp^3$,geometry is tetrahedral.$4$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,no pairing. Hybridization is $sp^3$,geometry is tetrahedral.Thus,$[Ni(CN)_4]^{2-}$ has a square planar geometry,while the others are tetrahedral.

Which of the following compounds has a different geometry from the others?

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