Which of the following pairs of compounds is | vedclass

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(C) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ has $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $dsp^2$ hybr

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$[Ni(CN)_4]^{2-}, [Rh(CO)_2(PPh_3)_2]^+$

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(C) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ has $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $dsp^2$ hybridization,resulting in a square planar geometry.$2$. $[Rh(CO)_2(PPh_3)_2]^+$: $Rh^+$ has $d^8$ configuration. Being a $4d$ series metal,it forms square planar complexes with strong field ligands like $CO$ and $PPh_3$. Thus,it is square planar.$3$. $[NiCl_4]^{2-}$ is tetrahedral $(sp^3)$.$4$. $[Ni(CO)_4]$ is tetrahedral $(sp^3)$.$5$. $[PtCl_4]^{2-}$ is square planar,but $[Ni(CN)_4]^{3-}$ is not a common stable species (usually $Ni(I)$ or $Ni(II)$ complexes are considered).Therefore,the pair $[Ni(CN)_4]^{2-}$ and $[Rh(CO)_2(PPh_3)_2]^+$ consists of two square planar complexes.

List-$I$ (Complex/Species)	List-$II$ (Shape and magnetic moment)
$A$. $[Ni(CO)_4]$	$I$. Tetrahedral,$2.8 \ BM$
$B$. $[Ni(CN)_4]^{2-}$	$II$. Square planar,$0 \ BM$
$C$. $[NiCl_4]^{2-}$	$III$. Tetrahedral,$0 \ BM$
$D$. $[MnBr_4]^{2-}$	$IV$. Tetrahedral,$5.9 \ BM$

Which of the following pairs of compounds is planar?

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