Question

Select correct option :

• A. $[\text{Ni}(\text{CN})_4]^{2-}$ and $[\text{Ni}(\text{CO})_4]$ both are diamagnetic while $[\text{NiCl}_4]^{2-}$ is paramagnetic
• B. $[\text{Ni}(\text{CN})_4]^{2-}$ and $[\text{NiCl}_4]^{2-}$ both are diamagnetic while $[\text{Ni}(\text{CO})_4]$ is paramagnetic
• C. $[\text{NiCl}_4]^{2-}$ and $[\text{Ni}(\text{CO})_4]$ both are diamagnetic while $[\text{Ni}(\text{CN})_4]^{2-}$ is paramagnetic
• D. Only $[\text{Ni}(\text{CN})_4]^{2-}$ is diamagnetic while both $[\text{NiCl}_4]^{2-}$ and $[\text{Ni}(\text{CO})_4]$ are paramagnetic

Hint:

For Ni(II) ($d^8$): SFL $\to$ Square Planar, Diamagnetic. WFL $\to$ Tetrahedral, Paramagnetic. Ni(0) with SFL is always Tetrahedral, Diamagnetic.

Answer 1

The Correct Option is A

$[\text{Ni}(\text{CN})_4]^{2-}$: $\text{Ni}^{2+} (3d^8)$ with Strong Field Ligand ($\text{CN}^-$). Electrons pair up. Configuration becomes $dsp^2$ (Square Planar). All paired $\implies$ Diamagnetic.
$[\text{Ni}(\text{CO})_4]$: $\text{Ni}^0 (3d^8 4s^2)$. $\text{CO}$ is SFL. $s$-electrons shift to $d$, making $3d^{10}$. Hybridization $sp^3$ (Tetrahedral). All paired $\implies$ Diamagnetic.
$[\text{NiCl}_4]^{2-}$: $\text{Ni}^{2+} (3d^8)$ with Weak Field Ligand ($\text{Cl}^-$). No pairing. $sp^3$ (Tetrahedral). Two unpaired electrons in $3d$ orbitals. Paramagnetic.