Identify, from the following, the diamagnetic, tetrahedral complex

Question

Aicharya · Accepted Answer

(a) $\mathrm{Ni}$ in $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$ exist as $\mathrm{Ni}^{2+}$ ion. $\because \quad \mathrm{C} \ell$ is a weak field ligand (high spin). It will not cause pairing of electrons. Hence, configuration of $\left[\mathrm{Ni}(\mathrm{C} \ell)_4\right]^{2-}$ is

Tetrahedral with two unpaired electrons (i.e., paramagnetic) and has $s p^3$ hybridisation. (b) In $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right],\left(\mathrm{C}_2 \mathrm{O}_4\right)_3$ is a bidentate ligand thus, give octahedral structure. (c) In $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$, $\mathrm{Ni}$ exist as $\mathrm{Ni}^{2+}$ ion. $\because \quad \mathrm{CN}^{-}$is a strong field ligand (low spin), causes pairing of electrons of $3 d$ orbital.

Hence, the structure of $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is square planar and it is diamagnetic. (d) In $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$, $\mathrm{Ni}$ has zero oxidation state. i.e.,

$\mathrm{CO}$ is a strong field ligand causes rearrangement and pairing of electrons of $3 d$ and $4 s$ orbital.

Structure of $\mathrm{Ni}(\mathrm{CO})_4$ is tetrahedral with $s p^3$ hybridisation and it is diamagnetic. Hence, (d) is the correct answer.

Identify, from the following, the diamagnetic, tetrahedral complex

Solution