Identify, from the following, the diamagnetic, tetrahedral complex

(a) Ni in $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$ exist as $\mathrm{Ni}^{2+}$ ion. Cl is a weak field ligand (high spin). It will not cause pairing of electrons. Hence, configuration of $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$ is Tetrahedral with two unpaired electrons (i.e., paramagnetic) and has $s p^3$ hybridisation. (b) In $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-},\left(\mathrm{C}_2 \mathrm{O}_4\right)_3^{2-}$ is a bidentate ligand thus, give octahedral structure. (c) In $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$, Ni exist as $\mathrm{Ni}^{2+}$ ion. $\mathrm{CN}^{-}$is a strong field ligand (low spin). Causes pairing of electrons of $3 d$ orbital. Hence, the structure of is square planar and it is diamagnetic. (d) In $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$, Ni has zero oxidation state. i.e., CO is a strong field ligand causes rearrangement and pairing of electrons of $3 d$ and $4 s$ orbital. Structure of $\mathrm{Ni}\left(\mathrm{CO}_4\right)$ is tetrahedral with $s p^3$ hybridisation and it is diamagnetic. Hence, (4) is the correct answer.