$[\mathrm{Ni(CO)_4}]$ is tetrahedral and diamagnetic. The oxidation state of nickel in this complex is

1. $[\mathrm{Ni(CO)_4}]$ is a NEUTRAL molecule. 2. CO (carbonyl) is a neutral ligand — contributes 0 charge. 3. So: $x + 4(0) = 0 \Rightarrow x = 0$. Nickel is in the ZERO oxidation state. 4. This is a METAL CARBONYL — an important class of coordination compounds with low or zero oxidation state metals stabilised by $\pi$-backbonding from filled metal d-orbitals into empty CO $\pi^*$ orbitals. 5. Ni$^0$ has configuration $3d^{10}$ (taking 4s electrons into 3d) — fully paired, so DIAMAGNETIC, consistent with experiment. 6. The complex is famous as the volatile intermediate in the Mond process for nickel purification. 7. Options B–D would require ligand-derived charge that CO doesn't have. _Source: syllabus Chemistry Part 1, Ch 5, §5.7 (Bonding in metal carbonyls — Ni(CO)$_4$ example), p. 16._