The correct statement about [Ni(CO)4] and [Ni(CN)4]2- is:

🧠 Same Metal, Same d-Count, Different Geometries

Both complexes have $\mathrm{Ni}$ in oxidation state 0 (with CO) or +2 (with $\mathrm{CN^-}$). The crucial trick is that two different ligand environments produce two different geometries for the same metal:

• $[\mathrm{Ni(CO)_4}]$: $\mathrm{Ni}$ is $\mathrm{d^{10}}$, $\mathrm{CO}$ is a strong-field $\pi$-acceptor → sp³ tetrahedral, all electrons paired → diamagnetic.
• $[\mathrm{Ni(CN)_4}]^{2-}$: $\mathrm{Ni^{2+}}$ is $\mathrm{d^8}$, $\mathrm{CN^-}$ is strong-field → forces pairing in d-orbitals → dsp² square planar → diamagnetic.

Both diamagnetic, but geometries differ — that's option (2).

🗺️ Verify Each

$[\mathrm{Ni(CO)_4}]$. Ni: $[\mathrm{Ar}]\,3\mathrm{d}^8\,4\mathrm{s}^2$. With CO bonding, Ni promotes to $\mathrm{d^{10}}\,\mathrm{s^0}$ configuration ("zero-valent"). All ten d-electrons paired. Hybridisation sp³. Diamagnetic.

$[\mathrm{Ni(CN)_4}]^{2-}$. $\mathrm{Ni^{2+}}$: $\mathrm{d^8}$. $\mathrm{CN^-}$ pairs the d-electrons → vacant 3d orbital → dsp² hybridisation → square planar. Diamagnetic.

⚡ The "Strong Field, Specific Geometry" Pair

Memorise these two together:

• $\mathrm{Ni}^0$ + $\mathrm{CO}$ → tetrahedral ($\mathrm{d^{10}}$).
• $\mathrm{Ni^{2+}}$ + $\mathrm{CN^-}$ → square planar ($\mathrm{d^8}$, low-spin).

JEE asks variants of this every other year.

⚠️ The "Both Tetrahedral" Reflex

Option (3) ("both diamagnetic and tetrahedral") is bait. Students who only learned "Ni → tetrahedral" miss that $\mathrm{d^8}$ with strong-field ligands flips to square planar. The d-count and ligand strength together determine geometry — never just metal identity.

$\boxed{\text{Answer: (2) Both diamagnetic; } [\mathrm{Ni(CO)_4}] \text{ tetrahedral, } [\mathrm{Ni(CN)_4}]^{2-} \text{ square planar}}$