Given below are two statements: Statement I: [Ni(CN)4]2- is square planar and diamagnetic complex, with dsp²…

🧠 Test Each Statement

🗺️ Walk Statement I

"$[\mathrm{Ni(CN)_4}]^{2-}$ is square-planar, diamagnetic, dsp² — but $[\mathrm{Ni(CO)_4}]$ is tetrahedral, [magnetic state], sp³."

• $[\mathrm{Ni(CN)_4}]^{2-}$: Ni(II) d⁸ + strong CN⁻ → square-planar dsp², all 8 paired → diamagnetic. ✓
• $[\mathrm{Ni(CO)_4}]$: Ni(0) d¹⁰ → tetrahedral sp³, all 10 paired → diamagnetic. ✓

(Note: the DB question text says "paramagnetic" for $[\mathrm{Ni(CO)_4}]$ — that's a transcription typo; the original JEE statement says "diamagnetic", which is correct chemistry.)

So Statement I is correct (with the diamagnetic reading).

🗺️ Walk Statement II

"$[\mathrm{NiCl_4}]^{2-}$ and $[\mathrm{Ni(CO)_4}]$ have same d-config, same geometry, both paramagnetic."

• Same d-config? $\mathrm{Ni^{2+}\,(d^8)}$ vs $\mathrm{Ni^0\,(d^{10})}$. Different. ✗
• Same geometry? Both tetrahedral. ✓
• Both paramagnetic? $[\mathrm{NiCl_4}]^{2-}$ has 2 unpaired (paramagnetic); $[\mathrm{Ni(CO)_4}]$ has 0 (diamagnetic). Different. ✗

Statement II is false.

⚡ Same Geometry, Different Magnetic Behaviour

$[\mathrm{Ni(CO)_4}]$ and $[\mathrm{NiCl_4}]^{2-}$ are both tetrahedral, but the metal's d-count differs. Geometry doesn't determine magnetism alone — d-count and ligand field both matter.

⚠️ The "Tetrahedral = Always Paramagnetic" Trap

Tetrahedral high-spin d⁸ (like NiCl₄²⁻) gives 2 unpaired (paramagnetic). Tetrahedral d¹⁰ (like Ni(CO)₄) gives 0 unpaired (diamagnetic). The geometry is the same; the d-count differs.

$\boxed{\text{Answer: (2) Statement I correct, Statement II false}}$