Question

[NiCl₄]^2- is paramagnetic, while Ni(CO)₄ is diamagnetic, although both are tetrahedral. Why?

Hint:

Compare the ligand field strength: Cl^- is weak so Ni²⁺ (d⁸) keeps 2 unpaired electrons, while CO is strong and pairs all electrons to 3d¹⁰. Both remain sp³ tetrahedral.

Answer 1

Step 1: Magnetic behaviour is decided by the number of unpaired electrons n through the spin-only moment \(\mu = \sqrt{n(n+2)}\) BM. If n = 0 the species is diamagnetic; if n > 0 it is paramagnetic.
Step 2: [NiCl₄]^2-. Nickel is +2 (d⁸). Chloride sits low in the spectrochemical series, so its field is too weak to beat the electron pairing energy. The d⁸ arrangement keeps two unpaired electrons, n = 2, giving \(\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83\) BM. Bonding is sp³ (tetrahedral) and the ion is paramagnetic.
Step 3: Ni(CO)₄. Nickel is in the zero oxidation state (3d⁸4s²). Carbon monoxide lies at the strong end of the spectrochemical series, so it drives all electrons together to give a 3d¹⁰ core with empty 4s and 4p orbitals. These accept the four CO lone pairs through sp³ hybridisation. With n = 0, \(\mu = 0\) and the molecule is diamagnetic though still tetrahedral.
Step 4: Hence identical tetrahedral geometry can still show opposite magnetism, because it is the ligand field strength, not the shape, that fixes the number of unpaired electrons.