The total number of unpaired electrons present in the complex K₃[Cr(oxalate)₃] is _____________.

Solution

Chemistry ma’am · 4 steps

Get chromium's oxidation state first: Cr³⁺ is d³, and three t_(2g) electrons never pair.

1. Oxidation state. Oxalate, C₂O₄²⁻, carries a charge of -2, and each K⁺ carries +1. Setting the total charge to zero gives 3 + x + 3(-2) = 0, so x = +3.
2. Electron count. Cr is [Ar] 3d⁵ 4s¹. Removing three electrons leaves Cr³⁺ as [Ar] 3d³.
3. Fill the orbitals. Three bidentate oxalate ions make an octahedral complex, so the d orbitals split into t_(2g) (lower) and e_g (upper). The three electrons go one each into the three t_(2g) orbitals, giving t_(2g)³ e_g⁰. Pairing is not needed whether oxalate is a weak or a strong field ligand.
4. So. All three d electrons stay unpaired, so the complex has 3 unpaired electrons.

The idea to keep. In an octahedral complex, d¹ to d³ (and d⁸ to d¹⁰) give the same unpaired count in weak and strong fields. Ligand strength matters only for d⁴ to d⁷.

Answer from the NTA final answer key. Solution by PrepWiser.