The number of unpaired d-electrons in [Co(H₂ O)₆]³⁺ is ________.

Solution

Chemistry ma’am · 3 steps

Water is a weak ligand, but Co(III) is a special case: its d6 electrons pair up in t_(2g).

1. Find the d count. Water is neutral, so cobalt carries the full +3 charge. Co is [Ar] 3d⁷ 4s², so Co³⁺ is 3d⁶.
2. Decide the spin. In an octahedral field the six electrons can go high spin, t_(2g)⁴ e_g², or low spin, t_(2g)⁶ e_g⁰. A +3 charge on a 3d metal makes the splitting Δₒ large. For Co³⁺ it is large enough even with H₂O to beat the pairing energy, so the complex is low spin.

   Co³⁺ (d⁶), Δₒ > P: t_(2g)⁶ e_g⁰

3. So. All three t_(2g) orbitals are full and both e_g orbitals are empty. There are 0 unpaired electrons, and [Co(H₂O)₆]³⁺ is diamagnetic. The answer is (a).

The idea to keep. Do not judge spin from the ligand alone. Co³⁺ complexes are low spin with almost every ligand, water included. Only F⁻ gives high-spin [CoF₆]³⁻.

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