The ratio of spin-only magnetic moment values μ_(eff)[Cr(CN)₆]³⁻ / μ_(eff)[Cr(H₂ O)₆]³⁺ is _________.

Solution

Chemistry ma’am · 3 steps

Cr³⁺ is d³, so three electrons sit singly in three t_(2g) orbitals with either ligand.

1. Oxidation state. In [Cr(CN)₆]³⁻, x + 6(-1) = -3 gives x = +3. In [Cr(H₂O)₆]³⁺, water is neutral, so Cr is again +3. Cr is [Ar] 3d⁵ 4s¹, so Cr³⁺ is 3d³ in both complexes.
2. Fill the orbitals. In an octahedral field the three electrons go into the three lower t_(2g) orbitals, one in each. This happens with strong-field CN⁻ and with weak-field H₂O, because pairing only becomes a choice from the fourth electron onwards. Both complexes are t_(2g)³ e_g⁰ with n = 3 unpaired electrons.

   μ = √n(n+2) = √(3 × 5) = √15 ≈ 3.87 BM

3. So. Both complexes have μ = √15 BM, so the ratio is √15/√15 = 1.

The common slip: 0.45. Pairing electrons in [Cr(CN)₆]³⁻ because CN⁻ is strong field leaves 1 unpaired electron and gives √3/√15 ≈ 0.45. Three electrons in three t_(2g) orbitals stay unpaired in any field.

The idea to keep. For d¹, d² and d³ ions in an octahedral field, ligand strength cannot change the number of unpaired electrons. High spin versus low spin only matters for d⁴ to d⁷.

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