Considering that Δ₀ > P, the magnetic moment
(in BM) of [Ru(H₂O)₆]²⁺ would be _________.

Solution

Chemistry ma’am · 3 steps

Get the d-count of Ru²⁺ first, then let Δ₀ > P pair every electron in the t₂g set.

1. Count d electrons. Ruthenium is [Kr] 4d⁷ 5s¹. Removing two electrons, the 5s electron first and then one 4d electron, leaves Ru²⁺ as 4d⁶.
2. Use Δ₀ > P. When the splitting Δ₀ is larger than the pairing energy P, it costs less to pair electrons in t_(2g) than to promote them to e_g. So all six electrons fill the lower set: t_(2g)⁶ e_g⁰, a low-spin complex.

   μ = √n(n+2) BM = √(0 × 2) = 0

3. So. With t_(2g)⁶ every electron is paired, so n = 0 and the magnetic moment is 0 BM. The complex is diamagnetic.

The common slip: 4.9. Filling the orbitals high-spin as t_(2g)⁴ e_g² leaves 4 unpaired electrons and gives √24 ≈ 4.9 BM, which ignores the given Δ₀ > P.

The idea to keep. Whenever you are told Δ₀ > P, fill t_(2g) completely before putting anything in e_g. For 4d and 5d metals this low-spin case is the usual one, even with water as the ligand.

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