Considering that \(\Delta_0 \gt P\), the magnetic moment (in BM) of \(\mathrm{[Ru(H_2O)_6]^{2+}}\) would be _________.
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Correct Answer: 0
Solution and Explanation
Ru (Z = 44) = [Kr] 4d75s1; \(\mathrm{Ru^{2+}}\) = 4d6.
Given \(\Delta_0 \gt P\), pairing is favoured: configuration \(t_{2g}^6\,e_g^0\) (low spin), so unpaired electrons n = 0.
\(\mu = \sqrt{n(n+2)} = \sqrt{0} = 0\) BM.
Answer: 0
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