Considering that $$\Delta_0 > P$$, the magnetic moment (in BM) of $$[\text{Ru}(\text{H}_2\text{O})_6]^{2+}$$ would be

First, we note the atomic number of ruthenium is $$44$$, so the ground-state electronic configuration of a neutral $$\text{Ru}$$ atom is $$[\text{Kr}]\,4d^7\,5s^1$$.

The complex under consideration is $$[\text{Ru}(\text{H}_2\text{O})_6]^{2+}$$. The oxidation state of ruthenium in this species is obtained by letting the total charge be the algebraic sum of the metal ion’s charge and the charges of the ligands. Since each $$\text{H}_2\text{O}$$ ligand is neutral, the metal must provide the overall $$+2$$ charge. Hence, the metal centre is $$\text{Ru}^{2+}$$.

To obtain the d-electron count of $$\text{Ru}^{2+}$$, we remove two electrons from the neutral atom’s configuration. The $$5s$$ orbital is higher in energy than $$4d$$, so the two electrons are removed first from $$5s$$ and then from $$4d$$ if needed:

$$ \text{Ru}: [\text{Kr}]\,4d^7\,5s^1 \\ \text{Ru}^{2+}: [\text{Kr}]\,4d^6 $$

Thus, $$\text{Ru}^{2+}$$ is a $$4d^6$$ ion.

The complex is octahedral, so the $$4d$$ orbitals split into a lower-energy $$t_{2g}$$ set and a higher-energy $$e_g$$ set. We are told that $$\Delta_0 > P$$, where $$\Delta_0$$ is the octahedral crystal-field splitting energy and $$P$$ is the pairing energy. The condition $$\Delta_0 > P$$ means that it is energetically favourable for electrons to pair up in the lower $$t_{2g}$$ orbitals rather than occupy the higher $$e_g$$ orbitals. Therefore the complex will be a low-spin complex.

For a $$d^6$$ ion in a low-spin octahedral field, all six electrons occupy the lower $$t_{2g}$$ orbitals, giving the configuration $$t_{2g}^6\,e_g^0$$. In this arrangement every electron is paired, so the number of unpaired electrons is

$$ n = 0. $$

The spin-only formula for magnetic moment is stated next:

$$ \mu = \sqrt{n(n+2)} \text{ BM}, $$

where $$n$$ is the number of unpaired electrons.

Substituting $$n = 0$$, we obtain

$$ \mu = \sqrt{0(0+2)} = \sqrt{0} = 0 \text{ BM}. $$

Hence, the correct answer is Option A.