Considering that ${\mathrm{\Delta }}_0>P$, the magnetic moment (in BM) of ${\left[\mathrm{Ru}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$ would be_____.

Outer electronic configuration of Ru is $4{\mathrm{d}}^{6}5{\mathrm{s}}^2.$ Oxidation state of Ru in the given complex is +2 and its outer electronic configuration becomes $4{\mathrm{d}}^6.$

 As ${\mathrm{\Delta }}_0>\mathrm{P}$electrons prefer to pair up in lower energy ${\mathrm{t}}_{2\mathrm{g}}$ of  octahedral complex then a low spin complex made of configuration, ${{\mathrm{t}}_{2\mathrm{g}}}^6{{\mathrm{e}}_{\mathrm{g}}}^0$.

Magnetic moment formula is:

 $$\begin{gathered} \mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2)} \\ \mathrm{n}=\mathrm{number} \mathrm{of} \mathrm{unpaired} \mathrm{electrons} \end{gathered}$$    

In the given complex, all electrons are paired, hence value of n is zero, $\mathrm{\mu }=0.$