Question

Considering that ${∆}_{°}>\mathrm{P}$, the magnetic moment (in BM) of ${\left[\mathrm{Ru}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+2}$ would be

Answer 1

The magnetic moment(in BM) of ${\left[\mathrm{Ru}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+2}$ is calculated as:

Step 1:

${\left[\mathrm{Ru}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+2}$

${\mathrm{Ru}}^{+2}=3{\mathrm{d}}^6(∆\mathrm{o}>\mathrm{P})$

Step 2:

Filling of ${\mathrm{Ru}}^{+2}$ electrons in the above complex according to Crystal field splitting theory:

$={{\mathrm{t}}_{2\mathrm{g}}}^6{{\mathrm{e}}_{\mathrm{g}}}^0$

Hence, Number of unpaired electrons($\mathrm{n}$) $=0$

Step 3:

Spin only magnetic moment($\mathrm{\mu }$) $=\sqrt{\mathrm{n}(\mathrm{n}+2)}$

$$\begin{gathered} =\sqrt{0(0+2)} \\ \mathrm{\mu }=0 \end{gathered}$$

Therefore, the magnetic moment(in BM) is calculated to be $0$.