Question

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

Answer 1

Answer: 00

Magnetic moment (in B.M.) of ${\left[Ru{\left(H_{2}O\right)}_6\right]}^{2+}$

would be; while considering that ${\Delta }_0>P$

$R{u}_{(44)};[Kr]4d^{7}5s^1$ (in ground state)

$\Rightarrow$ In $R{u}^{2+}$

$\Rightarrow 4d^6$

$\Rightarrow {\left(t_{2}g\right)}^6(eg{)}^0$



$\Rightarrow$ Here number of unpaired electrons in

$R{u}^{2+}={\left(t_{2}g\right)}^6(eg{)}^0=0$ and Hence

${\mu }_m=\sqrt{n(n+2)B\cdot M}=0B.M$