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

Magnetic moment (in B.M.) of $\left[ Ru \left( H _{2} O \right)_{6}\right]^{2+}$

would be; while considering that $\Delta_{0}> P$

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

$\Rightarrow $ In $Ru ^{2+} $

$\Rightarrow 4 d ^{6} $

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



$\Rightarrow $ Here number of unpaired electrons in

$Ru ^{2+}=\left( t _{2} g \right)^{6}( eg )^{0}=0$ and Hence

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