Considering that Delta_{0} P,the magnetic m | vedclass

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(A) The electronic configuration of $Ru$ $(Z=44)$ is $[Kr] 4d^{7} 5s^{1}$.For $Ru^{2+}$,the configuration is $4d^{6}$.Given that $\Delta_{0} > P$,the

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(A) The electronic configuration of $Ru$ $(Z=44)$ is $[Kr] 4d^{7} 5s^{1}$.For $Ru^{2+}$,the configuration is $4d^{6}$.Given that $\Delta_{0} > P$,the complex is a low-spin complex.In an octahedral field,the $6$ electrons in the $4d$ orbital will occupy the $t_{2g}$ orbitals as $(t_{2g})^{6} (e_{g})^{0}$.Since all electrons are paired,the number of unpaired electrons $(n)$ is $0$.The magnetic moment $\mu = \sqrt{n(n+2)} \ BM = \sqrt{0(0+2)} \ BM = 0 \ BM$.

Considering that $\Delta_{0} > P$,the magnetic moment (in $BM$) of $[Ru(H_{2}O)_{6}]^{2+}$ would be

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