The spin-only magnetic moment of [Mn(H_2O)_6] | vedclass

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(B) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. In the complex $[Mn(H_2O)_6]^{2+}$,$Mn$ is in the $+2$ oxidation state,so its c

Vedclass · Accepted Answer

$6$

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(B) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. In the complex $[Mn(H_2O)_6]^{2+}$,$Mn$ is in the $+2$ oxidation state,so its configuration is $3d^5$. Since $H_2O$ is a weak field ligand,the electrons remain unpaired in the $d$-orbitals,resulting in $n=5$ unpaired electrons. The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$. Substituting $n=5$,we get $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \ B.M.$. The nearest integer value is $6$.

The spin-only magnetic moment of $[Mn(H_2O)_6]^{2+}$ complex is $........ \ B.M.$ (Nearest integer) (Given: Atomic number of $Mn$ is $25$)

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