The magnetic moment of the central metal ion | vedclass

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(B) In $K_2MnO_4$,the oxidation state of $Mn$ is calculated as: $2(+1) + x + 4(-2) = 0$,which gives $x = +6$. The electronic configuration of $Mn$ $(Z

Vedclass · Accepted Answer

$1.73$

Vedclass · Answer

(B) In $K_2MnO_4$,the oxidation state of $Mn$ is calculated as: $2(+1) + x + 4(-2) = 0$,which gives $x = +6$. The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$. The electronic configuration of $Mn^{6+}$ is $[Ar] 3d^1$. Since there is $1$ unpaired electron $(n=1)$,the magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$.

The magnetic moment of the central metal ion in $K_2MnO_4$ is ........ $B.M.$

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