The spin-only magnetic moment value of M^{3+} | vedclass

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(A) The standard electrode potential $E^{0}_{M^{3+}/M^{2+}}$ is negative only for the $Cr^{3+}/Cr^{2+}$ pair $(E^{0} = -0.41 \, V)$.The electronic con

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The standard electrode potential $E^{0}_{M^{3+}/M^{2+}}$ is negative only for the $Cr^{3+}/Cr^{2+}$ pair $(E^{0} = -0.41 \, V)$.The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^{3}$.The number of unpaired electrons $(n)$ in $Cr^{3+}$ is $3$.The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \, B.M.$$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, B.M.$The nearest integer value is $4$.

The spin-only magnetic moment value of $M^{3+}$ ion (in gaseous state) from the pairs $Cr^{3+}/Cr^{2+}$,$Mn^{3+}/Mn^{2+}$,$Fe^{3+}/Fe^{2+}$,and $Co^{3+}/Co^{2+}$ that has a negative standard electrode potential,is $B.M.$ (Nearest integer).

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