Which cation from the following exhibits no m | vedclass

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(B) The magnetic moment of an ion is determined by the number of unpaired electrons,calculated as $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number o

Vedclass · Accepted Answer

$Sc^{3+}$

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(B) The magnetic moment of an ion is determined by the number of unpaired electrons,calculated as $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $Cr^{3+}$ $([Ar] 3d^3)$,$n = 3$,so it is paramagnetic.$2$. For $Sc^{3+}$ $([Ar] 3d^0)$,$n = 0$,so it has no unpaired electrons and exhibits no magnetic moment (diamagnetic).$3$. For $Cu^{2+}$ $([Ar] 3d^9)$,$n = 1$,so it is paramagnetic.$4$. For $V^{3+}$ $([Ar] 3d^2)$,$n = 2$,so it is paramagnetic.Therefore,$Sc^{3+}$ exhibits no magnetic moment.

Which cation from the following exhibits no magnetic moment?

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