The magnetic moment of a complex ion is 2.83 | vedclass

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Question

(A) The spin-only magnetic moment is given by the formula $\mu_{eff} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu_{eff

Vedclass · Accepted Answer

$[V(H_2O)_6]^{3+}$

Vedclass · Answer

(A) The spin-only magnetic moment is given by the formula $\mu_{eff} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu_{eff} = 2.83 \ BM$,we have $\sqrt{n(n+2)} = 2.83$,which implies $n(n+2) \approx 8$,so $n = 2$.In $[V(H_2O)_6]^{3+}$,the oxidation state of $V$ is $+3$. The electronic configuration of $V^{3+}$ is $[Ar]3d^2$,which has $2$ unpaired electrons.In $[Cr(H_2O)_6]^{3+}$,$Cr^{3+}$ is $[Ar]3d^3$ ($3$ unpaired electrons).In $[Cu(CN)_4]^{3-}$,$Cu^+$ is $[Ar]3d^{10}$ ($0$ unpaired electrons).In $[MnCl_4]^{2-}$,$Mn^{2+}$ is $[Ar]3d^5$ ($5$ unpaired electrons).Therefore,the complex with $2$ unpaired electrons is $[V(H_2O)_6]^{3+}$.Hence,the correct option is $A$.

The magnetic moment of a complex ion is $2.83 \ BM$. The complex ion is

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