The magnetic moment of a complex ion is 2.83 | vedclass

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(A) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 2.83 \ BM$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 2.83 \ BM$,we have $\sqrt{n(n+2)} = 2.83$,which implies $n(n+2) \approx 8$,so $n = 2$.Let us analyze the number of unpaired electrons in each complex:$A$: In $[V(H_2O)_6]^{3+}$,$V^{3+}$ is $3d^2$,so $n = 2$.$B$: In $[Cr(H_2O)_6]^{3+}$,$Cr^{3+}$ is $3d^3$,so $n = 3$.$C$: In $[Cu(CN)_4]^{2-}$,$Cu^{2+}$ is $3d^9$,so $n = 1$.$D$: In $[MnCl_4]^{2-}$,$Mn^{2+}$ is $3d^5$ (high spin),so $n = 5$.Thus,the complex with $2$ unpaired electrons is $[V(H_2O)_6]^{3+}$.

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

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