A chloro compound of Vanadium has a spin-only | vedclass

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(B) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n + 2)} \, BM$,where $n$ is the number of unpaired electrons.Given $\mu = 1.7

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$VCl_4$

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(B) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n + 2)} \, BM$,where $n$ is the number of unpaired electrons.Given $\mu = 1.73 \, BM$,we have $1.73 = \sqrt{n(n + 2)}$.Squaring both sides,$3 = n(n + 2)$,which gives $n^2 + 2n - 3 = 0$.Solving for $n$,we get $(n + 3)(n - 1) = 0$,so $n = 1$.Vanadium $(Z = 23)$ has the electronic configuration $[Ar] 3d^3 4s^2$.For $n = 1$ (one unpaired electron),Vanadium must be in the $+4$ oxidation state $(V^{4+})$,which has the configuration $[Ar] 3d^1$.Therefore,the formula of the chloride is $VCl_4$.

$A$ chloro compound of Vanadium has a spin-only magnetic moment of $1.73 \, BM$. This Vanadium chloride has the formula:

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