If the spin-only magnetic moment of VCl_x is | vedclass

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(C) 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 \

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$VCl_4$

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(C) 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 $\sqrt{n(n+2)} = 1.73$,which implies $n(n+2) = 3$,so $n = 1$.Vanadium $(V)$ has the atomic number $23$ and electronic configuration $[Ar] 3d^3 4s^2$.For $VCl_4$,the oxidation state of $V$ is $+4$,resulting in the configuration $[Ar] 3d^1$.Since there is $1$ unpaired electron $(n=1)$,the magnetic moment is $\sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.Therefore,the correct formula is $VCl_4$.

If the spin-only magnetic moment of $VCl_x$ is $1.73 \ BM$,then the correct formula is:

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