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 = 1.73 \

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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 $n = 1$ (since $n$ cannot be negative).Vanadium $(V)$ has the atomic number $23$ and electronic configuration $[Ar] 3d^3 4s^2$.For $n=1$,the vanadium ion must have one unpaired electron.In $VCl_4$,vanadium is in the $+4$ oxidation state $(V^{4+})$.The configuration of $V^{4+}$ is $[Ar] 3d^1$,which contains $1$ unpaired electron.Thus,the formula is $VCl_4$.

Column-$I$	Column-$II$
$(a) \ Co^{3+}$	$(i) \ \sqrt{8} \ B.M.$
$(b) \ Cr^{3+}$	$(ii) \ \sqrt{35} \ B.M.$
$(c) \ Fe^{3+}$	$(iii) \ \sqrt{3} \ B.M.$
$(d) \ Ni^{2+}$	$(iv) \ \sqrt{24} \ B.M.$
	$(v) \ \sqrt{15} \ B.M.$

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

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