Chloro compound of Vanadium has only spin magnetic moment of $1.73 \mathrm{BM}$. This Vanadium chloride has…
chloride has the formula:
- $\mathrm{VCl}_{2}$
- $\mathrm{VCl}_{4}$
- $\mathrm{VCl}_{3}$
- $\mathrm{VCl}_{5}$
Solution
$1.73=\sqrt{n(n+2)}$
On calculating the value of $n$ we find $n=1$ No. of unpaired electrons $=1$ hence its configuration will be $\mathrm{V}(23)=[\mathrm{Ar}] 3 d^{3} 4 s^{2}$
$\mathrm{V}^{4+}=[\mathrm{Ar}] 3 d^{1}$
$\therefore$ Its chloride has the formula $\mathrm{VCl}_{4}$ ^
Asked in: JEE-TOPICTESTS-CHEMISTRY
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