A compound of vanadium chloride has spin only magnetic moment of 1 . 73 BM. Its formula is

1. VCl₂
2. VCl₅
3. VCl₄
4. VCl₃

Solution

Spin only Magnetic Moment

$\text{μ}=\sqrt{\text{n}\left(\text{n}+2\right)}\text{ BM}=1\text{.}73=\sqrt{3}$

$\therefore n=1 \;$ (unpaired e^-)

Electronic configuration of $\text{V}={\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{3}}{\text{4s}}^{\text{2}}$



to get 1 unpaired electron $\mathrm{V}$ has to lose 4 electrons to form ${\text{V}}^{4+}.$

So, formula of chloride is ${\text{VCl}}_{\text{4}}$

Asked in: JEE Main 2014 (09 Apr Online)

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