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Q.
The compound of vanadium has magnetic moment of $1.73$ BM. The vanadium chloride has the formula
Classification of Elements and Periodicity in Properties
Solution:
$V (Z=23)=[A r] 4 s^2 3 d^3$
$V ^{ x +}=?$
If there are $N$ unpaired electrons then
Magnetic moment $=\sqrt{N(N+2)} B M=1.73$
$\therefore N(N+2)=3$
$\therefore N=1$
Thus, vanadium exists as $V^{4+}$
$V^{4+}=[A r] 3 d^1$
Thus, vanadium chloride is $VCl _4$
