A) \[VC{{l}_{2}}\]
B) \[VC{{l}_{4}}\]
C) \[VC{{l}_{3}}\]
D) \[VC{{l}_{5}}\]
Correct Answer: B
Solution :
[b] \[\mu =\sqrt{n(n+2)}\] \[1.73=\sqrt{n(n+2)}\] On calculating the value of n we find n = 1 No. of unpaired electrons = 1 hence its configuration will be \[V\left( 23 \right)=\left[ Ar \right]3{{d}^{3}}4{{s}^{2}}\] \[{{V}^{4+}}=[Ar]3{{d}^{1}}\] \[\therefore \] Its chloride has the formula \[VC{{l}_{4}}\]
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