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