question_answer

The compound of vanadium has magnetic moment \[1.73\] B.M. The vanadium chloride has the formula

A)  \[VC\,{{l}_{2}}\]                     

B)  \[VC\,{{l}_{3}}\]

C)  \[VC\,{{l}_{4}}\]                     

D)  \[VC\,{{l}_{5}}\]

Correct Answer: C

Solution :

\[\mu =1.73\,B.M\] \[\mu =\sqrt{n(n+2)}\] \[\therefore \,\,n=1\Rightarrow \] no. of unpaired \[{{e}^{-}}\] present in V is 1             \[_{23}V-................4{{s}^{2}}3{{d}^{3}}\] \[{{V}^{+4}}-...............4{{s}^{0}}3{{d}^{1}}\] \[\therefore \] the vanadium chloride has the formula \[VC{{l}_{4}}\].