Chloro compound of vanadium has only spin magnetic moment of 1.73 BM. This vanadium chloride has the formula: (at.no. of V=23)

Options

A. `VCl_2`,
B. `VCl_4`,
C. `VCl_3`,
D. `VCl_5`

Answer

B

Explanation

`mu = sqrt(n(n+2)) " BM,"mu_s =1.73 "B.M"` is obtained when n=1 thus one unpaired electron must be present in chloride of vanadium.
`V = [Ar] 3d^3 4s^2`
For one unpair electron vanadium must be present in `+4 (VCl_4)` oxidation state: `V^(4+) + 3d^1`