Question

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

• A. $VCl_2$
• B. $VCl_4$
• C. $VCl_3$
• D. $VCl_5$

Answer 1

The Correct Option is B

To determine the formula of the vanadium chloride compound with a spin magnetic moment of 1.73\, BM, we need to understand the relationship between the spin magnetic moment and the number of unpaired electrons. The formula for calculating the spin-only magnetic moment is given by:

\mu_s = \sqrt{n(n+2)}\, BM

where n is the number of unpaired electrons.

According to the question, the spin magnetic moment is 1.73\, BM. Let's calculate the number of unpaired electrons.

1.73 = \sqrt{n(n+2)}

Solving this equation:

1. 1.73^2 = n(n+2)
2. 2.9929 \approx n(n+2)

The nearest integer solution to n = 1 because:

1. {1}(1+2) = 3 which approximately equals 2.9929.

Therefore, the compound has 1 unpaired electron. Now, let's consider the electronic configuration of vanadium:
Vanadium in its ground state: [Ar] \, 3d^3 \, 4s^2

To find the oxidation state of vanadium in VCl_4:
Assume vanadium in VCl_4 has oxidation state +4:
[Ar] \, 3d^1 as it loses four electrons (3d electrons).

Thus, there is 1 unpaired electron in the 3d orbital, which corresponds to a spin-only magnetic moment of 1.73 BM.

Therefore, the correct formula for the vanadium chloride with the given magnetic moment is VCl_4.