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_(4)`
B. `VCl_(3)`
C. `VCl_(2)`
D. `VCl_(5)`

Answer

To determine the formula of the chloro compound of vanadium with a spin magnetic moment of 1.73 Bohr magneton (BM), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Magnetic Moment**:
The spin-only magnetic moment (μ) is given by the formula:
\[ μ = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

2. **Setting Up the Equation**:
Given that the magnetic moment is 1.73 BM, we can set up the equation:
\[ 1.73 = \sqrt{n(n + 2)} \]

3. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ (1.73)^2 = n(n + 2) \]
Calculating \( (1.73)^2 \):
\[ 1.73^2 = 2.9929 \approx 3 \]
Thus, we have:
\[ 3 = n(n + 2) \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ n^2 + 2n - 3 = 0 \]

5. **Factoring the Quadratic**:
We can factor this quadratic equation:
\[ (n + 3)(n - 1) = 0 \]
This gives us two possible solutions for \( n \):
\[ n = -3 \quad \text{(not possible)} \quad \text{or} \quad n = 1 \]

6. **Identifying the Oxidation State**:
Since \( n = 1 \), this means there is 1 unpaired electron. The ground state electron configuration of vanadium (atomic number 23) is:
\[ [Ar] 4s^2 3d^3 \]
To have 1 unpaired electron, vanadium must lose 4 electrons, resulting in the configuration:
\[ 3d^1 \]
This indicates that vanadium is in the +4 oxidation state.

7. **Determining the Formula**:
Since vanadium in the +4 oxidation state will combine with 4 chloride ions (Cl⁻), the formula of the vanadium chloride will be:
\[ \text{VCl}_4 \]

### Final Answer:
The formula of the chloro compound of vanadium is **VCl₄**.