A compound of vanadium chloride has spin only magnetic moment of 1.73 BM. Its formula is

Options

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

Answer

To determine the formula of the vanadium chloride compound with a spin-only magnetic moment of 1.73 BM, we will follow these steps:

### Step 1: Understand the Magnetic Moment Formula
The magnetic moment (\( \mu \)) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \text{ BM} \]
where \( n \) is the number of unpaired electrons.

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

### Step 3: Square Both Sides
To eliminate the square root, we square both sides of the equation:
\[ (1.73)^2 = n(n + 2) \]
Calculating \( (1.73)^2 \):
\[ 1.73^2 = 2.9929 \approx 3 \]
So, we have:
\[ 3 = n(n + 2) \]

### Step 4: Solve the Quadratic Equation
This simplifies to the quadratic equation:
\[ n^2 + 2n - 3 = 0 \]
Factoring the quadratic:
\[ (n + 3)(n - 1) = 0 \]
Thus, the solutions for \( n \) are:
\[ n = -3 \quad \text{(not possible)} \quad \text{or} \quad n = 1 \]

### Step 5: Determine the Oxidation State of Vanadium
Since \( n = 1 \), this means there is 1 unpaired electron. We need to find the oxidation state of vanadium in the compound. The outer electronic configuration of vanadium is \( [Ar] 3d^3 4s^2 \).

### Step 6: Analyze Possible Compounds
1. **Vanadium(II) Chloride (VCl2)**:
- Oxidation state: +2
- Configuration: \( 3d^3 \) (3 unpaired electrons) → Not suitable.

2. **Vanadium(III) Chloride (VCl3)**:
- Oxidation state: +3
- Configuration: \( 3d^2 \) (2 unpaired electrons) → Not suitable.

3. **Vanadium(IV) Chloride (VCl4)**:
- Oxidation state: +4
- Configuration: \( 3d^1 \) (1 unpaired electron) → Suitable.

4. **Vanadium(V) Chloride (VCl5)**:
- Oxidation state: +5
- Configuration: \( 3d^0 \) (0 unpaired electrons) → Not suitable.

### Step 7: Conclusion
The only compound with 1 unpaired electron and a magnetic moment of 1.73 BM is **Vanadium(IV) Chloride (VCl4)**.

### Final Answer:
The formula of the compound is \( \text{VCl}_4 \).

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