Chloro compound of Vanadium has only spin magnetic moment of 1.73 BM. This Vanadium chloride has the formula

Options

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

Explanation

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

### Step 2: Set Up the Equation
Given that the spin 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:
\[ (1.73)^2 = n(n + 2) \]
Calculating \( (1.73)^2 \):
\[ 2.9929 \approx n(n + 2) \]

### Step 4: Solve for \( n \)
Now we need to find integer values of \( n \) that satisfy the equation:
\[ n(n + 2) \approx 2.9929 \]
Testing \( n = 1 \):
\[ 1(1 + 2) = 1 \times 3 = 3 \]
This is close to 2.9929, so we can conclude that \( n = 1 \) is correct.

### Step 5: Determine the Oxidation State of Vanadium
Vanadium has an atomic number of 23, with the electron configuration of:
\[ \text{[Ar]} 3d^3 4s^2 \]
To have 1 unpaired electron, we need to remove 4 electrons from the outer shell. This suggests that vanadium is in the +4 oxidation state (removing 2 from 4s and 2 from 3d).

### Step 6: Determine the Formula of the Compound
Since we have vanadium in the +4 oxidation state, we can combine it with chloride ions (Cl⁻). The chloride ion has a -1 charge. Therefore, to balance the +4 charge of vanadium, we need 4 chloride ions:
\[ \text{V}^{4+} + 4 \text{Cl}^- \rightarrow \text{VCl}_4 \]

### Conclusion
The formula for the chloro compound of vanadium with a spin magnetic moment of 1.73 BM is:
\[ \text{VCl}_4 \]

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