A compound of vanadium has a magneitc moment `(mu)` of `1.73 BM`. If the vanadium ion in the compound is present as `V^(x+)`, then, the value of `x` is ?

Options

A. `1`
B. `2`
C. `3`
D. `4`

Answer

To determine the oxidation state of vanadium in the compound with a magnetic moment of 1.73 Bohr Magneton (BM), we can follow these steps:

### Step 1: Understand the relationship between magnetic moment and unpaired electrons
The magnetic moment (μ) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

### Step 2: Set up the equation using the given magnetic moment
We know the magnetic moment is 1.73 BM. Therefore, we can set up the equation:
\[ 1.73 = \sqrt{n(n + 2)} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives us:
\[ 1.73^2 = n(n + 2) \]
Calculating \( 1.73^2 \):
\[ 1.73^2 \approx 2.9929 \]
So we have:
\[ n(n + 2) = 2.9929 \]

### Step 4: Solve the quadratic equation
This leads to the quadratic equation:
\[ n^2 + 2n - 2.9929 = 0 \]
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -2.9929 \):
\[ n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-2.9929)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 2^2 + 4 \cdot 2.9929 = 4 + 11.9716 = 15.9716 \]
Now calculating \( n \):
\[ n = \frac{-2 \pm \sqrt{15.9716}}{2} \]
\[ n = \frac{-2 \pm 3.99}{2} \]
Taking the positive root:
\[ n \approx \frac{1.99}{2} \approx 0.995 \]
Since \( n \) must be a whole number, we round it to \( n = 1 \).

### Step 5: Determine the oxidation state of vanadium
Vanadium (V) has an atomic number of 23, and its ground state electronic configuration is:
\[ [Ar] 3d^3 4s^2 \]
In the +4 oxidation state, vanadium would lose 4 electrons (2 from 4s and 2 from 3d), resulting in:
\[ 3d^1 \]
This configuration has 1 unpaired electron, which corresponds to the magnetic moment of 1.73 BM.

### Conclusion
Thus, the oxidation state \( x \) of vanadium in the compound is:
\[ x = 4 \]

### Final Answer
The value of \( x \) is 4.

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