A compound of vanadium possesses a magnetic moment of `1.73BM`. The oxidation state of vanadium in this compounds is:

Options

A. 1
B. 2
C. 4
D. can 't be predicted

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
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
Squaring both sides gives us:
\[ (1.73)^2 = n(n + 2) \]
Calculating \( (1.73)^2 \):
\[ 2.9929 \approx n(n + 2) \]

### Step 4: Solve for \( n \)
Now we need to solve the quadratic equation:
\[ n^2 + 2n - 2.9929 = 0 \]
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 2 \), and \( c = -2.9929 \).

Calculating the discriminant:
\[ b^2 - 4ac = 2^2 - 4(1)(-2.9929) = 4 + 11.9716 = 15.9716 \]
Now applying the quadratic formula:
\[ n = \frac{-2 \pm \sqrt{15.9716}}{2} \]
Calculating \( \sqrt{15.9716} \approx 3.99 \):
\[ n = \frac{-2 \pm 3.99}{2} \]
Calculating the two possible values:
1. \( n = \frac{1.99}{2} \approx 0.995 \) (not valid since \( n \) must be a whole number)
2. \( n = \frac{-5.99}{2} \approx -2.995 \) (not valid)

Since the only valid integer solution is \( n = 1 \), we conclude that there is 1 unpaired electron.

### Step 5: Determine the oxidation state of vanadium
Vanadium (V) has the electronic configuration:
\[ \text{[Ar]} 3d^3 4s^2 \]
To have only 1 unpaired electron, vanadium must lose 4 electrons (2 from the 4s and 2 from the 3d). This gives us:
\[ \text{[Ar]} 3d^1 \]
Thus, the oxidation state of vanadium in this compound is +4.

### Final Answer
The oxidation state of vanadium in the compound is **+4**.

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