A compound of vanadium has a magnetic moment of `1.73BM`. Work out the electronic configuration of vanadium in the compound

Answer

To determine the electronic configuration of vanadium in a compound with a magnetic moment of 1.73 Bohr magnetons (BM), we can follow these steps:

### Step 1: Understand the Relationship Between Magnetic Moment and Unpaired Electrons
The magnetic moment (μ) is related to the number of unpaired electrons (n) in an atom. The formula for calculating the magnetic moment is given by:
\[ \mu = \sqrt{n(n + 2)} \]

### Step 2: Substitute the Given Magnetic Moment
We know 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 \( (1.73)^2 \):
\[ 2.9929 \approx 3 \]
So, we have:
\[ 3 = n(n + 2) \]

### Step 4: Solve for n
Now we need to solve the quadratic equation:
\[ n^2 + 2n - 3 = 0 \]
Factoring the equation:
\[ (n + 3)(n - 1) = 0 \]
This gives us two possible solutions:
\[ n = -3 \quad \text{(not possible, as n must be non-negative)} \quad \text{or} \quad n = 1 \]
Thus, we find that:
\[ n = 1 \]
This means there is 1 unpaired electron in the compound.

### Step 5: Determine the Electronic Configuration of Vanadium
Vanadium (V) has an atomic number of 23, and its ground state electronic configuration is:
\[ \text{1s}^2 \, \text{2s}^2 \, \text{2p}^6 \, \text{3s}^2 \, \text{3p}^6 \, \text{3d}^3 \, \text{4s}^2 \]

### Step 6: Determine the Oxidation State of Vanadium
Since we have determined that there is 1 unpaired electron, we need to find the oxidation state of vanadium in the compound. The presence of 1 unpaired electron suggests that vanadium has lost some electrons.

For vanadium in the +4 oxidation state (V^4+), the electronic configuration would be:
- From the ground state configuration, we lose 2 electrons from the 4s orbital and 1 from the 3d orbital:
\[ \text{V}^{4+} \Rightarrow \text{3d}^1 \]

### Step 7: Final Electronic Configuration
Thus, the electronic configuration of vanadium in the +4 oxidation state is:
\[ \text{1s}^2 \, \text{2s}^2 \, \text{2p}^6 \, \text{3s}^2 \, \text{3p}^6 \, \text{3d}^1 \]

### Conclusion
The electronic configuration of vanadium in the compound with a magnetic moment of 1.73 BM is:
\[ \text{1s}^2 \, \text{2s}^2 \, \text{2p}^6 \, \text{3s}^2 \, \text{3p}^6 \, \text{3d}^1 \]