To solve the problem, we need to determine the number of unpaired electrons and the electronic configuration of vanadium in a compound where its magnetic moment is given as 1.73 BM (Bohr Magneton).
### Step-by-Step Solution:
1. **Understand the Magnetic Moment Formula**:
The magnetic moment (μ) can be calculated using the formula:
\[
\mu = \sqrt{n(n + 2)}
\]
where \( n \) is the number of unpaired electrons.
2. **Set Up the Equation**:
Given that the magnetic moment is 1.73 BM, we can set up the equation:
\[
1.73 = \sqrt{n(n + 2)}
\]
3. **Square Both Sides**:
To eliminate the square root, square both sides of the equation:
\[
(1.73)^2 = n(n + 2)
\]
This simplifies to:
\[
2.9929 = n(n + 2)
\]
4. **Rearrange the Equation**:
Rearranging gives us a quadratic equation:
\[
n^2 + 2n - 2.9929 = 0
\]
5. **Solve the Quadratic Equation**:
We can use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -2.9929 \):
\[
n = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-2.9929)}}{2 \cdot 1}
\]
\[
n = \frac{-2 \pm \sqrt{4 + 11.9716}}{2}
\]
\[
n = \frac{-2 \pm \sqrt{15.9716}}{2}
\]
\[
n = \frac{-2 \pm 3.99}{2}
\]
This gives two possible values for \( n \):
\[
n \approx 0.995 \quad \text{(not valid, as n must be a whole number)}
\]
\[
n \approx 1 \quad \text{(valid)}
\]
6. **Determine the Electronic Configuration of Vanadium**:
Vanadium (V) has an atomic number of 23. The ground state electronic configuration of vanadium is:
\[
\text{V: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^3
\]
7. **Determine the Oxidation State**:
Since we found that \( n = 1 \), this indicates that in the compound, vanadium must lose electrons to have only one unpaired electron. The oxidation state of vanadium in this compound is +4 (V^4+).
8. **Electronic Configuration of V^4+**:
When vanadium is in the +4 oxidation state, it loses 4 electrons (2 from 4s and 2 from 3d):
\[
\text{V}^{4+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^0 \, 3d^1
\]
In this configuration, there is 1 unpaired electron in the 3d orbital.
### Final Answers:
- **Number of unpaired electrons**: 1
- **Electronic configuration of vanadium in this compound**: \( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^0 \, 3d^1 \)
