1. Identify the most basic oxide of vanadium

For transition metal oxides, basic character decreases with increase in oxidation state.

Given oxides:

• $V_2O_3$ : vanadium is in $+3$ oxidation state
• $V_2O_4$ : vanadium is in $+4$ oxidation state
• $V_2O_5$ : vanadium is in $+5$ oxidation state

Hence, the most basic oxide is: $V_2O_3$

2. Find the electronic configuration of vanadium in $V_2O_3$

In $V_2O_3$, each vanadium is $V^{3+}$.

Atomic number of vanadium = 23

Neutral vanadium: $V: [Ar] \, 3d^3 4s^2$

For $V^{3+}$, remove two $4s$ electrons and one $3d$ electron: $V^{3+}: [Ar] \, 3d^2$

So, number of unpaired electrons: $n = 2$

3. Calculate spin-only magnetic moment

Formula: $\mu = \sqrt{n(n+2)} \, \text{B.M.}$

Substitute $n=2$: $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2\sqrt{2} \approx 2.83 \, \text{B.M.}$

Nearest integer: $3$

4. Final answer

The spin-only magnetic moment is: $\boxed{3}$