1. Identify oxidation states and compare oxidising ability

We need the species with least oxidising ability among:

• $\mathrm{VO_2^+}$
• $\mathrm{MnO_4^-}$
• $\mathrm{Cr_2O_7^{2-}}$

Known strong oxidising agents in acidic medium:

• $\mathrm{MnO_4^-}$ is a very strong oxidising agent.
• $\mathrm{Cr_2O_7^{2-}}$ is also a strong oxidising agent, but weaker than permanganate.
• $\mathrm{VO_2^+}$ corresponds to vanadium in high oxidation state, but it is less oxidising than the above two.

So, the species with least oxidising ability is: $\boxed{\mathrm{VO_2^+}}$

2. Find oxidation state of V in $\mathrm{VO_2^+}$

Let oxidation state of V be $x$.

$x + 2(-2) = +1$ $x - 4 = 1$ $x = +5$

So vanadium is in $+5$ oxidation state.

3. Find electronic configuration of $\mathrm{V^{5+}}$

Atomic number of V = 23.

Neutral V: $\mathrm{V}: [Ar] \, 3d^3 4s^2$

For $\mathrm{V^{5+}}$, remove 5 electrons:

• first 2 from $4s$
• next 3 from $3d$

Thus: $\mathrm{V^{5+}}: [Ar] \, 3d^0$

So number of unpaired electrons: $n = 0$

4. Calculate spin-only magnetic moment

Spin-only magnetic moment is: $\mu = \sqrt{n(n+2)} \text{ BM}$

With $n=0$: $\mu = \sqrt{0(0+2)} = 0 \text{ BM}$

Thus the required magnetic moment is: $\boxed{0 \text{ BM}}$

5. Compare with stored correct answer

Stored correct answer = $0$

Our derived answer also is $0$, so it agrees.