Identify which ion is a strong oxidising agent in aqueous solution

We compare the given ions: $\mathrm{Ti}^{2+},\ \mathrm{V}^{2+},\ \mathrm{Co}^{3+},\ \mathrm{Cr}^{2+}$

In aqueous solution:

• $\mathrm{Ti}^{2+}$ is a reducing agent (easily oxidised to $\mathrm{Ti}^{3+}$).
• $\mathrm{V}^{2+}$ is also a reducing agent.
• $\mathrm{Cr}^{2+}$ is a strong reducing agent.
• $\mathrm{Co}^{3+}$ is a strong oxidising agent in aqueous solution because it is readily reduced to the more stable $\mathrm{Co}^{2+}$ state.

Hence, the required ion is: $\mathrm{Co}^{3+}$

Find electronic configuration of $\mathrm{Co}^{3+}$

Atomic number of Co is 27.

Neutral cobalt: $\mathrm{Co}: [\mathrm{Ar}]\,3d^7 4s^2$

For $\mathrm{Co}^{3+}$, remove two electrons from $4s$ and one from $3d$: $\mathrm{Co}^{3+}: [\mathrm{Ar}]\,3d^6$

Determine number of unpaired electrons

Since the ion is given as a free ion (no ligand field specified), use the high-spin/free-ion count for $3d^6$: $n = 4$ unpaired electrons.

Calculate spin-only magnetic moment

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

Substituting $n=4$: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\ \text{BM}$

Near integer: $\boxed{5\ \text{BM}}$

Comparison with stored answer

Derived answer = $5$

Stored correct answer = $5$

Therefore, they agree.