1. We need to identify which metal pair among $Cr^{3+}/Cr^{2+},\ Mn^{3+}/Mn^{2+},\ Fe^{3+}/Fe^{2+},\ Co^{3+}/Co^{2+}$ has negative standard electrode potential for the reduction: $M^{3+}+e^- \rightarrow M^{2+}$

2. Recall the standard reduction potentials:

   • $Cr^{3+}+e^- \rightarrow Cr^{2+},\ E^\circ = -0.41\,\text{V}$
   • $Mn^{3+}+e^- \rightarrow Mn^{2+},\ E^\circ = +1.51\,\text{V}$
   • $Fe^{3+}+e^- \rightarrow Fe^{2+},\ E^\circ = +0.77\,\text{V}$
   • $Co^{3+}+e^- \rightarrow Co^{2+},\ E^\circ = +1.82\,\text{V}$

3. The only pair with negative standard electrode potential is: $Cr^{3+}/Cr^{2+}$

4. So we need the spin-only magnetic moment of $Cr^{3+}$ in gaseous state.

5. Electronic configuration of chromium: $Cr: [Ar]3d^5 4s^1$

   Therefore, $Cr^{3+}: [Ar]3d^3$

6. Number of unpaired electrons in $Cr^{3+}$ is: $n=3$

7. Spin-only magnetic moment is: $\mu = \sqrt{n(n+2)}\,\text{B.M.}$

   Substituting $n=3$: $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\,\text{B.M.}$

8. Nearest integer: $4$

Therefore, the required answer is 4 B.M.