Find the electronic configurations of the ions

• Chromium: $Z=24$

  Neutral Cr has the configuration: $\mathrm{Cr} = [Ar] \, 3d^5 4s^1$

  For $\mathrm{Cr^{2+}}$, remove electrons first from $4s$ and then from $3d$: $\mathrm{Cr^{2+}} = [Ar] \, 3d^4$

• Manganese: $Z=25$

  Neutral Mn has the configuration: $\mathrm{Mn} = [Ar] \, 3d^5 4s^2$

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

Hence, both $\mathrm{Cr^{2+}}$ and $\mathrm{Mn^{3+}}$ are $d^4$ species.

Check option C

Option C says both $\mathrm{Cr^{2+}}$ and $\mathrm{Mn^{3+}}$ exhibit $d^4$ configuration.

This is correct.

Check option A: $\mathrm{Cr^{2+}}$ is a reducing agent

$\mathrm{Cr^{2+}}$ can easily lose one electron to form $\mathrm{Cr^{3+}}$: $\mathrm{Cr^{2+} \rightarrow Cr^{3+} + e^-}$

Now, $\mathrm{Cr^{3+}} = [Ar] \, 3d^3$

Since $\mathrm{Cr^{2+}}$ gets oxidized itself and causes reduction of another species, it acts as a reducing agent.

Therefore, A is correct.

Check option B: $\mathrm{Mn^{3+}}$ is an oxidizing agent

$\mathrm{Mn^{3+}}$ can gain one electron to form $\mathrm{Mn^{2+}}$: $\mathrm{Mn^{3+} + e^- \rightarrow Mn^{2+}}$

And, $\mathrm{Mn^{2+}} = [Ar] \, 3d^5$

The $d^5$ configuration is especially stable, so $\mathrm{Mn^{3+}}$ readily gets reduced to $\mathrm{Mn^{2+}}$. Hence it acts as an oxidizing agent.

Therefore, B is correct.

Check option D: when $\mathrm{Cr^{2+}}$ is used as a reducing agent, the chromium ion attains $d^5$ configuration

As a reducing agent, $\mathrm{Cr^{2+}}$ is oxidized to $\mathrm{Cr^{3+}}$: $\mathrm{Cr^{2+} \rightarrow Cr^{3+}}$

But, $\mathrm{Cr^{3+}} = [Ar] \, 3d^3$

It does not attain $d^5$ configuration.

So D is incorrect.

Final conclusion

Correct statements are: $\boxed{A,\ B,\ C}$

Comparison with stored answer

Stored correct answer: $A, B, C$

My derived answer matches the stored answer.