1. Rule for electronic configuration of transition metal ions

When transition metals form cations, electrons are removed first from the $4s$ orbital and then from the $3d$ orbital.

So we first write the neutral atom configurations.

2. Write configurations of the neutral atoms

• $Cr$ : $[Ar] \, 3d^5 4s^1$
• $Mn$ : $[Ar] \, 3d^5 4s^2$
• $Ni$ : $[Ar] \, 3d^8 4s^2$
• $Cu$ : $[Ar] \, 3d^{10} 4s^1$
• $V$ : $[Ar] \, 3d^3 4s^2$
• $Fe$ : $[Ar] \, 3d^6 4s^2$
• $Co$ : $[Ar] \, 3d^7 4s^2$

3. Find the ionic configurations in each option

Option A: $Cr^+$ and $Mn^{2+}$

• $Cr^+$: remove one electron from $4s$ $Cr^+ = [Ar] \, 3d^5$
• $Mn^{2+}$: remove two electrons from $4s$ $Mn^{2+} = [Ar] \, 3d^5$

Both have the same outer configuration: $3d^5$.

So, A is correct.

Option B: $Ni^{2+}$ and $Cu^+$

• $Ni^{2+}$: $Ni^{2+} = [Ar] \, 3d^8$
• $Cu^+$: $Cu^+ = [Ar] \, 3d^{10}$

These are different.

So, B is incorrect.

Option C: $V^{2+}$ and $Cr^+$

• $V^{2+}$: $V^{2+} = [Ar] \, 3d^3$
• $Cr^+$: $Cr^+ = [Ar] \, 3d^5$

These are different.

So, C is incorrect.

Option D: $Fe^{2+}$ and $Co^+$

• $Fe^{2+}$: $Fe^{2+} = [Ar] \, 3d^6$
• $Co^+$: remove one electron from $4s$ $Co^+ = [Ar] \, 3d^7 4s^1$

These are different.

So, D is incorrect.

4. Final conclusion

The pair having the same outermost electronic configuration is:

$\boxed{A \; (Cr^+ \text{ and } Mn^{2+})}$

5. Comparison with stored correct answer

Stored correct answer = A.

My derived answer = A.

So, they agree.