1. Find the electronic configuration of each ion

For transition metal ions, electrons are removed first from the $4s$ orbital and then from $3d$.

2. Option A: $\mathrm{Ti}^{3+}, \mathrm{Mn}^{2+}$

• Titanium: $Z=22$ $\mathrm{Ti} = [\mathrm{Ar}]\,3d^2 4s^2$ For $\mathrm{Ti}^{3+}$, remove two $4s$ electrons and one $3d$ electron: $\mathrm{Ti}^{3+} = [\mathrm{Ar}]\,3d^1$ Number of unpaired electrons $=1$.

• Manganese: $Z=25$ $\mathrm{Mn} = [\mathrm{Ar}]\,3d^5 4s^2$ For $\mathrm{Mn}^{2+}$, remove two $4s$ electrons: $\mathrm{Mn}^{2+} = [\mathrm{Ar}]\,3d^5$ Number of unpaired electrons $=5$.

So, option A does not match.

3. Option B: $\mathrm{Ti}^{2+}, \mathrm{Co}^{2+}$

• Titanium: $\mathrm{Ti}^{2+} = [\mathrm{Ar}]\,3d^2$ Number of unpaired electrons $=2$.

• Cobalt: $Z=27$ $\mathrm{Co} = [\mathrm{Ar}]\,3d^7 4s^2$ For $\mathrm{Co}^{2+}$: $\mathrm{Co}^{2+} = [\mathrm{Ar}]\,3d^7$ In free ion/high-spin form, number of unpaired electrons in $d^7$ is $3$.

So, option B does not match.

4. Option C: $\mathrm{Fe}^{3+}, \mathrm{Cr}^{2+}$

• Iron: $Z=26$ $\mathrm{Fe} = [\mathrm{Ar}]\,3d^6 4s^2$ For $\mathrm{Fe}^{3+}$: $\mathrm{Fe}^{3+} = [\mathrm{Ar}]\,3d^5$ Number of unpaired electrons $=5$.

• Chromium: $Z=24$ $\mathrm{Cr} = [\mathrm{Ar}]\,3d^5 4s^1$ For $\mathrm{Cr}^{2+}$, remove one $4s$ and one $3d$ electron: $\mathrm{Cr}^{2+} = [\mathrm{Ar}]\,3d^4$ Number of unpaired electrons $=4$.

So, option C does not match.

5. Option D: $\mathrm{V}^{2+}, \mathrm{Co}^{2+}$

• Vanadium: $Z=23$ $\mathrm{V} = [\mathrm{Ar}]\,3d^3 4s^2$ For $\mathrm{V}^{2+}$: $\mathrm{V}^{2+} = [\mathrm{Ar}]\,3d^3$ Number of unpaired electrons $=3$.

• Cobalt: $\mathrm{Co}^{2+} = [\mathrm{Ar}]\,3d^7$ Number of unpaired electrons $=3$.

So, option D does match.

6. Conclusion

The pair having the same number of unpaired electrons is: $\boxed{\mathrm{V}^{2+},\ \mathrm{Co}^{2+}}$ Hence, the correct option is: $\boxed{D}$