In which of the following pairs, the outermost electronic configuration will be the same?

We need to find which pair of ions has the same outermost electronic configuration by writing the configurations for each ion.

For Option 1: Chromium (Z = 24) has the neutral configuration $$[Ar]\,3d^5\,4s^1$$ (anomalous due to half-filled d-orbital stability). Removing one electron from the 4s orbital gives $$Cr^+ = [Ar]\,3d^5$$. Manganese (Z = 25) has the neutral configuration $$[Ar]\,3d^5\,4s^2$$. Removing two electrons (both from 4s) gives $$Mn^{2+} = [Ar]\,3d^5$$. Both ions have identical $$3d^5$$ configurations.

For Option 2: Nickel (Z = 28) gives $$Ni^{2+} = [Ar]\,3d^8$$ (removing two 4s electrons from $$[Ar]\,3d^8\,4s^2$$). Copper (Z = 29) has the neutral configuration $$[Ar]\,3d^{10}\,4s^1$$. Removing the 4s electron gives $$Cu^+ = [Ar]\,3d^{10}$$. Since $$3d^8 \neq 3d^{10}$$, these do not match.

For Option 3: Iron (Z = 26) gives $$Fe^{2+} = [Ar]\,3d^6$$. Cobalt (Z = 27) has the neutral configuration $$[Ar]\,3d^7\,4s^2$$. Removing one electron from 4s gives $$Co^+ = [Ar]\,3d^7\,4s^1$$. Since $$3d^6 \neq 3d^7\,4s^1$$, these do not match.

For Option 4: Vanadium (Z = 23) gives $$V^{2+} = [Ar]\,3d^3$$. We already found $$Cr^+ = [Ar]\,3d^5$$. Since $$3d^3 \neq 3d^5$$, these do not match.

Therefore, the correct answer is Option 1: $$Cr^+$$ and $$Mn^{2+}$$, both having the outermost electronic configuration $$3d^5$$.