1. Write the electronic configurations of the atoms:

• $\mathrm{V}(23): [\mathrm{Ar}]\,3d^3 4s^2$
• $\mathrm{Cr}(24): [\mathrm{Ar}]\,3d^5 4s^1$
• $\mathrm{Mn}(25): [\mathrm{Ar}]\,3d^5 4s^2$
• $\mathrm{Fe}(26): [\mathrm{Ar}]\,3d^6 4s^2$

2. Understand second ionization enthalpy:

Second ionization enthalpy is the energy required for:

So we must examine the electronic configuration of the monocation $\mathrm{M^+}$ and see how easy it is to remove one more electron.

3. Form the singly charged ions by removing the first electron (from $4s$ before $3d$ in transition elements):

• $\mathrm{V^+}: [Ar] \, 3d^3 4s^1$
• $\mathrm{Cr^+}: [Ar] \, 3d^5$
• $\mathrm{Mn^+}: [Ar] \, 3d^5 4s^1$
• $\mathrm{Fe^+}: [Ar] \, 3d^6 4s^1$

4. Now remove the second electron:

• For $\mathrm{V^+}$: removing $4s^1$ gives $\mathrm{V^{2+}} = [Ar]3d^3$
• For $\mathrm{Cr^+}$: removing an electron from $3d^5$ gives $\mathrm{Cr^{2+}} = [Ar]3d^4$
• For $\mathrm{Mn^+}$: removing $4s^1$ gives $\mathrm{Mn^{2+}} = [Ar]3d^5$
• For $\mathrm{Fe^+}$: removing $4s^1$ gives $\mathrm{Fe^{2+}} = [Ar]3d^6$

5. Compare the difficulty:

$\mathrm{Cr^+}$ has configuration $[Ar]3d^5$, which is a half-filled d-subshell, a particularly stable arrangement. Removing the second electron from chromium would destroy this especially stable $3d^5$ configuration.

By contrast:

• In $\mathrm{V^+}$, $\mathrm{Mn^+}$, and $\mathrm{Fe^+}$, the second electron removed is a $4s$ electron.
• For $\mathrm{Mn^+}$, removal of the second electron actually leads to the stable $[Ar]3d^5$ configuration in $\mathrm{Mn^{2+}}$, so it is relatively easier.

Therefore, the highest second ionization enthalpy should be for chromium.

6. Evaluate options:

• A: Cr — Correct
• B: Mn — No, because second ionization gives stable $3d^5$
• C: Fe — No
• D: V — No

Hence, the correct answer is: