Meaning of second ionization enthalpy

The second ionization enthalpy is the energy required for: $M^+ \rightarrow M^{2+} + e^-$ So we must compare the stability of the singly charged cations $V^+, Cr^+, Mn^+, Fe^+$.

Write the electronic configurations

Neutral atoms:

• $V(Z=23): [Ar]3d^34s^2$
• $Cr(Z=24): [Ar]3d^54s^1$
• $Mn(Z=25): [Ar]3d^54s^2$
• $Fe(Z=26): [Ar]3d^64s^2$

Form the singly charged ions

Electrons are removed first from the $4s$ orbital.

• $V^+: [Ar]3d^34s^1$
• $Cr^+: [Ar]3d^5$
• $Mn^+: [Ar]3d^54s^1$
• $Fe^+: [Ar]3d^64s^1$

Now consider removal of one more electron

The second ionization enthalpy corresponds to removing one electron from these $M^+$ ions.

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

Key stability argument

$Cr^+$ has configuration $[Ar]3d^5$, which is a half-filled d-subshell, especially stable.

Removing the second electron from $Cr^+$ means breaking this stable half-filled configuration: $Cr^+([Ar]3d^5) \rightarrow Cr^{2+}([Ar]3d^4) + e^-$ Since this destroys extra stability, the required energy is expected to be maximum among the given options.

In contrast:

• $V^+$, $Mn^+$, and $Fe^+$ still have a $4s$ electron to remove.
• Removing a $4s$ electron is generally easier than removing a $3d$ electron from a particularly stable $3d^5$ configuration.

Evaluate options

• A: Cr — highest second ionization enthalpy because $Cr^+$ is $3d^5$ (half-filled, very stable). ✅
• B: Mn — second ionization removes $4s$ electron, giving stable $3d^5$, so not the highest. ❌
• C: Fe — second ionization removes $4s$ electron. ❌
• D: V — second ionization removes $4s$ electron. ❌

Final answer

The element expected to have the highest second ionization enthalpy is: $\boxed{Cr}$