Read the following passage and answer the next five questions based on it. (37-41)
Transition Series Elements:

Sc Ti V Cr MnFeCoNiCuZn

YZrNbMoTcRuRhPdAgCd
La Hf Ta WReOsIrPtAuHg

In any transition series, as we move from left to right the d-orbitals are progressively filled and their properties vary accordingly.
f-block Elements:

Ce Pr Nd PmSmEuGdTbDyHoErTmYbLu
Th Pa UNpPuAmCmBkCfEsFmMdNoLr

The above are the two series of f-block elements in which the chemical properties won’t change much. The 5f-series elements are radioactive in nature and mostly are artificially synthesized in laboratories and thus much is not known about their chemical properties.

Which of the following is the correct order of second ionization enthalpy?

The correct option is V < Cr > Mn.

To understand why this order is correct, we need to analyze the electronic configurations of vanadium (V), chromium (Cr), and manganese (Mn), both in their neutral states and after losing one electron (first ionization state).

First, let us write the ground-state electronic configurations of these neutral transition elements (atomic numbers: V = 23, Cr = 24, Mn = 25):
- Vanadium (V): $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}4s^2$ or simply [Ar] $3d^{3}4s^2$
- Chromium (Cr): [Ar] $3d^{5}4s^1$ (due to the extra stability of a half-filled d-subshell)
- Manganese (Mn): [Ar] $3d^{5}4s^2$

The first ionization enthalpy (IE₁) involves removing the first electron, which typically comes from the outermost 4s orbital. The resulting configurations for the mono-positive ions are:
- V⁺: [Ar] $3d^{3}4s^1$
- Cr⁺: [Ar] $3d^5$
- Mn⁺: [Ar] $3d^{5}4s^1$

The second ionization enthalpy (IE₂) is the energy required to remove a second electron from these mono-positive ions:
- For V⁺, the electron is removed from the $4s$ orbital, leaving V²⁺ with a [Ar] $3d^3$ configuration.
- For Cr⁺, the electron must be removed from the $3d^5$ configuration. The $3d^5$ state is a half-filled d-subshell, which possesses exceptionally high stability due to symmetrical distribution of electron density and maximum exchange energy. Breaking this stable configuration requires an exceptionally large amount of energy.
- For Mn⁺, the electron is removed from the $4s$ orbital, which is relatively easy compared to disrupting a stable, half-filled $3d^5$ shell, leaving Mn²⁺ with a stable [Ar] $3d^5$ configuration.

Comparing the three:
1. Cr⁺ has a highly stable, half-filled $3d^5$ configuration, making its second ionization enthalpy (IE₂) remarkably high.
2. For V⁺ and Mn⁺, the second electron is removed from the outer 4s orbital. Since manganese has a higher nuclear charge (Z = 25) than vanadium (Z = 23), the electrostatic attraction on the 4s electron is stronger in manganese, making IE₂ of Mn greater than that of V.
Therefore, we have: V < Mn < Cr.

Comparing the options, this relative order is represented by: V < Cr > Mn.