Arrange the following 3d-series elements in decreasing order of their…

The correct answer is Option (4) → (C), (D), (B), (A)

The second ionisation enthalpy is the energy needed to remove one electron from the gaseous M⁺ ion. Along the 3d series it generally rises with the nuclear charge. It is unusually high when the electron must be taken from a stable half-filled (d⁵) or filled (d¹⁰) set of d orbitals, because such a set has extra stability (exchange energy).

To find the M⁺ ion, write the configuration of the atom and remove one 4s electron first. The second electron then comes from 4s if one is left, otherwise from 3d.

• (C) Cr — first (highest). Cr (Z = 24) is [Ar]3d⁵4s¹, so Cr⁺ is 3d⁵, with no 4s electron left. The second electron must come from the stable half-filled 3d⁵ set: 1592 kJ mol⁻¹.

• (D) Mn — second. Mn (Z = 25) is [Ar]3d⁵4s², so Mn⁺ is 3d⁵4s¹. The second electron is the remaining 4s electron, and its removal leaves the stable 3d⁵ set, so the value is lower than for Cr: 1509 kJ mol⁻¹.

• (B) V — third. V (Z = 23) is [Ar]3d³4s², so V⁺ is 3d³4s¹; no stable d⁵ set is broken. The value follows the normal trend: 1414 kJ mol⁻¹.

• (A) Ti — last (lowest). Ti (Z = 22) is [Ar]3d²4s², so Ti⁺ is 3d²4s¹. It has the smallest nuclear charge of the four and no special stability to overcome: 1309 kJ mol⁻¹.

NCERT (2026-27 edition):
Ionisation Enthalpies – “The lowest common oxidation state of these metals is +2. To form the M²⁺ ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required in addition to the enthalpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M⁺ ions have the d⁵ and d¹⁰ configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of one 4s electron which results in the formation of stable d¹⁰ configuration.”

Table 4.2 (part):