Four successive members of the first-row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionisation enthalpy?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Manganese (Z = 25)
The third ionisation enthalpy removes an electron from the M2+ ion, which has lost both 4s electrons and is a pure 3dn ion. Removing an electron from a stable half-filled 3d5 set is the hardest.
V (Z = 23): [Ar]3d34s2 → \( \ce{V^2+} \) 3d3; third ionisation: 3d3 → 3d2.
Cr (Z = 24): [Ar]3d54s1 → \( \ce{Cr^2+} \) 3d4; third ionisation: 3d4 → 3d3.
Mn (Z = 25): [Ar]3d54s2 → \( \ce{Mn^2+} \) 3d5; third ionisation: 3d5 → 3d4, which breaks the half-filled set.
Fe (Z = 26): [Ar]3d64s2 → \( \ce{Fe^2+} \) 3d6; third ionisation: 3d6 → 3d5, which gives the half-filled set, so it is easier.
So manganese has the highest third ionisation enthalpy of the four.
NCERT (2026-27 edition):
4.3.3 Ionisation Enthalpies – “… The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d5 (Mn2+) and d10 (Zn2+) ions.”