Assertion (A): The second ionisation enthalpy of zinc is lower than t…

The correct answer is Option (2) → Both (A) and (R) are true but (R) is not the correct explanation of (A).

Each ionisation enthalpy depends on which electron is removed and what configuration is left behind. Zinc is 3d¹⁰4s² and copper is 3d¹⁰4s¹.

Assertion (A): The second ionisation enthalpy of zinc is lower than that of copper. — True. Zn⁺ (3d¹⁰4s¹) loses its 4s electron and becomes Zn²⁺ with the stable d¹⁰ configuration, so the value is low (1734 kJ mol⁻¹).
Cu⁺ is already d¹⁰, and its second electron must come out of that stable set, so copper’s value is unusually high (1958 kJ mol⁻¹).

Reason (R): The third ionisation enthalpy of zinc is very high. — True. The third electron has to be removed from the stable d¹⁰ Zn²⁺ ion. This is why zinc does not show an oxidation state above +2.

Why R does not explain A?

• Assertion (A) is about the second step, Zn⁺ → Zn²⁺; Reason (R) is about the third step, Zn²⁺ → Zn³⁺. How hard the third step is says nothing about how easy the second one is.

• The correct reason for A is that the second ionisation of zinc removes the lone 4s electron and forms the stable d¹⁰ configuration.

NCERT (2026-27 edition):
4.3.3 Ionisation Enthalpies – “…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. 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. In general, the third ionisation enthalpies are quite high. Also the high values for third ionisation enthalpies of copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements.”