The third ionisation enthalpies of copper, nickel and zinc are high.…

The correct answer is Option (1) → It is difficult to obtain oxidation states greater than +2 for these elements.

The third ionisation enthalpy is the energy needed to remove a third electron, that is, to change M²⁺ into M³⁺. When this energy is very high, the +3 state and any higher state are hard to reach. This is why the typical species at the end of the series are Ni(II), Cu(I) and Cu(II), and Zn(II).

Option (1) It is difficult to obtain oxidation states greater than +2 for these elements. — Correct. A high third ionisation enthalpy means the M²⁺ ion holds its electrons strongly, so these metals stop at +2 (or +1 for copper).

Option (2) The +3 state is the most stable oxidation state of these elements. — Incorrect. Forming M³⁺ needs the very step that costs so much energy. For zinc, +2 is the only oxidation state.

Option (3) Their M²⁺ ions are readily oxidised to M³⁺ ions in aqueous solution. — Incorrect. Oxidising M²⁺ to M³⁺ means removing the third electron; a high third ionisation enthalpy makes this difficult, not easy.

Option (4) These elements commonly form compounds in oxidation states above +4. — Incorrect. States above +4 need even more electrons removed. After manganese there is an abrupt decrease in the stability of higher oxidation states.

NCERT (2026-27 edition):
4.3.3 Ionisation Enthalpies – “…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.”

4.3.4 Oxidation States – “…The maximum oxidation states of reasonable stability correspond in value to the sum of the s and d electrons upto manganese … followed by a rather abrupt decrease in stability of higher oxidation states, so that the typical species to follow are Fe^II,III, Co^II,III, Ni^II, Cu^I,II, Zn^II.”