Transition metals have higher (less negative) standard reduction potentials than the s-block metals and are weaker reducing agents. This is mainly due to their:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → High ionisation enthalpies
The standard electrode potential E°(M2+/M) measures how readily the solid metal turns into its hydrated ion. That change takes three energy terms: atomisation of the metal, ionisation of the gaseous atom, and hydration of the ion.
For the transition metals the sum of the first and second ionisation enthalpies is high. So much energy is needed to remove the electrons that forming the ion is less favourable, the E° value is less negative than for the s-block metals, and the metal is a weaker reducing agent.
Why the other options are incorrect
Option (1) High enthalpies of hydration of their ions — Incorrect. Hydration releases energy; a high hydration enthalpy favours ion formation and makes E° more negative (as for nickel), the opposite of the effect asked about.
Option (3) Low ionisation enthalpies — Incorrect. Low ionisation enthalpies would make the electrons easier to remove and the metals stronger reducing agents.
Option (4) Low enthalpies of atomisation — Incorrect. Transition metals have high enthalpies of atomisation; a low value would make them more reactive, not less.
NCERT (2026-27 edition):
4.3.8 Chemical Reactivity and E° Values – “… The Eo values for M2+/M … indicate a decreasing tendency to form divalent cations across the series. This general trend towards less negative Eo values is related to the increase in the sum of the first and second ionisation enthalpies.”