When the first electron gain enthalpy ( Heg) of oxygen is -141 kJ/mol, its second electron gain enthalpy is:

🧠 Adding a SECOND electron to an already-negative ion always costs energy — the negative ion repels the incoming electron First electron gain for O: $\ce{O(g) + e^- -> O^-(g)}$, $\Delta H = -141\,kJ/mol$ (exothermic). The neutral O happily accepts an electron. But for the second step, O is already $\ce{O^-}$, carrying a $-1$ charge. Pushing another electron into a negatively charged species means overcoming Coulomb repulsion — that ALWAYS requires energy input.

🗺️ Why second EA is positive $\ce{O^-(g) + e^- -> O^{2-}(g)}$, $\Delta H \approx +780\,kJ/mol$. The numerical value is large and positive. The same pattern holds for any element — the second electron gain enthalpy is always endothermic, no matter how reactive the element.

The only reason $\ce{O^{2-}}$ exists in compounds (like $\ce{MgO}$) is that lattice energy compensates for this $+780$ cost. In the gas phase, $\ce{O^{2-}}$ is unstable.

⚠️ The trap Option (c) "less negative than the first" tempts students who reason "well, it's harder but maybe still slightly favorable". No — the second EA is not just less favorable; it is genuinely positive (endothermic). The sign FLIPS, not just the magnitude. Always remember: monoanion → dianion is always endothermic, period.

$\boxed{\text{Answer: (a)}}$