The electron affinity values are negative for: A. Be - Be- B. N - N- C. O - O2- D. Na - Na- E. Al - Al-

🧠 A negative electron affinity means endothermic — it happens when adding the electron disrupts a stable arrangement "Negative" in this question's wording means $\Delta H_{eg} > 0$ (energy must be supplied). That happens in three classic cases: filled $ns^2$ (Be), half-filled $np^3$ (N), and the second-electron addition to an already-negative ion (O$^- \to$ O$^{2-}$). 🗺️ Check each case A. $\ce{Be -> Be^-}$: Be is $1s^2\,2s^2$. Extra electron is forced into the higher 2p subshell — unfavorable. Endothermic. B. $\ce{N -> N^-}$: N is $2p^3$ half-filled (extra stable). The new electron must pair up in a 2p orbital, facing repulsion. Endothermic. C. $\ce{O^- -> O^{2-}}$: pushing a negative electron into an already negative ion — strong electrostatic repulsion. Endothermic (about $+780\,kJ/mol$). D. $\ce{Na -> Na^-}$: Na is $3s^1$. Adding one electron gives $3s^2$ — a lower energy state. Exothermic. E. $\ce{Al -> Al^-}$: Al is $3s^2\,3p^1$. Extra electron joins the half-empty 3p. Slightly favorable. Exothermic.

So A, B, C are endothermic ("negative" EA). D and E are exothermic.

⚠️ The trap Students get C confused with neutral O. The first electron gain for O is exothermic ($\Delta H = -141\,kJ/mol$). It is the SECOND electron gain (going from $\ce{O^-}$ to $\ce{O^{2-}}$) that is endothermic. Always read whether the species starts as neutral or as an anion.

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