Question

The formation of the oxide ion $\left(O^{2-}\right(g\left)\right)$ required first an exothermic and then an endothermic step as shown below.
$O+e^{-}\to O^{-}{}_{\left(g\right)}+142kJmo{l}^{-1}$

$O^{-}+e^{-}\to O^{-2}{}_{\left(g\right)}-844kJmo{l}^{-1}$

This is because:

• A. oxygen is more electronegative
• B. oxygen has high electron affinity
• C. $O^{-}$ ion lends to resist the addition of another electron
• D. $O^{-}$ ion has comparatively larger size than oxygen atom.

Answer 1

Answer: (B), (C)

The correct options are
B oxygen has high electron affinity
C $O^{-}$ ion lends to resist the addition of another electron

An oxygen atom has a high affinity towards electron hence the energy is released while one electron is added to it. However, the $O^{-}$ ion repels the incoming electron and hence the second step is endothermic.

Hence answer for the given question is options B & C.