# Which atom has the greater electron affinity, "O"^(-) or "O"?

Aug 6, 2016

Well, if you look at the definition, electron affinity is just the propensity to accept electrons.

It seems at first that ${\text{O}}^{-}$ should have a greater electron affinity since it is closer to an octet.

Actually, since ${\text{O}}^{-}$ is negatively-charged, it would be more difficult to accept an electron, and significant coulombic repulsion would make the electron affinity of ${\text{O}}^{-}$ more positive than that of $\text{O}$ (which is negative).

$\text{EA"_(1,"O") = -"142 kJ/mol}$

${\text{O"(g) + e^(-) -> "O}}^{-} \left(g\right)$

$\text{EA"_(2,"O") = +"844 kJ/mol}$

${\text{O"^(-)(g) + e^(-) -> "O}}^{2 -} \left(g\right)$

(the second electron affinity of $\text{O}$ is the first electron affinity of ${\text{O}}^{-}$)

If electron affinity is favorable, energy is released upon adding an electron, meaning it's negatively-signed. However, since ${\text{EA}}_{2}$ is positive, it means it requires energy to add that electron to ${\text{O}}^{-}$!

Therefore, the "greater" electron affinity (really, the more favorable electron affinity) is with $\text{O}$.