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

1 Answer
Truong-Son N.
Aug 6, 2016

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

It seems at first that #"O"^(-)# should have a greater electron affinity since it is closer to an octet.

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

#"EA"_(1,"O") = -"142 kJ/mol"#

#"O"(g) + e^(-) -> "O"^(-)(g)#

#"EA"_(2,"O") = +"844 kJ/mol"#

#"O"^(-)(g) + e^(-) -> "O"^(2-)(g)#

(the second electron affinity of #"O"# is the first electron affinity of #"O"^(-)#)

If electron affinity is favorable, energy is released upon adding an electron, meaning it's negatively-signed. However, since #"EA"_2# is positive, it means it requires energy to add that electron to #"O"^(-)#!

Therefore, the "greater" electron affinity (really, the more favorable electron affinity) is with #"O"#.

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