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

Answer 1

According to the definition, electron affinity is simply the tendency to take in 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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Answer 2

Natalie Anton

The "O" atom has greater electron affinity than "O"^(-).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.