Imagine an arrangement of atomic orbitals in an alternate universe, in which the s subshell contains 2 orbitals instead of 1, and the p subshell contains 4 orbitals rather than 3. In which groups would you expect the anomalously low electron affinities to occur?

My thoughts are groups 3 and 7, due to the fact that they are "one away" from filling up their subshell. The groups would look like:

s: _ _ _ _ ... p: _ _ _ _ _ _ _ _

However, the answer says 4 and 8. Is there something I'm missing?

Edit: in terms of electron configuration blanks, it would be s: _ _ ... p: _ _ _ _

  • $\begingroup$ I am confused. Are you assuming that the energy ordering is the same? In other words, can you assume that it is 1s2s2p3s3p4s3d...etc? $\endgroup$ Jan 4, 2014 at 5:45
  • $\begingroup$ Yes, but there's only one "shell" in this question. The order would still be s then p so ex. 2s4 2p7 for group 11. $\endgroup$
    – halcyon
    Jan 4, 2014 at 19:00

1 Answer 1


After rediscovering my own question, I'm posting it as an answer for anyone else to reference or to correct.

Electron affinity (EA) refers to "the energy released when an atom in the gas phase accepts an electron." (Burdge, Atoms First, p.118)

Lower EA means that the process of accepting an electron is not energetically favorable. In other words, lower EA = less likely to accept an electron.

Groups 4 and 8, in this case, already have filled subshells. Adding an electron would place it in a new subshell that is higher in energy. The EA would be lower for these groups because of the "reluctance" of the atom to accept an additional electron.

Groups 3 and 7 have the anomalously high EAs, which might have been where the confusion was.

  • $\begingroup$ Since the periodic table (only maingroups) would then contain of 12 Groups, you are forgetting 2 (half s) and 12 (full p). $\endgroup$ Mar 31, 2014 at 8:04

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