[k42490]

I have a question that asks why are changes in electron affinity as one goes down a group not as large as the changes in ionization energy going down a group?

I know that ion. energy increases down a group because the atomic radius and distance from nucleus are increasing also elec. affinity decreases...then I am stuck.

[shelanachium]

Ionisation energy tends to decrease, not increase down a group as the atoms increase in size and the outermost electron gets further fron the nucleus. When an atom ionises two oppositely charged species are formed, a negative electron and a positive ion. These strongly attract electrostatically, so considerable energy is needed to separate them, even in easily-ionised atoms like the alkali metals.

Electron affinity can be regarded as the converse of the 'ionisation energy' of a negative ion. The species formed are a neutral atom and a free electron. Once these are separated there is no strong electrostatic attraction between them, so the energy required to separate an electron from a neutral atom is less than that needed to separate an electron and a positive ion.

Consequently electron affinities are in general lower than ionisation potentials, and the differences between them correspondingly less.

[k42490]

thanks...that makes more sense now

[ARGOS++]

Dear Shelanachium,

You gave a very good answer why both are not of a comparable Amount/Strength but it is not giving the answer why both forces don’t decrease in a “parallel order” from group to group.

I was also searching and got an answer but it is not really satisfying me either.
It tells that the different effect should com from the fact, that the number of nucleuses increases faster then the radius of the atom increases, and therefore the effect decreases differently.

Maybe somebody else has found an even better answer.

Good Luck!
                    ARGOS⁺⁺

[shelanachium]

A lot to do with the details of electronic structure. Thus B and O have lower ionisation potentials than 'expected' because B+ has an 1s2 2s2 configuration, i.e. a full 2s subshell, and O+ has 1s2 2s2 2p3 with a half-full p shell. Removal of an electron from O is relatively easy because there are two mutually repelling electrons in the same orbital; the half-full 2p orbitals in O+ are relatively stable (cf Mn2+ in the first transition series with a half-full set of d orbitals - 1s2 2s2 2p6 3s2 3p6 4d5).

Electron affinities are similarly affected e.g C on adding an electron attains a 1s2 2s2 2p3 configuration like O+ (and N) so its electron affinity is relatively high.

[ARGOS++]

Dear Shelanachium,

Your argumentation is still conclusive if you do always one step ahead or with other words if you travel horizontally in the atom table.
But the argumentation is even still not answering the initial question or with other words if you move vertical always a “whole shell”.

Only one example is the sequence:

F ~ Cl ~ Br ~ J ~ At      with E_A:  328 ~ 349 ~ 343 ~ 295 ~ 270   [kJ/mole]

Herein Cl and also Br are clearly two Spoilsports especially as Cl has the largest value of the whole table and the sequence is not “monotone” decreasing.

For such situations I have not found any satisfying answer and that’s in congruence with the initial question of K42490.

I’m still searching for it.

Good Luck!
                   ARGOS⁺⁺

[shelanachium]

In electron affinities F<Cl>Br>I>At

Here F<Cl seems puzzling. It is said to be due to the small size of F- making the ion too 'charge-dense' so the electron is less tightly bound. However the increased lattice energies due to the small size of F- more than compensate for its lower electron affinity so fluorides are more exothermic than chlorides.

With groups 13 onwards the relationship is complicated by the fact that the underlying electron shells change in character following the transition elements. Thus B is 1s2 (2s2 2p1) and Al 1s2 2s2 2p6 (3s2 3p1). In both these the valence shell (bracketed) is underlain by a full p subshell. However gallium is 1s2 2s2 2p6 3s2 3p6 3d10 (4s2 4p1) with a full d shell underlying the valence electrons.

As in Ga, in In and Tl the valence shell is also underlain by a more weakly shielding d shell (its electrons spend less time near the nucleus than those of a p shell, so shield its charge less; electrons in the next orbital are therefore more tightly bound). So Ga, In and also Ge, Sn are more electronegative than 'expected' by simply assuming electronegativity and electron affinities fall as one descends a period.

If this is starting to look complicated, there really is no very simple answer; the mathematics needed to fully understand is formidable and well betond my reach!

For example, one of the highest electron affinities is to be found in gold, a metal from which salts of Au- have been made. This is because the outermost orbital, an s orbital, is underlain by weakly-shielding d and f shells, and this electron's mass and hence binding-energy are increased by a relativistic effect. This is because it moves so fast when near the nucleus - and every s electron spends some time near the nucleus, unlike p,d, f etc.