## Electronegativity, Electron Affinity and Ionization Energy

$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$

ERIKTORRESDisc3C
Posts: 18
Joined: Wed Sep 21, 2016 2:58 pm

### Electronegativity, Electron Affinity and Ionization Energy

How do Ionization Energy and Electron Affinity affect Electronegativity? I also don't understand electronegativity trend over the Periodic Table. Does it follow the same trend that ionization energy follows on the periodic table?

Eljie_2F
Posts: 19
Joined: Fri Jun 17, 2016 11:28 am

### Re: Electronegativity, Electron Affinity and Ionization Energy

Ionization energy is the energy that it takes to remove the outermost valence electron from an atom. Since atoms with high ionization energies are less prone to losing electrons, this usually means that these elements are more electronegative and are instead prone to accepting electrons. This is a loose connection, not a real causation for one or the other. Electron affinity is an atom's likelihood of gaining an electron. Generally, the higher and atom's electron affinity the higher its electronegativity. All three have the same trend of increasing as you move up and to the right on the periodic table.

Britney Pheng 1L
Posts: 22
Joined: Wed Sep 21, 2016 2:56 pm
Been upvoted: 1 time

### Re: Electronegativity, Electron Affinity and Ionization Energy

Hi Erik!

To add on to what Eljie said:

The trends of ionization energy, electron affinity, and electronegativity on the periodic table are the same. From left to right AND down to up on the table, the elements will have an increase in these trends.

So if the ionization energy of an atom is high, the atom is more reluctant in giving up an electron. This means that the atom will also be more electronegative because it would rather attract an electron than give one up.

If the electron affinity of an atom is high, that means more energy is needed to remove an electron from that atom. Instead, attaching an electron would be more energetically favored. Again, this goes with the idea that the atom will be more electronegative since it will want to attract an electron rather than release one.

Elements with both high I.E. and E.A. are reluctant to lose their electrons, therefore they tend to gain them instead (highly electronegative).

There’s also this equation from the textbook (Section 3.12): X = ½(I + E$_{a}$)
Where X = Electronegativity, I = Ionization Energy, E$_{a}$ = Electron Affinity.
This equation shows the positive correlation between all three trends.

Hope this helps!