How do you explain the trend of ionization energies as you go from left to right along the second row of the Periodic Table?

Question

Wyatt Atkins · Accepted Answer

I explain it by the stability of filled and half-filled sub orbitals and by the increasing number of protons in the nucleus.

Here is a graph of the ionization energies for the elements of Period 2.

As the number of protons goes up, the attraction for the electrons increases and the ionization energy generally increases. The exceptions are the places where the atoms are more stable because of filled and half-filled subshells.

The I A atoms have the lowest first ionization energy. The lone electron in the #"s"# orbital is easily removed.

The II A atoms have a filled #"s"# subshell, which is more stable, so the ionization energy goes up.

The III A and IV A have more protons in their nuclei, so their ionization energies go up.

The V A atoms have three electrons in the #"p"# orbitals, creating a half filled #"p"# subshell. The half filled #"p"# subshell has a high degree of stability so the ionization energy goes up.

The VI A atoms have four electrons in the #"p"# orbitals, and this creates a #"p"# orbital with 2 electrons. Electrons "prefer" not to be paired, so the fourth electron in the #"p"# orbitals is more easily removed and the ionization energy goes down.

The VII A atoms have an extra proton in the nucleus, so the ionization energy goes up.

The VIII A atoms have a filled shell. These are the most stable of all the family of atoms.

Cameron Andrews · Answer

undefined The trend of ionization energies as you go from left to right along the second row of the Periodic Table generally increases.