The energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous 1+ ions.
The energy required to remove 1 mole of electrons from 1 mole of gaseous 1+ ions to form 1 mole of gaseous 2+ ions.
So on and so forth for the definitions of every other ionization energy.
A larger number of protons yields a higher ionization energy since the attraction between protons and electrons is stronger.
Something else to note is that the nuclear charge increases for every successive ionization energy (i.e. the 3rd ionization energy deals with a higher nuclear charge than the 1st). Again, there are less electrons at this stage, so the attraction between protons and electrons will be stronger, resulting in a greater required ionization energy to remove electrons.
As you move away from the nucleus, the attraction between protons and electrons weakens. Hence, a greater atomic radius results in less required ionization energy. Inverse proportion.
Inner shells are more attracted to the nucleus than outer shells. Hence, the more inner shells you have, the weaker the attraction is between the nucleus and the outer shells. The outer shells are "shielded" from the attraction of the nucleus, so to speak. More shielding results in less required ionization energy. Inverse proportion.
Greater ionization energy is required to remove electrons from fully or half-filled orbitals.
Looking at a graph of the ionization energies of period 3, we can see that the ionization energies of aluminium and sulfur are less than the elements before them.
In the case of aluminium, a new subshell of the same energy level as magnesium is introduced. Since this new subshell is further away from the nucleus (more shielded), less ionization energy is required (Mg ends with a 3s^2 subshell, Al ends with a 3p^1 subshell). In more general terms, the outermost electron of Al is more shielded than Mg.
In the case of sulfur, we have a newly filled orbital. Since we're dealing with a 3p^4 subshell, one of the orbitals is full and the other two are half-full. In the full orbital, both electrons experience repulsion towards each other, resulting in it being easier to remove an electron from that orbital. Phosphorus deals with 3p^3, in which all orbitals are half-full. This repulsion doesn't exist there. Hence, a lesser ionization energy for sulfur.
These hold true for the rest of the periods. There exists a dip in ionization for oxygen too, but the reasoning remains the same.
This article was written on 16/12/2023. If you have any thoughts, feel free to send me an email with them. Have a nice day!