Plots of first ionization energy across Periods 2 and 3 show a clear overall rise, but the graphs are not perfectly smooth. The dips in specific places are important evidence about electron arrangement.

When explaining any point on the graph, connect the visible pattern to the electron being removed and the forces acting on it.

Shape of the plots

A plot of first ionization energy against atomic number for Period 2 or Period 3 has a similar overall pattern:

• it starts at a relatively low value

• it rises across the period

• it reaches a maximum at the noble gas

• it contains small deviations from the steady increase

For Period 2, the elements are Li to Ne.
For Period 3, the elements are Na to Ar.

First ionisation energy across Period 3 (Na to Ar), plotted against atomic number. The overall upward trend is visible, with the characteristic deviations at Al (electron removed from a higher-energy $3p$ subshell than $3s$) and at S (electron–electron repulsion from a paired $3p$ electron). Source

The graph is not a straight line because the outer electron is not always removed from an identical type of orbital. Changes in subshell and changes in electron pairing produce local drops.

If you compare the two periods, the pattern of rises and dips is similar. However, the absolute first ionization energies in Period 3 are generally lower than the equivalent values in Period 2 because the outer electrons in Period 3 are farther from the nucleus and experience more shielding.

Why the overall trend rises across a period

Across each period, proton number increases. This means the nucleus has a greater positive charge.

At the same time:

• electrons are being added to the same main shell

• shielding does not increase very much across the period

• atomic radius decreases slightly

• attraction between the nucleus and the outer electron becomes stronger

As a result, more energy is needed to remove the outer electron, so the first ionization energy generally increases.

This is the main explanation for the upward trend in both Period 2 and Period 3. In exam answers, it is best to state all parts of the reasoning clearly:

• greater nuclear charge

• similar shielding

• smaller radius or stronger attraction

• therefore more energy required

A common mistake is to say only that there are “more protons.” That is not enough unless you link it to the stronger attraction for the electron being removed.

Significant deviations in Period 2

There are two important dips in Period 2 that students must be able to explain.

The dip from Be to B

The first ionization energy of B is lower than that of Be, even though B has one more proton.

This happens because:

• in Be, the electron removed is from a 2s subshell

• in B, the electron removed is from a 2p subshell

• a 2p electron is at slightly higher energy than a 2s electron

• it is also slightly farther from the nucleus on average

• so it is easier to remove

Therefore, the move from Be to B produces a drop in first ionization energy.

The key idea is that the overall increase across the period is interrupted when the electron removed comes from a higher-energy subshell.

The dip from N to O

The first ionization energy of O is lower than that of N.

This is because:

• N has three electrons occupying the three 2p orbitals singly

• O has four electrons in the 2p subshell

• in O, one of the 2p orbitals contains a pair of electrons

• repulsion between paired electrons makes one of them easier to remove

So, although O has a higher nuclear charge than N, the extra electron-electron repulsion in the paired orbital lowers the energy needed to remove an electron.

This dip is evidence that electrons occupy orbitals singly before pairing, and that paired electrons repel each other.

Orbital box diagrams for N and O showing how the three $2p$ orbitals fill according to Hund’s rule. Nitrogen has three unpaired $2p$ electrons (one in each orbital), whereas oxygen has one paired set in a $2p$ orbital, increasing electron–electron repulsion and lowering the first ionisation energy for O. Source

Significant deviations in Period 3

Period 3 shows the same type of pattern as Period 2.

The dip from Mg to Al

The first ionization energy of Al is lower than that of Mg.

The explanation matches the Be to B pattern:

• Mg loses an electron from a 3s subshell

• Al loses an electron from a 3p subshell

• a 3p electron is higher in energy than a 3s electron

• it is therefore easier to remove

So the graph dips from Mg to Al instead of continuing to rise smoothly.

The dip from P to S

The first ionization energy of S is lower than that of P.

The explanation matches the N to O pattern:

• P has three electrons in the 3p subshell, occupying separate orbitals

• S has four electrons in the 3p subshell

• one 3p orbital in S contains a paired set of electrons

• repulsion within that pair makes removal easier

This extra repulsion slightly reduces the first ionization energy of S.

Interpreting exam plots accurately

When you are asked to interpret or recall a plot, focus on the shape as well as the reasons.

Useful points to state are:

• there is an overall increase across each period

• the increase is due to increasing nuclear charge

• shielding remains roughly similar across the period

• there is a dip after Group 2

• there is another dip after Group 15

For Period 2, the important deviations are:

• Be to B

• N to O

For Period 3, the important deviations are:

• Mg to Al

• P to S

Avoid vague phrases such as “the pattern is irregular.” The deviations are not random. They occur for clear structural reasons:

• change from s to p subshell

• pairing of electrons in a p orbital

In high-mark answers, precise language matters. Use terms like nuclear charge, shielding, subshell, orbital, electron repulsion, and easier to remove.