First ionisation energy across period 3
Learning outcomes
After studying this page, you should be able to:
- describe and explain the trend in first ionisation energy across period 3.
First ionisation energy
| Element | Symbol | Atomic number | First ionisation energy /kJ mol–1 |
|---|---|---|---|
| Sodium | Na | 11 | 496 |
| Magnesium | Mg | 12 | 738 |
| Aluminium | Al | 13 | 578 |
| Silicon | Si | 14 | 789 |
| Phosphorus | P | 15 | 1012 |
| Sulfur | S | 16 | 1000 |
| Chlorine | Cl | 17 | 1251 |
| Argon | Ar | 18 | 1521 |
The graph shows how the first ionisation energy varies across period 3.
First ionisation energy generally increases across period 3. However, the trend needs a more detailed consideration than the trend in group 2. This is because first ionisation energies:
- decrease from magnesium to aluminium then increase again, and
- decrease from phosphorus to sulfur then increase again.
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Explanation
General increase across the period
Going across period 3:
- there are more protons in each nucleus so the nuclear charge increases …
- therefore the force of attraction between the nucleus and outer electrons is increased, and …
- there is a negligible increase in shielding because each successive electron enters the same shell …
- so more energy is needed to remove an outer electron.
Magnesium to aluminium
Look at their electronic configurations:
- Magnesium: 1s2 2s2 2p6 3s2
- Aluminium: 1s2 2s2 2p6 3s2 3p1
The outer electron in magnesium is in an s sub-shell. However, the outer electron in aluminium is in a p sub-shell, so it is higher in energy than the outer electron in magnesium. This means that less energy is needed to remove it.
Phosphorus to sulfur
Look at their electronic configurations:
- Phosphorus: 1s2 2s2 2p6 3s2 3p3
- Sulfur: 1s2 2s2 2p6 3s2 3p4
It’s not immediately obvious what’s going on until you look at the arrangements of the electrons. The 3p electrons in phosphorus are all unpaired. In sulfur, however, two of the 3p electrons are paired. There is some repulsion between paired electrons in the same sub-shell, so the force of their attraction to the nucleus is reduced. This means that less energy is needed to remove one of these paired electrons than is needed to remove an unpaired electron from phosphorus.
It may help your understanding if you look at the diagrams below.
