A trend with two steps down
The first ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms, one electron from each atom. You met it in the lecture on ionisation energies, where it was used to reveal shells. This lecture compares it across the periodic table.
Here is the first ionisation energy of every element in Period 3, from sodium to argon:
The overall shape is easy to predict from the last lecture. But look closely. Twice, the next element needs less energy, not more. Aluminium has more protons than magnesium, and it is easier to ionise. Sulfur has more protons than phosphorus, and it is easier to ionise too.
These two small steps down are the most interesting points on the graph. They are not errors in the data. They are direct evidence that shells are divided into subshells and orbitals. By the end of this lecture, you will be able to explain both, and to predict where they appear in any period.
The same tug-of-war as atomic radius
The last lecture explained atomic radius with a tug-of-war: the nucleus pulls the outer electrons in, while distance and shielding weaken that pull. Ionisation energy measures the same contest from a different side. It asks: how much energy does it take to pull the outer electron away completely? A strongly held electron needs a lot of energy. A weakly held one needs little.
Across a period, each element has one more proton, and the new electron goes into the same outer shell. Electrons in the same shell shield each other very little. So the rough effective nuclear charge rises: about 1 for sodium, up to about 8 for argon. The outer electrons are held more tightly, the atoms are smaller, and the first ionisation energy rises.
Down a group, each element has one more shell. The outer electron is further from the nucleus, and the extra inner electrons cancel most of the extra nuclear charge. So the outer electron is held less tightly, and the first ionisation energy falls.
| Group 1 atom | Outer shell | First IE (kJ/mol) |
|---|---|---|
| Li | 2 | 520 |
| Na | 3 | 496 |
| K | 4 | 419 |
| Rb | 5 | 403 |
| Cs | 6 | 376 |
So the two trends mirror atomic radius. Small atoms hold their outer electrons tightly and have high ionisation energies. Large atoms hold them loosely and have low ones. Caesium, the largest atom in the last lecture, is the easiest of these to ionise.
One more point about the end of each period. Argon has the highest first ionisation energy in Period 3. Some books say this is because a full outer shell is "stable". The clearer reason is the one you already have: argon has the most protons of any Period 3 element, with the same 10 inner electrons shielding them. Its outer electrons feel the strongest pull in the period.
The first step down: magnesium to aluminium
Now the dips. They both come from the subshells inside a shell, which you met in the lecture on electron configuration.
Magnesium is [Ne] 3s². The electron removed first is a 3s electron. Aluminium is [Ne] 3s² 3p¹. The electron removed first is the single 3p electron.
A 3p electron has higher energy than a 3s electron. On average it is slightly further from the nucleus, and the two 3s electrons shield it a little. So aluminium's 3p electron is held less tightly than magnesium's 3s electrons, even though aluminium's nucleus has one more proton. Less energy is needed to remove it: 578 kJ/mol, against 738 kJ/mol for magnesium.
Read that carefully. Aluminium's nucleus pulls harder than magnesium's. The dip happens because the electron being removed is a different kind of electron: the first one in a new, higher-energy subshell.
The second step down: phosphorus to sulfur
Phosphorus is [Ne] 3s² 3p³. By Hund's rule, its three 3p electrons occupy the three 3p orbitals singly: ↑ | ↑ | ↑. Sulfur is [Ne] 3s² 3p⁴. The fourth 3p electron must share an orbital with another electron: ↑↓ | ↑ | ↑.
Two electrons in the same orbital occupy the same region of space. Both are negative, so they repel each other. This repulsion pushes one of them away a little, so it is easier to remove. Sulfur's first ionisation energy, 1000 kJ/mol, is therefore slightly below phosphorus's, 1012 kJ/mol.
That is it. That is the whole picture. The general rise comes from the nuclear charge. The two dips come from which electron is removed: the first electron in a p subshell (Group 2 to Group 13), and the first paired electron in a p subshell (Group 15 to Group 16). Everywhere else in the period, the rise continues.
Notice the sizes. The dip from Mg to Al is 160 kJ/mol. The dip from P to S is only 12 kJ/mol. Both are real, but the pairing effect is much smaller.
Does the dip at aluminium mean that aluminium's nucleus attracts electrons less strongly than magnesium's?
Show the answer
No. Aluminium has 13 protons and magnesium has 12, so aluminium's nucleus attracts more strongly. The dip happens because a different electron is removed. In magnesium it is a 3s electron. In aluminium it is a 3p electron, which has higher energy and is easier to remove. Ionisation energy measures how tightly one particular electron is held, not how strong the nucleus is overall.
Build the pattern yourself
Period 2 has exactly the same pattern, because its subshells fill in the same order: 2s, then 2p. Below, build either period one element at a time. Before each bar appears, decide whether it will be higher or lower than the one before.
Choose a period. For each new element, decide whether its first ionisation energy will be higher or lower than the element before it. The bar appears with the reason.
The first ionisation energies of Period 2 (Period 3 is in the bar chart above), in kJ/mol, with the reason for each step:
| Atom | First IE | Compared with the one before |
|---|---|---|
| Li | 520 | start of the period |
| Be | 900 | higher: one more proton, same 2s subshell |
| B | 801 | lower: first electron in the higher-energy 2p subshell |
| C | 1087 | higher: one more proton, unpaired 2p electrons |
| N | 1402 | higher: three 2p electrons, one per orbital |
| O | 1314 | lower: the fourth 2p electron is paired, and repulsion makes it easier to remove |
| F | 1681 | higher: one more proton |
| Ne | 2081 | higher: the most protons, same inner shielding |
In both periods, the dips fall in the same two places: Group 2 to Group 13, and Group 15 to Group 16. That is because both periods fill an s subshell and then a p subshell in the same way. (Some books number the groups I to VIII. In that system, these are Groups II to III and V to VI.)
Worked examples
Explain why the first ionisation energy rises from sodium (496 kJ/mol) to magnesium (738 kJ/mol).
Sodium is [Ne] 3s¹ and magnesium is [Ne] 3s². In both, the electron removed is a 3s electron, in the same shell and the same subshell. Magnesium has one more proton, and the extra 3s electron hardly shields the other one, because they are at the same distance from the nucleus. So magnesium's outer electrons feel a stronger pull, and more energy is needed to remove one.
Explain why the first ionisation energy of oxygen (1314 kJ/mol) is lower than that of nitrogen (1402 kJ/mol).
Nitrogen is 1s² 2s² 2p³. Its three 2p electrons are in separate orbitals: ↑ | ↑ | ↑. Oxygen is 1s² 2s² 2p⁴, so one 2p orbital holds a pair: ↑↓ | ↑ | ↑. The two electrons in that orbital repel each other. This repulsion makes one of them easier to remove, even though oxygen has one more proton. So oxygen's first ionisation energy is lower.
Notice what the answer does not say. Both outer electrons are in the 2p subshell, so the "higher-energy subshell" reason does not apply here. That reason is for boron.
Arrange Na, Cl, Ar and K in order of increasing first ionisation energy.
K is below Na in Group 1, so K is lower than Na. Na, Cl and Ar are in Period 3, and none of the steps from Na to Cl or from Cl to Ar is one of the two dips. So the general rise applies: Na < Cl < Ar. The order is K < Na < Cl < Ar.
Check with the data: 419, 496, 1251 and 1521 kJ/mol.
Which has the higher first ionisation energy: sulfur or chlorine?
Both are in Period 3. Sulfur is in Group 16 and chlorine is in Group 17. The dips are only at Group 2 to Group 13 and Group 15 to Group 16. The step from Group 16 to Group 17 is an ordinary step. So the general rise applies, and chlorine is higher: 1251 kJ/mol against 1000 kJ/mol.
Sulfur is a "dip" element only when it is compared with phosphorus. Against chlorine, it is simply one step to the left.
The answer below confuses the two dips. Find the wrong line.
A student explains why oxygen's first ionisation energy is lower than nitrogen's. One line is wrong. Tap it.
The student's answer, with the mistake marked:
- Line 1. Nitrogen is 1s² 2s² 2p³ and oxygen is 1s² 2s² 2p⁴. (Correct.)
- Line 2. Oxygen has one more proton, which on its own would raise its first ionisation energy. (Correct.)
- Line 3. Oxygen's outer electron is in the 2p subshell, which is higher in energy than nitrogen's 2s subshell. The mistake: nitrogen's outer electrons are also in 2p. This reason explains boron, not oxygen.
- Line 4. Nitrogen's three 2p electrons are in separate orbitals. (Correct.)
- Line 5. In oxygen, one 2p orbital holds a pair of electrons. (Correct.)
- Line 6. The paired electrons repel each other, so one of them is easier to remove. (Correct.)
Three classic mistakes
Group 2 to Group 13 is about a new subshell: the first p electron is higher in energy. Group 15 to Group 16 is about pairing: two electrons in one orbital repel. Each dip has its own reason, and examiners check that you use the right one.
Aluminium and sulfur each have one more proton than the element before them. Their nuclei pull harder. The dips happen because the electron removed is easier to remove for another reason.
There are only two dips in each of Periods 2 and 3. Every other step to the right is an ordinary rise, including Group 16 to Group 17. Before calling a step a dip, check which two groups it joins.
Work each one on paper first. Open a hint only when you are stuck.
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Arrange Mg, Na, Ar and Cl in order of increasing first ionisation energy.
Hint 1
All four are in Period 3. Are any of the steps between them one of the two dips?
Hint 2
The dips are at Mg to Al and P to S. None of these four elements is on the right-hand side of a dip.
Full answer
The general rise applies: Na < Mg < Cl < Ar. Check: 496, 738, 1251 and 1521 kJ/mol.
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The first ionisation energy of beryllium is 900 kJ/mol, and that of boron is 801 kJ/mol. Explain why boron's is lower.
Hint 1
Write both configurations. Which subshell does the electron removed come from in each?
Hint 2
Beryllium is 1s² 2s². Boron is 1s² 2s² 2p¹.
Full answer
In beryllium, the electron removed is a 2s electron. In boron, it is the single 2p electron. The 2p subshell is higher in energy than 2s: the 2p electron is slightly further out on average and slightly shielded by the 2s electrons. So it is easier to remove, even though boron has one more proton. This is the Group 2 to Group 13 dip, in Period 2.
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Element X is in Period 2. Its first ionisation energy is lower than that of the element just before it, and an atom of X has two unpaired electrons. Identify X.
Hint 1
In Period 2, the two dips are at boron and oxygen.
Hint 2
Count the unpaired electrons: boron is 2p¹, oxygen is 2p⁴ (↑↓ | ↑ | ↑).
Full answer
X must be boron or oxygen. Boron has one unpaired electron. Oxygen, ↑↓ | ↑ | ↑, has two. So X is oxygen. Check: oxygen's 1314 kJ/mol is lower than nitrogen's 1402 kJ/mol.
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Argon (18 protons) has a first ionisation energy of 1521 kJ/mol. Potassium (19 protons) is the very next element, with 419 kJ/mol, less than a third as much. Explain this huge fall, and explain why potassium's value is also lower than sodium's (496 kJ/mol).
Hint 1
Write the configurations of argon and potassium. Where does potassium's outer electron go?
Hint 2
Use the rough effective charge: protons minus inner electrons. For argon, the inner electrons are 10. For potassium, they are 18.
Full answer
Argon is [Ne] 3s² 3p⁶: its outer electrons are in shell 3, with 10 inner electrons, so the rough effective charge is 18 − 10 = 8. Potassium is [Ar] 4s¹: its outer electron starts a new shell, shell 4, further from the nucleus, with 18 inner electrons, so the rough effective charge is 19 − 18 = 1. A far weaker pull at a greater distance means far less energy is needed, so the value falls from 1521 to 419. Compared with sodium, potassium's outer electron is one shell further out, with about the same effective charge of 1, so it is held less tightly: 419 against 496. One extra proton made almost no difference; one extra shell made all the difference.
A graph of measured energies, made in laboratories, has two small steps down in exactly the places that the subshell and orbital model predicts. That is how chemists know the model is more than a drawing. The s, p and orbital boxes are not only a way to write configurations. They describe something that shows up in real measurements.
The same tug-of-war continues in the next lecture. Electronegativity asks how strongly an atom attracts the electrons it shares in a bond. Like ionisation energy, it generally rises across a period and falls down a group.
- First ionisation energy generally rises across a period, because the nuclear charge rises while the inner shielding stays the same.
- First ionisation energy falls down a group, because each new shell puts the outer electron further away, and extra inner electrons cancel most of the extra nuclear charge.
- The dip from Group 2 to Group 13 happens because the electron removed is the first in a p subshell, which is higher in energy than the s subshell.
- The dip from Group 15 to Group 16 happens because the electron removed is one of a pair sharing an orbital, and the pair repel each other.
- The noble gas at the end of each period has the highest first ionisation energy, because it has the most protons with the same inner shielding.