Ionisation energies — evidence for electron shells — Chemistry

Before this

This leads to

How do we know the shells are there?

For two lectures, we have drawn electrons in shells: sodium is 2, 8, 1. Here is a question that should bother you. Nobody has ever seen a shell. An electron is far too small to photograph. So why should you believe any of it?

A good scientist does not accept a model because a teacher draws it. She asks for evidence. This lecture gives you some of the strongest evidence there is, and it comes from a simple idea. Pull the electrons off an atom one at a time, and measure how much energy each one costs. If electrons really sit in shells, the costs should show it.

They do. By the end of this lecture, you will be able to look at a list of numbers and read the shell structure of an atom straight out of it.

Ionisation energy: the price of one electron

Removing an electron from an atom is called ionisation, because the atom becomes a positive ion. The electron is attracted to the positive nucleus, so pulling it away always costs energy. That energy is the ionisation energy.

First ionisation energy

The first ionisation energy of an element is the energy needed to remove one mole of electrons from one mole of gaseous atoms, forming one mole of gaseous 1+ ions.

For sodium: $Na(g) \to Na^{+} (g) + e^{-}$ first ionisation energy $= 495.8 kJ/mol$

Every part of that definition has a reason.

Gaseous atoms. In a solid or a liquid, atoms touch and attract each other. In a gas, each atom is on its own. So the energy measured is only the cost of pulling the electron away from its own nucleus, and nothing else.
One mole. One electron needs a tiny amount of energy. Chemists measure in moles, so the value is per mole of atoms, in kJ/mol.
Always positive. Energy must be put in, because the electron is attracted to the nucleus. No atom gives up an electron and releases energy in this process.

One electron after another

We do not have to stop at one electron. The second ionisation energy removes a second electron from the 1+ ion:

Na^{+} (g) \to Na^{2 +} (g) + e^{-}

The third removes a third electron from the 2+ ion, and so on, until the nucleus is bare. These are the successive ionisation energies. Each one starts from the ion the previous step made.

Successive ionisation energies always increase. The reason is simple. Each time an electron leaves, the same number of protons is left pulling on fewer electrons. The remaining electrons repel each other less, and the ion is more positive. So each electron is held more tightly than the one before.

If that were the whole story, the values would rise smoothly. Before you see sodium's real data, commit to a guess.

Predict, then reveal — sodium's eleven electrons

Sodium is 2, 8, 1. Its eleven successive ionisation energies are measured. What pattern do you expect?

Sodium's eleven successive ionisation energies, on a log scale (each grid line is ten times the one below):

One low value, then eight that rise gently, then two very high ones: groups of 1, 8 and 2. There are two jumps, after the 1st electron and after the 9th.

Here are the real values, from data tables, with each one divided by the one before it:

Electron removed	Ionisation energy (kJ/mol)	× previous	Comes from
1st	495.8	—	shell 3
2nd	4 562	9.2	shell 2
3rd	6 910	1.5	shell 2
4th–8th	9 543 → 25 496	1.2 to 1.4	shell 2
9th	28 932	1.1	shell 2
10th	141 362	4.9	shell 1
11th	159 076	1.1	shell 1

Read the pattern: 1 easy electron, then 8 harder ones, then 2 very hard ones. That is 1, 8, 2. It is sodium's configuration, 2, 8, 1, read from the outside in. That is it. That is the whole idea: the measured energies fall into exactly the groups that the shell model predicts.

Why the jumps happen

A jump means the next electron is held much more tightly. Two things decide how tightly an electron is held: how close it is to the nucleus, and how much of the nucleus's pull reaches it.

Inner electrons sit between the nucleus and the outer ones, and they cancel part of the nucleus's pull. This is called shielding. Sodium's outer electron is in shell 3. It is far from the nucleus, and 10 inner electrons shield it from the 11 protons. So it feels only a weak pull, roughly that of a single proton. That makes it cheap to remove.

The 2nd electron comes from shell 2. It is much closer to the nucleus, and only the 2 electrons in shell 1 shield it. It feels a far stronger pull. So the price jumps by a factor of about 9. The same thing happens again at the 10th electron, which comes from shell 1, right next to the nucleus, with nothing shielding it at all.

Stop and think

Sodium's 2nd ionisation energy is about 9 times its 1st. A student says: "That is because Na⁺ is positive, so it holds its electrons more tightly." Is that the main reason?

Show the answer

No. Look at the next step. Going from Na⁺ to Na²⁺, the ion also becomes one charge more positive, but the 3rd ionisation energy is only about 1.5 times the 2nd. So a rising charge alone gives a small step. The ×9.2 step happens because the 2nd electron comes from a different shell: closer to the nucleus and much less shielded. The change of shell, not the charge, makes the jump.

This also explains something from the last lecture: why sodium forms Na⁺ and never Na²⁺. The first electron costs about 496 kJ/mol. The second costs about 4 562 kJ/mol, more than nine times as much. In chemical reactions, the energy to pay that second price is not available. So sodium stops at 1+.

Reading jumps: ratios, not differences

Here is a detail that most explanations skip, and it catches many students. Look at sodium's numbers again. The step from the 1st to the 2nd ionisation energy is about 4 066 kJ/mol. The step from the 10th to the 11th is about 17 714 kJ/mol, more than four times bigger. Yet the first is a change of shell, and the second is not. Both of the last two electrons come from shell 1.

So the size of a difference can mislead you, because the later numbers are all huge. What matters is the ratio: how many times bigger each value is than the one before. Inside a shell, the ratio is usually small, often between 1.1 and 2. At a change of shell, it is much bigger, usually around 4 or more. When you are unsure, compare the ratios with each other and choose the biggest.

This is why data books and exam papers often plot successive ionisation energies on a log scale, where each grid line is ten times the one below. On a log scale, equal ratios make equal steps. So a big ratio shows up as a big step, whether the numbers are small or huge. Try both scales below.

Explore — the same data on two scales

Choose an element and a scale. Then step through the electrons and watch the ratio to the previous value.

Electron removed: 2

Magnesium (2, 8, 2) on an ordinary scale and on a log scale. On the ordinary scale, the jump after the 2nd electron is invisible. On the log scale, both jumps are clear.

Top: ordinary scale. Bottom: log scale, where each grid line is ten times the one below.

Element	Big ratios (jumps)	Groups of electrons
Sodium	×9.2 after the 1st · ×4.9 after the 9th	1, 8, 2
Magnesium	×5.3 after the 2nd · ×4.8 after the 10th	2, 8, 2
Aluminium	×4.2 after the 3rd · ×4.7 after the 11th	3, 8, 2

On the ordinary scale, the first nine or ten values of every element look like a flat row of small bars, and only the innermost shell stands out. On the log scale, every change of shell shows up as a clear step. The rule for reading the data is now simple.

Reading the shells from the data

Find the big ratios. The number of electrons removed before the first big ratio is the number of electrons in the outer shell. Each later big ratio marks the next shell inwards.

Worked examples

Worked Example 1 — Predict magnesium, then check

Magnesium is 2, 8, 2. Predict the pattern of its 12 successive ionisation energies, then check it against the data.

Prediction: 2 low values (shell 3), a jump, 8 middle values (shell 2), a jump, and 2 very high values (shell 1). The groups should be 2, 8, 2.

Data: 737.7, 1 450.7, 7 732.7, … , 35 458, 169 988, 189 368 kJ/mol. The ratios are ×2.0 from the 1st to the 2nd (same shell), ×5.3 from the 2nd to the 3rd (new shell), ×4.8 from the 10th to the 11th (new shell). The prediction holds.

Notice the ×2.0 between the 1st and 2nd values. It is bigger than the ×1.1 to ×1.5 steps inside shell 2, but far smaller than a change of shell. Removing an electron from a 1+ ion is harder than from a neutral atom, so the first step inside the outer shell is often around ×2.

Worked Example 2 — Identify an unknown element

An element in period 3 has these first five ionisation energies, in kJ/mol: 577.5, 1 816.7, 2 744.8, 11 577, 14 842. Which group is it in, and what is it?

Work out each ratio: $1816.7 \div 577.5 \approx 3.1$ , $2744.8 \div 1816.7 \approx 1.5$ , $11577 \div 2744.8 \approx 4.2$ , $14842 \div 11577 \approx 1.3$ .

The biggest ratio, ×4.2, comes after the 3rd electron. So 3 electrons are in the outer shell. In period 3, the element with 3 outer electrons is in Group 13. (Some books number the groups I to VIII, and call this Group III.) It is aluminium, 2, 8, 3.

Check: the ×3.1 after the 1st electron is big for a step inside one shell. It is still clearly smaller than ×4.2, so the shell boundary is after the 3rd electron. Always compare the ratios with each other, and choose the biggest.

Worked Example 3 — Equations and energy for an ion

Write the equation for the second ionisation energy of magnesium. Then find the energy needed to turn one mole of gaseous magnesium atoms into gaseous Mg²⁺ ions.

Mg^{+} (g) \to Mg^{2 +} (g) + e^{-}

The second ionisation energy starts from the 1+ ion, not from the atom. To go all the way from Mg(g) to Mg²⁺(g), both steps happen, one after the other. So the energies add:

737.7 + 1450.7 = 2188.4 kJ/mol

Check: the third electron alone would cost 7 732.7 kJ/mol, more than three times this total. That is why magnesium forms Mg²⁺ but not Mg³⁺.

Three classic mistakes

Trap 1 — Equations without (g), or from the wrong start

Every ionisation equation needs state symbols, and every species is a gas: $Na(g) \to Na^{+} (g) + e^{-}$ . The second ionisation energy starts from the 1+ ion. Writing $Mg(g) \to Mg^{2 +} (g) + 2 e^{-}$ describes the first two steps added together, not the second ionisation energy.

Trap 2 — Looking for the biggest difference

Later ionisation energies are huge, so their differences are huge too. Sodium's step from the 10th to the 11th value is about 17 714 kJ/mol, but it is not a change of shell. Divide each value by the one before, and look for the biggest ratio.

Trap 3 — Reading the shells in the wrong order

The first group of electrons removed comes from the outer shell. Sodium's data give groups of 1, 8, 2, which means the configuration 2, 8, 1. Read the groups from the outside in.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Write equations, with state symbols, for (a) the first ionisation energy of potassium, K, and (b) the third ionisation energy of aluminium, Al.

Hint 1
Every species in the equation is a gas, and one electron is removed in each equation.

Hint 2
The third ionisation energy starts from the ion that has already lost two electrons.

Full answer
(a) $K(g) \to K^{+} (g) + e^{-}$ . (b) $Al^{2 +} (g) \to Al^{3 +} (g) + e^{-}$ . Check: in each equation the charges balance. In (b), 2+ on the left equals 3+ and 1− on the right.
Calcium's first four ionisation energies are 589.8, 1 145.4, 4 912.4 and 6 491 kJ/mol. How many electrons are in its outer shell, and which group is it in?

Hint 1
Divide each value by the one before it.

Hint 2
The ratios are about 1.9, 4.3 and 1.3. Where is the biggest?

Full answer
The ratios are $1145.4 \div 589.8 \approx 1.9$ , $4912.4 \div 1145.4 \approx 4.3$ and $6491 \div 4912.4 \approx 1.3$ . The big ratio comes after the 2nd electron, so calcium has 2 outer electrons and is in Group 2. Check: calcium is 2, 8, 8, 2, which has 2 electrons in its outer shell.
Silicon's first five ionisation energies are 786.5, 1 577.1, 3 231.6, 4 355.5 and 16 091 kJ/mol. Find the number of outer electrons, the group, and silicon's configuration in shells.

Hint 1
The first three ratios are all close to 2 or below. Look at the last one.

Hint 2
$16091 \div 4355.5 \approx 3.7$ .

Full answer
The ratios are about 2.0, 2.0, 1.3 and 3.7. The biggest, ×3.7, comes after the 4th electron. So silicon has 4 outer electrons and is in Group 14. Its atomic number is 14, so its configuration is 2, 8, 4. Check: $2 + 8 + 4 = 14$ electrons.
Lithium has only three ionisation energies: 520.2, 7 298.1 and 11 815.0 kJ/mol. (a) Explain why there are only three. (b) Find the shell structure from the data. (c) The jump ratio for lithium is about ×14, but for sodium's first jump it is only about ×9.2. Suggest why lithium's jump is bigger.

Hint 1
Lithium's atomic number is 3. Find the two ratios.

Hint 2
After lithium's first electron, the next electron comes from shell 1. After sodium's first electron, the next one comes from shell 2. Which is closer to the nucleus, and less shielded?

Full answer
(a) Lithium has 3 protons and so 3 electrons. Once all three are removed, nothing is left to remove. (b) The ratios are $7298.1 \div 520.2 \approx 14.0$ and $11815.0 \div 7298.1 \approx 1.6$ . The big ratio comes after the 1st electron, so the groups are 1 and 2: lithium is 2, 1. (c) Lithium's 2nd electron comes from shell 1, right next to the nucleus, with no inner electrons shielding it. Sodium's 2nd electron comes from shell 2, which is further out and is still shielded by the 2 electrons in shell 1. So the change in how tightly the electron is held is bigger for lithium. Check: $520.2 \times 14.0 \approx 7283$ , close to 7 298.1.

The bigger idea

Nobody can see a shell, but we can measure what shells do. That is how much of science works. A model makes a prediction, here that electrons should come off in groups. Then an experiment checks it. When the numbers fall into exactly the groups 1, 8, 2 for sodium and 2, 8, 2 for magnesium, the model earns our trust.

The same data hold more than shells. With careful measurements, smaller steps inside a shell reveal the subshells from the last lecture too. The next topic compares the first ionisation energies of many elements across the periodic table, and those small steps become clear.

Key concepts in this lesson

The first ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions, for example $Na(g) \to Na^{+} (g) + e^{-}$ .
Successive ionisation energies always increase, because each electron is removed from an ion with the same protons but fewer electrons.
A large jump happens when the next electron comes from a shell closer to the nucleus, where it is less shielded and held much more tightly.
Jumps are found by comparing ratios of successive values, not differences, which is why the data are often drawn on a log scale.
The number of electrons removed before the first big jump is the number of electrons in the outer shell, which gives the group.