3.1.1.3 - Electron Configuration

Shells, sub-shells, and orbitals

Electrons occupy main energy levels, often called shells. Within each shell there are sub-shells, and within each sub-shell there are orbitals. For this specification, the important sub-shells are s, p, and d.

An s sub-shell contains 1 orbital, so it can hold 2 electrons. A p sub-shell contains 3 orbitals, so it can hold 6 electrons. A d sub-shell contains 5 orbitals, so it can hold 10 electrons. Up to $Z = 36$, the filling order you need is:

This order matters because electrons fill the lowest available energy orbital first. Orbitals of the same energy fill singly before pairing, and no orbital can hold more than two electrons.

The diagram below lays out the orbital boxes and the filling order up to 4p.

[DIAGRAM: asset_name: 1.1.3 - Electron Configuration - Diagram 1; asset_slug: 1.1.3 - Electron Configuration - Diagram 1; recommended_method: deterministic_code; description: Orbital capacities and filling order; nitrogen 2p electrons occupy separate orbitals, and transition-metal ions lose 4s electrons first.]

For example:

• Sodium, $Z = 11$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{1}$
• Calcium, $Z = 20$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{4s}^{2}$
• Vanadium, $Z = 23$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{3} \mathrm{4s}^{2}$
• Krypton, $Z = 36$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{10} \mathrm{4s}^{2} \mathrm{4p}^{6}$

You should be able to translate between shell language and sub-shell language. Calcium is 2,8,8,2 in shells, but $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{4s}^{2}$ in sub-shell notation.

The simple filling sequence has two important exceptions within this range: chromium is $[\ce{Ar}] \mathrm{3d}^{5} \mathrm{4s}^{1}$ and copper is $[\ce{Ar}] \mathrm{3d}^{10} \mathrm{4s}^{1}$. Both still lose 4s electrons before 3d electrons when positive ions form.

Writing configurations for ions

To write the electron configuration of an ion, start from the neutral atom and then add or remove electrons.

For negative ions, add electrons to the next available orbital:

• Chloride ion, $\ce{Cl^-}$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6}$
• Oxide ion, $\ce{O^{2-}}$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6}$

For positive ions, remove electrons from the outer shell first. That means the electrons in the highest main energy level are lost first.

• Sodium ion, $\ce{Na^+}$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6}$
• Calcium ion, $\ce{Ca^{2+}}$: $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6}$

This rule is especially important for transition-metal ions. The simple filling rule places 4s before 3d, but the relative energies change as the 3d sub-shell fills. In transition-metal atoms the 4s electrons are removed first when positive ions form. Do not use the neutral-atom filling sequence as a removal sequence.

Worked example: iron has the neutral configuration $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{6} \mathrm{4s}^{2}$. Therefore $\ce{Fe^{2+}}$ is $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{6}$, not $\mathrm{3d}^{4} \mathrm{4s}^{2}$.

Start by working out how many electrons the ion actually has, then write the configuration. The proton number stays the same, but the charge changes the electron count.

First and successive ionisation energies

Ionisation energy is about removing electrons from gaseous species.

So the first ionisation energy of sodium is represented by:

Successive ionisation energies remove more electrons from the ions formed:

• second ionisation energy: $\ce{Na^+(g)} \rightarrow \ce{Na^{2+}(g)} + e^-$
• third ionisation energy: $\ce{Na^{2+}(g)} \rightarrow \ce{Na^{3+}(g)} + e^-$

Each successive ionisation energy is larger than the one before it because the same nucleus is attracting fewer electrons. The ion becomes more positive, the remaining electrons are held more strongly, and more energy is needed to remove the next one.

A very large jump between two successive ionisation energies is the key pattern to spot. It means the next electron would come from a lower shell, closer to the nucleus.

Worked example: magnesium has configuration $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2}$. Its first two electrons are removed from the 3s sub-shell, but the third electron would have to come from the $n = 2$ shell. So there is a large jump between the second and third ionisation energies.

This is how successive ionisation energy data gives evidence for shells: large jumps show when you start removing electrons from a new shell.

Evidence for shells from data

Successive ionisation energies are powerful because they let you work backwards to an electron arrangement. If an atom shows a very large jump after the second ionisation energy, it has two outer electrons. That is exactly what you expect for a Group 2 element.

For sodium, the huge jump after the first ionisation energy shows that sodium has one outer electron. The next eight electrons are much harder to remove because they are in the second shell, and the final two are hardest of all because they are in the first shell. That pattern supports the shell arrangement 2,8,1.

For magnesium, a huge jump after the second ionisation energy supports 2,8,2. Across Period 3, the position of the big jump changes from Na to Ar because the number of outer-shell electrons changes across the period.

The diagram below shows that characteristic jump for magnesium very clearly.

[DIAGRAM: asset_name: 1.1.3 - Electron Configuration - Diagram 2; asset_slug: 1.1.3 - Electron Configuration - Diagram 2; recommended_method: deterministic_code; description: Successive ionisation energies of magnesium, using rounded data in kJ per mole; the large jump follows removal of its two outer electrons.]

First ionisation energies down Group 2 also support the idea of shells. From Be to Ba, the first ionisation energy decreases overall even though nuclear charge increases. The reason is that the outer electron is in a higher shell each time, so it is further from the nucleus and more shielded by inner electrons.

That means Group 2 first ionisation energy data is evidence that electrons occupy different main energy levels, not all the same distance from the nucleus.

You can use the same logic in reverse: find the big jump, then count how many electrons were removed before it.

A logarithmic vertical axis must be converted before quoting an energy. If a graph gives $\log_{10}(IE)=3.50$, then $IE=10^{3.50}=3162\ldots$, or $3.16\times10^3\ \mathrm{kJ\,mol^{-1}}$ when those are the energy units specified. The graph coordinate 3.50 is not itself an ionisation energy. On a logarithmic axis, equal vertical gaps represent equal ratios.

So, successive ionisation energy patterns tell you how electrons are grouped into shells, while Group 2 first ionisation energy trends show that outer electrons sit in shells of increasing distance and shielding.

Evidence for sub-shells in Period 3

Across Period 3, first ionisation energy increases overall from Na to Ar because proton number increases while electrons are added to the same main shell. The attraction between the nucleus and the outer electron becomes stronger, so more energy is usually needed to remove it.

However, the graph is not perfectly smooth, and those small dips matter. They are evidence for sub-shells and electron pairing.

The diagram below combines the overall Period 3 trend with the two structural explanations for the dips.

[DIAGRAM: asset_name: 1.1.3 - Electron Configuration - Diagram 3; asset_slug: 1.1.3 - Electron Configuration - Diagram 3; recommended_method: deterministic_code; description: First ionisation energies across Period 3 using rounded data; the aluminium and sulfur dips show subshell and electron-pairing effects.]

The first dip is from magnesium to aluminium. Magnesium ends in $\mathrm{3s}^{2}$, but aluminium ends in $\mathrm{3p}^{1}$. A 3p electron is in a higher-energy sub-shell than a 3s electron, so it is easier to remove. That is why aluminium has a lower first ionisation energy than magnesium.

The second dip is from phosphorus to sulfur. Phosphorus has $\mathrm{3p}^{3}$, with one electron in each 3p orbital. Sulfur has $\mathrm{3p}^{4}$, so one 3p orbital contains a pair of electrons. Repulsion between paired electrons makes one of them slightly easier to remove, so sulfur has a lower first ionisation energy than phosphorus.

These two anomalies are important evidence:

• the Mg to Al dip shows that 3s and 3p are different sub-shells with different energies
• the P to S dip shows that electrons can pair within an orbital and that this affects the energy needed to remove one

Putting the whole topic together: electron configurations predict where electrons are, and ionisation-energy data tests that model. Large jumps in successive ionisation energies show shells, while the small dips across Period 3 show sub-shells and pairing effects.