Describe the structure of the atom in terms of protons, neutrons and electrons, deduce electronic configurations using s, p and d subshells, and explain successive and periodic ionisation energy trends in terms of nuclear charge, shielding and subshell energies

What this dot point is asking

SEAB wants you to describe the atom in terms of its three subatomic particles, write electronic configurations using the $s$, $p$ and $d$ subshell notation (including the well-known anomalies), and use ionisation energy data as quantitative evidence for the shell and subshell model. This is the foundation topic of Physical Chemistry, and the ionisation-energy reasoning appears every year.

The answer

Subatomic particles

An atom has a tiny, dense nucleus of protons and neutrons, surrounded by electrons in shells.

The proton (atomic) number $Z$ defines the element. The nucleon (mass) number $A$ is protons plus neutrons. Isotopes are atoms of the same element with different numbers of neutrons.

Subshells and the filling order

Electrons occupy energy levels (shells, principal quantum number $n$) divided into subshells: $s$ (holds 2), $p$ (holds 6), $d$ (holds 10), $f$ (holds 14). Subshells fill in order of increasing energy:

Note that $4s$ fills before $3d$ because the $4s$ subshell is at slightly lower energy when empty. Two rules complete the picture:

• Aufbau principle: electrons fill the lowest-energy subshell first.
• Hund's rule: within a subshell, electrons singly occupy each orbital with parallel spins before pairing.

Writing configurations

Iron ($Z = 26$): $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^6\,4s^2$, often written $[\text{Ar}]\,3d^6\,4s^2$.

Two anomalies you must know:

• Chromium ($Z = 24$): $[\text{Ar}]\,3d^5\,4s^1$, not $3d^4\,4s^2$. A half-filled $3d^5$ with a single $4s$ electron is more stable.
• Copper ($Z = 29$): $[\text{Ar}]\,3d^{10}\,4s^1$, not $3d^9\,4s^2$. A full $3d^{10}$ is favoured.

When transition elements ionise, the $4s$ electrons are lost first: $\text{Fe}^{2+}$ is $[\text{Ar}]\,3d^6$, not $[\text{Ar}]\,3d^4\,4s^2$.

Ionisation energy defined

The first ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms:

It depends on three factors: nuclear charge (more charge means harder to remove), distance of the electron from the nucleus (further means easier), and shielding by inner electrons (more shielding means easier).

Successive ionisation energies as evidence for shells

Successive ionisation energies always increase, because each electron is pulled from an increasingly positive ion. A large jump appears whenever the next electron must come from a shell closer to the nucleus. Counting electrons removed before the big jump gives the number of valence electrons, hence the group.

First ionisation energy across a period

Across Period 3, first ionisation energy rises overall (nuclear charge increases, same shell), but there are two dips:

• Al below Mg: the electron removed from Al is a higher-energy, shielded $3p$ electron, versus a $3s$ electron in Mg.
• S below P: in sulfur, the fourth $3p$ electron pairs in an orbital, and inter-electron repulsion makes it easier to remove than the half-filled $3p^3$ arrangement in phosphorus.

Examples in context

Example 1. Identifying an unknown from ionisation data. A Junior College student is given the first five ionisation energies of element Z as 496, 4562, 6912, 9544, 13352 kJ per mole. The huge jump after the 1st ionisation energy (a factor of about 9) shows just one easily-removed electron, so Z is in Group 1. The configuration $1s^2\,2s^2\,2p^6\,3s^1$ identifies Z as sodium. This kind of reasoning is the standard way SEAB tests the shell model without naming the element.

Example 2. Predicting an ion's configuration. For a structured Paper 2 question on transition chemistry, $\text{Cu}^{2+}$ is needed. Copper is $[\text{Ar}]\,3d^{10}\,4s^1$. Remove the $4s$ electron first, then one $3d$ electron, giving $\text{Cu}^{2+} = [\text{Ar}]\,3d^9$. The single unpaired $d$ electron is consistent with the colour and paramagnetism of copper(II) complexes met later in Inorganic Chemistry.

Try this

Q1. Write the full electronic configuration of (a) sulfur and (b) the $\text{Fe}^{3+}$ ion. [2 marks]

• Cue. (a) $1s^2\,2s^2\,2p^6\,3s^2\,3p^4$. (b) $[\text{Ar}]\,3d^5$ (lose two $4s$ then one $3d$).

Q2. Explain why the first ionisation energy of sulfur is lower than that of phosphorus. [3 marks]

• Cue. In S the fourth $3p$ electron is paired; inter-electron repulsion in the paired orbital makes it easier to remove than from the stable half-filled $3p^3$ of P.

Q3. The successive ionisation energies of element Q (kJ per mole) are 578, 1817, 2745, 11577, 14842. (a) Deduce the group of Q. (b) Write its electronic configuration. [2+2 marks]

• Cue. (a) Largest jump after the 3rd electron, so Group 13. (b) Q is aluminium, $1s^2\,2s^2\,2p^6\,3s^2\,3p^1$.