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Atomic Structure and Periodicity

Atomic Structure and Periodicity

This lesson covers the detailed model of atomic structure required at A-Level, including sub-shells and orbitals, electron configurations using s, p, d, f notation, ionisation energies, periodic trends, mass spectrometry, and the determination of relative atomic mass from isotopic data. A thorough understanding of these concepts underpins the whole of A-Level Chemistry.


Sub-Shells and Orbitals

At GCSE, you learnt that electrons occupy energy levels (shells). At A-Level, each shell is divided into sub-shells, and each sub-shell contains one or more orbitals. An orbital is a region of space around the nucleus where there is a high probability of finding an electron.

Sub-shell Number of orbitals Maximum electrons
s 1 2
p 3 6
d 5 10
f 7 14

Key Definition: An orbital is a region of space around the nucleus that can hold a maximum of two electrons with opposite spins.

Each orbital can hold a maximum of two electrons with opposite spins (the Pauli exclusion principle). Electrons fill orbitals in order of increasing energy — this is known as the Aufbau principle. Within a sub-shell, electrons occupy orbitals singly before pairing up — this is Hund's rule.

The order of filling is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p ...

graph LR
    A["1s"] --> B["2s"] --> C["2p"] --> D["3s"] --> E["3p"] --> F["4s"] --> G["3d"] --> H["4p"] --> I["5s"] --> J["4d"] --> K["5p"]

Note that 4s fills before 3d because it has a lower energy in neutral atoms. However, once the 3d sub-shell is occupied, the 3d energy level drops below 4s, which is why 4s electrons are removed first when forming ions.

Shapes of Orbitals

s orbitals are spherical in shape, centred on the nucleus. As the principal quantum number increases (1s, 2s, 3s ...), the orbital gets larger.

p orbitals are dumbbell-shaped (two lobes either side of the nucleus). There are three p orbitals in each sub-shell, oriented at right angles to each other along the x, y, and z axes: px, py, and pz.

d orbitals have more complex shapes — most have four lobes arranged in a clover-leaf pattern, while one (the dz² orbital) has two lobes along the z-axis with a ring (torus) around the middle. There are five d orbitals in each sub-shell.


Electron Configuration

The full electron configuration of an atom describes the distribution of its electrons among orbitals. For example:

  • Carbon (Z = 6): 1s² 2s² 2p²
  • Iron (Z = 26): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²
  • Chromium (Z = 24): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ (an exception — a half-filled 3d sub-shell is particularly stable)
  • Copper (Z = 29): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ (another exception — a fully filled 3d sub-shell is especially stable)

When transition metals form positive ions, electrons are lost from the 4s sub-shell before the 3d sub-shell. For example, Fe²⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ (not 3d⁴ 4s²).

Exam Tip: Always remember that when forming ions, 4s electrons are removed first even though 4s fills before 3d. This is a very common exam question. The reason is that once the 3d sub-shell is occupied, 3d becomes lower in energy than 4s.

Electron-in-Boxes Diagrams

An electron-in-boxes diagram represents each orbital as a box and each electron as an arrow (upward for spin +½, downward for spin −½). For example, nitrogen (Z = 7) would be shown as:

1s: [↑↓] 2s: [↑↓] 2p: [↑] [↑] [↑]

The three 2p electrons are placed one per box before any pairing occurs (Hund's rule), and all have the same spin direction (parallel spins). For oxygen (Z = 8):

1s: [↑↓] 2s: [↑↓] 2p: [↑↓] [↑] [↑]

One 2p orbital now contains a pair of electrons with opposite spins, whilst the other two each contain a single electron. This pairing in oxygen is important when explaining the lower ionisation energy of oxygen compared to nitrogen.


Mass Spectrometry

A mass spectrometer is used to identify isotopes and determine the relative atomic mass of an element. The process involves four stages: ionisation, acceleration, deflection (or separation), and detection.

Key Definition: Relative atomic mass (Ar) is the weighted mean mass of an atom of an element compared to one-twelfth of the mass of a carbon-12 atom.

Time-of-Flight (TOF) Mass Spectrometry

In a time-of-flight mass spectrometer, the sample is ionised (typically by electrospray ionisation, where the sample is dissolved in a volatile solvent and passed through a charged needle to form ions with minimal fragmentation, or by electron impact, where a beam of high-energy electrons knocks an electron from each particle). All ions are then accelerated through an electric field so they all have the same kinetic energy. Because KE = ½mv², lighter ions travel faster than heavier ones. The ions then pass through a field-free drift region (the flight tube) and reach the detector at different times. The time of flight is measured and used to calculate the mass-to-charge ratio (m/z).

graph LR
    A["Sample<br/>Introduction"] --> B["Ionisation<br/>(Electrospray or<br/>Electron Impact)"]
    B --> C["Acceleration<br/>(Electric Field)<br/>All ions gain<br/>same KE"]
    C --> D["Flight Tube<br/>(Field-Free Drift)<br/>Lighter ions<br/>travel faster"]
    D --> E["Detection<br/>(Time of arrival<br/>measured)"]
    E --> F["Data Processing<br/>(m/z calculated<br/>from time of flight)"]

The key equation for TOF is:

KE = ½mv² and v = d/t

Therefore: m/z = 2(KE)t² / d²

where d is the length of the flight tube and t is the time of flight.

Using Mass Spectrometry to Identify Isotopes and Calculate Ar

A mass spectrum of an element shows peaks at different m/z values, each corresponding to a different isotope. The height (or area) of each peak is proportional to the relative abundance of that isotope.

Worked Example: Calculating Relative Atomic Mass

A sample of boron contains two isotopes. The mass spectrum shows peaks at m/z = 10 (relative abundance 19.9%) and m/z = 11 (relative abundance 80.1%). Calculate the relative atomic mass of boron.

Solution:

Ar = Σ (isotopic mass × relative abundance) / Σ relative abundances

Ar = (10 × 19.9 + 11 × 80.1) / (19.9 + 80.1)

Ar = (199 + 881.1) / 100

Ar = 1080.1 / 100

Ar = 10.8 (to 3 s.f.)

Worked Example: Determining Isotopic Abundances from Ar

Chlorine has a relative atomic mass of 35.5 and has two isotopes: ³⁵Cl and ³⁷Cl. Calculate the percentage abundance of each isotope.

Solution:

Let the abundance of ³⁵Cl = x %, so the abundance of ³⁷Cl = (100 − x) %.

35.5 = (35x + 37(100 − x)) / 100

3550 = 35x + 3700 − 37x

3550 = 3700 − 2x

2x = 150

x = 75

Therefore ³⁵Cl has 75% abundance and ³⁷Cl has 25% abundance.

Exam Tip: In mass spectrometry questions, always check whether the question gives abundances as percentages (summing to 100) or as relative intensities (where you must sum them yourself before dividing). This is a common source of error.


Ionisation Energy

Key Definition: The first ionisation energy is the energy required to remove one mole of the most loosely held electrons from one mole of gaseous atoms to form one mole of gaseous unipositive ions: X(g) → X⁺(g) + e⁻

Factors that affect ionisation energy:

  1. Nuclear charge — more protons means a greater attraction for electrons, increasing ionisation energy.
  2. Atomic radius (distance) — a larger distance between the nucleus and the outer electron reduces the attractive force, decreasing ionisation energy.
  3. Electron shielding — inner electrons repel outer electrons, reducing the effective nuclear charge felt by the outermost electron.
  4. Electron pairing — two electrons in the same orbital experience mutual repulsion, making it slightly easier to remove one of them.

Trends in Ionisation Energy

  • Across a period, ionisation energy generally increases because nuclear charge increases whilst shielding remains roughly constant and atomic radius decreases.
  • Down a group, ionisation energy decreases because the outermost electron is in a higher energy level, further from the nucleus, with more inner electron shielding.

There are two anomalies across Period 3:

  1. Mg → Al drop: Aluminium's outer electron is in a 3p orbital, which is higher in energy and slightly further from the nucleus than the 3s orbital of magnesium, so it is easier to remove.
  2. P → S drop: Sulfur has a paired electron in one of its 3p orbitals. The electron–electron repulsion within this pair makes it easier to remove than an unpaired electron in phosphorus.

Successive Ionisation Energies

Successive ionisation energies increase because each electron is removed from an increasingly positive ion (greater net nuclear charge per remaining electron). A large jump in ionisation energy indicates that the next electron is being removed from a shell closer to the nucleus (i.e. an inner shell with much less shielding from the nucleus). By examining the pattern of successive ionisation energies, you can determine the group of an element.

Worked Example: Determining Group from Successive Ionisation Energies

An element X has the following successive ionisation energies (in kJ mol⁻¹):

IE 1st 2nd 3rd 4th 5th 6th 7th
Value 578 1817 2745 11578 14831 18378 23293

Determine the group of element X and explain your reasoning.

Solution:

Look for the biggest jump between successive ionisation energies:

  • 1st to 2nd: 1817 − 578 = 1239 kJ mol⁻¹
  • 2nd to 3rd: 2745 − 1817 = 928 kJ mol⁻¹
  • 3rd to 4th: 11578 − 2745 = 8833 kJ mol⁻¹ ← very large jump
  • 4th to 5th: 14831 − 11578 = 3253 kJ mol⁻¹

The largest jump occurs between the 3rd and 4th ionisation energies. This means the first three electrons are relatively easy to remove (they are in the outer shell), but the fourth electron is much harder to remove (it is in the inner shell, closer to the nucleus, with less shielding).

Therefore, element X is in Group 3 (it has 3 electrons in its outer shell). The element is aluminium.

Exam Tip: When analysing successive ionisation energy data, look for where the largest percentage increase (or absolute jump) occurs. The number of electrons removed before this jump equals the group number. Remember: the jumps get larger generally, but the really big jump indicates a new inner shell.


Periodicity — Trends across Period 3

Atomic Radius

Atomic radius decreases across Period 3 (Na to Ar) because the nuclear charge increases while electrons are added to the same shell, so there is a greater pull on the outer electrons drawing them closer to the nucleus.

Melting Points across Period 3

Element Na Mg Al Si P₄ S₈ Cl₂ Ar
Structure Giant metallic Giant metallic Giant metallic Giant covalent Simple molecular Simple molecular Simple molecular Monatomic
Melting point trend Low Higher Higher Very high Low Low Very low Very low
  • Na, Mg, Al: Giant metallic structures. Melting point increases Na → Mg → Al because the number of delocalised electrons increases (1, 2, 3 respectively), the ionic radius decreases, and so the metallic bonding strengthens.
  • Si: Giant covalent structure (like diamond) with very strong covalent bonds throughout — very high melting point.
  • P₄, S₈, Cl₂, Ar: Simple molecular structures held together by weak London forces. The strength of London forces depends on molecular size: S₈ > P₄ > Cl₂ > Ar.

Electronegativity

Key Definition: Electronegativity is the ability of an atom to attract the bonding pair of electrons in a covalent bond towards itself.

Electronegativity generally increases across a period (increasing nuclear charge) and decreases down a group (increasing atomic radius and shielding). Fluorine is the most electronegative element. The Pauling scale is the most commonly used scale; on this scale, fluorine has a value of 4.0, oxygen 3.4, nitrogen 3.0, and carbon 2.6.


Summary

  • Electrons occupy orbitals within sub-shells (s, p, d, f), following the Aufbau principle, Hund's rule, and the Pauli exclusion principle.
  • Electron-in-boxes diagrams show individual orbitals with arrows representing electron spin.
  • Electron configurations can be written in full or abbreviated using noble gas core notation.
  • Time-of-flight mass spectrometry separates ions by their speed in a flight tube; lighter ions arrive first.
  • Relative atomic mass is calculated as the weighted mean from isotopic masses and abundances.
  • Ionisation energy depends on nuclear charge, atomic radius, shielding, and electron pairing.
  • Successive ionisation energies reveal the electron configuration and group number of an element.
  • Periodic trends in ionisation energy, atomic radius, melting point, and electronegativity can all be explained by considering atomic structure.