Atomic Structure — A-Level Chemistry Study Guide

For: A-Level Chemistry candidates sitting A-Level Chemistry.

Covers: Subatomic particle properties, isotopes and mass spectrometry calculations, s/p/d orbital electron configurations, ionisation energy periodic trends, and links between electronic structure and periodic table groupings.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Chemistry style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Atomic Structure?

Atomic structure is the study of the subatomic components that make up all matter, their relative properties, and their arrangement within individual atoms, forming the foundation of all chemical reactivity and macroscopic material properties. It is the first core topic in the A-Level Chemistry syllabus, tested across both AS and A Level papers, and underpins all subsequent topics including bonding, energetics, and organic chemistry. Synonyms include atomic architecture, subatomic arrangement, and electronic structure (a subcomponent of the broader topic).

2. Subatomic particles — proton, neutron, electron

All atoms are made of three core subatomic particles, whose counts and arrangement define the identity and behaviour of an element:

• Proton: Positively charged particle located in the dense central nucleus of the atom. The number of protons in an atom defines its atomic number ($Z$), a unique identifier for each element.
• Neutron: Neutral (uncharged) particle also located in the nucleus, contributing to the total mass number ($A$) of the atom.
• Electron: Negatively charged particle occupying discrete orbitals in the electron cloud surrounding the nucleus, responsible for all chemical bonding and reactivity.

The key properties of these particles are summarised below, using the standard scale where $\frac{1}{12}$ the mass of a neutral carbon-12 atom = 1:

Worked Example 1

A neutral phosphorus atom has an atomic number $Z=15$ and mass number $A=31$. State the number of protons, neutrons and electrons it contains.

• Solution: Number of protons = $Z$ = 15. Number of neutrons = $A-Z$ = $31-15=16$. Number of electrons = number of protons (for neutral atoms) = 15.
• Exam tip: For charged ions, only the electron count changes: add 1 electron for each negative charge, subtract 1 electron for each positive charge, while proton and neutron counts remain identical to the neutral atom.

3. Isotopes and mass spectrometry

Isotopes are atoms of the same element (same number of protons, same $Z$) with different numbers of neutrons (different $A$ values). They have identical chemical properties (same electron arrangement) but different physical properties (e.g. boiling point, density, mass).

A mass spectrometer is an instrument used to measure the relative atomic mass ($A_r$) of an element by separating its isotopes based on their mass-to-charge ($m/z$) ratio. The formula for calculating relative atomic mass from mass spectrometry data is: $$A_r=\frac{\sum (\text{isotopic mass}\times \text{percentage abundance})}{100}$$

Worked Example 2

A sample of naturally occurring chlorine has two isotopes: ${}^{35}\text{Cl}$ (75.8% abundance, mass 35.0) and ${}^{37}\text{Cl}$ (24.2% abundance, mass 37.0). Calculate the relative atomic mass of chlorine to 1 decimal place.

• Solution: $$A_r=\frac{(35.0\times 75.8)+(37.0\times 24.2)}{100}=\frac{2653+895.4}{100}=35.5$$
• Exam tip: If abundances are given as decimal fractions instead of percentages, omit the division by 100 in the calculation.

4. Electron configuration in $s,p,d$ orbitals

Electrons occupy discrete energy levels (shells, labelled $n=1,n=2,n=3$ etc.) around the nucleus. Each shell is split into subshells: $s$, $p$, $d$, $f$, each containing a fixed number of orbitals (regions of space where an electron is likely to be found). The Pauli Exclusion Principle states that each orbital can hold a maximum of 2 electrons with opposite spin.

• $s$ subshell: 1 orbital, max 2 electrons
• $p$ subshell: 3 degenerate (equal energy) orbitals, max 6 electrons
• $d$ subshell: 5 degenerate orbitals, max 10 electrons

Electrons fill subshells following three core rules:

1. Aufbau Principle: Fill lower energy subshells first, order of filling: $1s<2s<2p<3s<3p<4s<3d<4p$ (note $4s$ fills before $3d$ and empties first when ions form)
2. Hund's Rule: Fill degenerate orbitals singly with parallel spin before pairing electrons to minimise repulsion
3. Stability rule: Half-filled ($d^5$) and fully filled ($d^{10}$) d subshells are extra stable, leading to exceptions for chromium ($Z=24$) and copper ($Z=29$)

Worked Example 3

Write the full electron configuration of: (a) Zinc ($Z=30$) (b) $C{u}^{2+}$ ion ($Z=29$)

• Solution: (a) Zn: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^2$ (note $3d$ is written before $4s$ in final configurations for transition metals) (b) Neutral Cu has configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^1$. When forming ions, $4s$ electrons are lost first, so $C{u}^{2+}$: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^9$

5. Ionisation energies and trends

The first ionisation energy (1st IE) is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions, with units ${\text{kJ mol}}^{-1}$. For a generic element X, the equation is: $$X_{(g)}\to X_{(g)}^{+}+e^{-}\Delta H=\text{positive (endothermic)}$$

Three factors control ionisation energy:

1. Nuclear charge: Higher proton count = stronger attraction to outer electrons = higher IE
2. Atomic radius: Larger distance between nucleus and outer electrons = weaker attraction = lower IE
3. Electron shielding: Inner shell electrons block the nuclear charge from outer electrons = more shielding = lower IE

Key Trends

1. Across a period: IE generally increases, as nuclear charge rises, atomic radius shrinks, and shielding stays approximately constant. There are two consistent dips:

• Between Group 2 and 3: The highest energy electron for Group 3 elements occupies a higher energy $p$ orbital, so it is easier to remove
• Between Group 5 and 6: The highest energy electron for Group 6 elements is paired in a $p$ orbital, so repulsion makes it easier to remove

2. Down a group: IE decreases, as atomic radius and shielding increase, outweighing the effect of higher nuclear charge

Large jumps in successive ionisation energies indicate that an electron is being removed from an inner shell, so you can deduce the group number of an element from its successive IE values.

Worked Example 4

The first 4 ionisation energies of an element are: 578, 1817, 2745, 11577 ${\text{kJ mol}}^{-1}$. State which group the element is in.

• Solution: There is a large jump between the 3rd and 4th IE, meaning the 4th electron is removed from an inner shell. The element has 3 outer electrons, so it is in Group 13.

6. Electronic structure and the periodic table

The periodic table is arranged by increasing atomic number, and its structure directly maps to electron configurations:

• Blocks: Correspond to the highest energy subshell being filled:
• $s$-block: Groups 1 and 2, highest energy electron in an $s$ subshell
• $p$-block: Groups 13 to 18, highest energy electron in a $p$ subshell
• $d$-block: Transition metals, highest energy electron in a $d$ subshell
• Period number: Equals the highest occupied electron shell number
• Group number (for s and p block): Equals the number of outer shell electrons (e.g. Group 17 elements have 7 outer electrons: 2 in $s$, 5 in $p$)

Exam tip: When asked to justify the position of an element in the periodic table, always reference its electron configuration, not just chemical properties, to get full marks.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Writing $4s$ before $3d$ in transition metal electron configurations, or keeping $4s$ electrons when writing transition metal ion configurations. Why: Students memorise the filling order but forget that $4s$ becomes higher energy than $3d$ once $3d$ starts filling. Correct move: Write $3d$ before $4s$ in final transition metal configurations, and remove $4s$ electrons first when forming positive ions.
• Wrong move: Stating that isotopes have different chemical properties. Why: Students confuse physical and chemical properties, linking mass to reactivity. Correct move: Isotopes have identical electron arrangements, so identical chemical properties; only physical properties (mass, density, boiling point) differ.
• Wrong move: Calculating relative atomic mass without dividing by 100 when using percentage abundances. Why: Rushing through calculations and forgetting the formula definition. Correct move: Double check if abundances are given as percentages or fractions; only divide by 100 for percentages.
• Wrong move: Claiming ionisation energy increases uniformly across a period, ignoring the dips at Group 3 and Group 6. Why: Students memorise the general trend but forget the exceptions caused by subshell energy levels and electron repulsion. Correct move: Always mention the two dips when explaining period trends, and link them to the relevant subshell structure.
• Wrong move: Applying s/p block group number rules to d-block elements. Why: Students assume all groups follow the same outer electron count rule. Correct move: For A-Level Chemistry, d-block elements are not assigned standard group numbers; state their block based on the highest energy subshell being filled.

8. Practice Questions (A-Level Chemistry Style)

Question 1

(a) Define the term relative atomic mass. [2 marks] (b) A sample of neon has two isotopes: ${}^{20}\text{Ne}$ and ${}^{22}\text{Ne}$, with a relative atomic mass of 20.2. Calculate the percentage abundance of each isotope. [3 marks]

Solution

(a) The weighted average mass of one atom of an element relative to $\frac{1}{12}$ the mass of one atom of carbon-12. (2 marks: 1 for "weighted average", 1 for reference to 1/12 C-12) (b) Let abundance of ${}^{20}\text{Ne}=x\%$, abundance of ${}^{22}\text{Ne}=(100-x)\%$ $$20.2=\frac{20x+22(100-x)}{100}$$ $$2020=20x+2200-22x$$ $$2x=180\to x=90$$ Abundance of ${}^{20}\text{Ne}=90\%$, ${}^{22}\text{Ne}=10\%$. (3 marks: 1 for correct equation setup, 1 for rearrangement, 1 for both correct abundances)

Question 2

(a) Write the full electron configuration of a $C{r}^{3+}$ ion ($Z=24$). [2 marks] (b) State which block of the periodic table chromium belongs to, and explain your answer. [2 marks]

Solution

(a) Neutral Cr has configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^1$ (exception for half-filled d subshell stability). Removing 3 electrons (first 1 from 4s, then 2 from 3d) gives $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^3$. (2 marks: 1 for correct electron count, 1 for no 4s electrons and correct 3d count) (b) Chromium is in the d-block, as its highest energy electron occupies a d orbital. (2 marks: 1 for d-block, 1 for correct explanation referencing highest energy subshell)

Question 3

Explain why the first ionisation energy of oxygen (Group 16, Period 2) is lower than that of nitrogen (Group 15, Period 2). [3 marks]

Solution

Nitrogen has an electron configuration of $1s^{2}2s^{2}2p^3$, with three unpaired electrons in separate p orbitals. Oxygen has a configuration of $1s^{2}2s^{2}2p^4$, with one paired electron in a p orbital. The repulsion between the paired electrons in oxygen's 2p subshell means less energy is required to remove one of these electrons, even though oxygen has a higher nuclear charge than nitrogen. (3 marks: 1 for reference to paired p electron in oxygen, 1 for mention of repulsion, 1 for linking to lower energy required for removal)

9. Quick Reference Cheatsheet

10. What's Next

Atomic structure is the foundational topic for all A-Level Chemistry content moving forward. A solid understanding of electron configurations will directly support your study of chemical bonding (ionic, covalent and metallic), as bonding relies on the transfer or sharing of outer shell electrons. Ionisation energy trends link directly to periodicity, Group 2 and Group 17 chemistry, and redox reactivity, while mass spectrometry is used across organic chemistry to identify unknown compounds and determine molecular formulas.

If you have questions about any of the content in this guide, or want to practise more exam-style questions tailored to your weak areas, reach out to Ollie, our AI tutor, for personalized support. You can also find more study guides, flashcards and past paper walkthroughs for A-Level Chemistry on the homepage to help you prepare for your exams.

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