Atomic Structure — IB Chemistry HL HL Study Guide

For: IB Chemistry HL candidates sitting IB Chemistry HL.

Covers: All core and HL content for Atomic Structure, including subatomic particles and isotopes, electron configuration and orbitals, ionisation energy trends, mass spectrometry interpretation, and emission spectra with the Bohr model.

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 IB Chemistry HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Atomic Structure?

Atomic structure is the study of the composition, arrangement, and behaviour of subatomic particles within atoms, the fundamental building blocks of all matter. It forms the conceptual foundation of every other topic in IB Chemistry HL, from bonding to organic reactivity, and is assessed directly in both Paper 1 (multiple choice) and Paper 2 (structured response), with 5-8 marks allocated to this topic per exam session. Examiners frequently combine atomic structure concepts with periodicity or organic analysis questions to test cross-topic application, so a strong grasp of this content is non-negotiable for high scores.

2. Subatomic particles, isotopes

All atoms are made of three core subatomic particles, whose properties are summarised below:

Elements are defined by their atomic number ($Z$), the number of protons in the nucleus, while the mass number ($A$) is the sum of protons and neutrons in the nucleus. Standard A/Z notation uses the format ${}_Z^{A}E$, where $E$ is the element symbol.

Isotopes are atoms of the same element (identical $Z$) with different mass numbers, meaning they have different numbers of neutrons. Isotopes have identical chemical properties, as their electron configuration is the same, but different physical properties (e.g. boiling point, diffusion rate) due to their different masses.

Worked Example: Calculate the number of protons, neutrons, and electrons in a ${}_8^{18}{O}^{2-}$ ion.

• Protons = $Z=8$
• Neutrons = $A-Z=18-8=10$
• Electrons = protons + 2 (for 2- charge) = 10

Exam tip: Examiners regularly test that you do not confuse mass number (whole number, for a single isotope) with relative atomic mass (weighted average of all isotope masses for an element).

3. Electron configuration and orbitals

While the simple Bohr shell model is sufficient for basic explanations, IB HL requires you to use the quantum mechanical model of the atom, where electrons occupy orbitals: regions of space where there is a 90% probability of finding an electron. Orbitals are grouped into subshells with distinct energy levels:

• s subshell: 1 spherical orbital, holds max 2 electrons
• p subshell: 3 dumbbell-shaped degenerate (equal energy) orbitals, holds max 6 electrons
• d subshell: 5 degenerate orbitals, holds max 10 electrons
• f subshell: 7 degenerate orbitals, holds max 14 electrons

Electrons fill orbitals following three rules:

1. Aufbau principle: Fill lowest energy orbitals first, following the order $1s<2s<2p<3s<3p<4s<3d<4p<5s<4d$ (note the 4s orbital fills before 3d, and empties before 3d when ions form)
2. Pauli exclusion principle: Each orbital holds a maximum of 2 electrons with opposite spin
3. Hund’s rule: Fill degenerate orbitals singly with parallel spins before pairing electrons to minimise repulsion

Worked Example: Write the full electron configuration for a Co³⁺ ion ($Z=27$)

• Neutral Co: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^7$
• Remove 4s electrons first for ion formation: Co³⁺ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^6$

Exam tip: Transition metal ion configurations are a very common HL question, and marks are always deducted if you leave 4s electrons in the ion configuration.

4. Ionisation energies trends

The first ionisation energy ($I{E}_1$) is the minimum energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous +1 ions, as shown in the equation: $$X(g)\to X^{+}(g)+e^{-}\Delta H=I{E}_1>0$$ All ionisation energies are endothermic, as energy is required to overcome the attraction between the negative electron and positive nucleus.

Key trends for first ionisation energy:

1. Across a period (e.g. Period 3): General increase, as nuclear charge increases and atomic radius decreases, strengthening the attraction between the nucleus and outer electron. Two HL-required exceptions:

• $I{E}_1(Mg)>I{E}_1(Al)$: Al’s outer electron is in the higher-energy 3p orbital, so less energy is required to remove it than Mg’s outer 3s electron
• $I{E}_1(P)>I{E}_1(S)$: S has a paired electron in its 3p subshell, and repulsion between the paired electrons makes it easier to remove one electron than P’s unpaired 3p electrons

2. Down a group: Decrease, as atomic radius increases and inner electron shells create a shielding effect that reduces the attraction between the nucleus and outer electron.

Successive ionisation energies show a large jump when an electron is removed from an inner, lower-energy shell, so the position of the jump tells you the group number of the element.

Worked Example: The successive ionisation energies of an element are 590, 1145, 4912, 6491 kJ mol⁻¹. What group is it in?

• Large jump between 2nd and 3rd IE, meaning the 3rd electron is removed from an inner shell. The element has 2 valence electrons, so it is in Group 2.

5. Mass spectrometry interpretation

Mass spectrometry is an analytical technique used to measure relative atomic mass, relative molecular mass, and the relative abundance of isotopes. The output is a mass spectrum, where the x-axis shows the mass/charge ($m/z$) ratio (equal to relative mass for +1 ions) and the y-axis shows relative abundance (percentage).

The relative atomic mass ($A_r$) of an element is calculated as a weighted average of its isotope masses using the formula: $$A_r=\frac{\sum (\text{isotope mass}\times \text{percentage abundance})}{100}$$

Worked Example: Bromine has two stable isotopes: ⁷⁹Br (50.7% abundance) and ⁸¹Br (49.3% abundance). Calculate its relative atomic mass to 1 decimal place. $$A_r=\frac{(79\times 50.7)+(81\times 49.3)}{100}=\frac{4005.3+3993.3}{100}=79.986=80.0$$

Exam tip: HL questions often ask you to interpret mass spectra of diatomic molecules, e.g. Br₂ will have peaks at 158 (⁷⁹Br⁷⁹Br), 160 (⁷⁹Br⁸¹Br), and 162 (⁸¹Br⁸¹Br) with a roughly 1:2:1 abundance ratio.

6. Emission spectra and Bohr model

When atoms are supplied with energy, electrons jump from lower to higher energy levels. When they fall back to lower levels, they emit photons of specific energy equal to the difference between the two levels: $$\Delta E=hf=\frac{hc}{\lambda }$$ Where $h$ = Planck’s constant ($6.626\times 10^{-34}$ J s), $c$ = speed of light ($3\times 10^8$ m s⁻¹), $f$ = frequency of radiation, and $\lambda$ = wavelength of radiation.

The hydrogen emission spectrum has discrete lines (not a continuous spectrum) because electron energy levels are quantised. The lines are grouped into series based on the final energy level of the falling electron:

• Lyman series: Transitions end at $n=1$, emit ultraviolet radiation
• Balmer series: Transitions end at $n=2$, emit visible radiation
• Paschen series: Transitions end at $n=3$, emit infrared radiation

The Bohr model accurately predicts the hydrogen emission spectrum using the assumption that electrons orbit the nucleus in fixed, quantised energy levels, but it has key limitations: it only works for single-electron species, cannot explain line splitting in magnetic fields, and does not account for the wave-particle duality of electrons.

Worked Example: A line in the Balmer series has a wavelength of 486 nm (blue). Calculate the energy difference between the two levels involved. $$\Delta E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{486\times 10^{-9}}=4.09\times 10^{-19}\text{ J per atom}$$ Multiply by Avogadro’s constant to get 246 kJ mol⁻¹ for 1 mole of transitions.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Writing transition metal ion configurations with 4s electrons retained after 3d electrons. Why: Students remember 4s fills first, but forget it is higher in energy than 3d once the 3d subshell starts filling. Correct move: Always remove 4s electrons before 3d electrons when writing transition metal ion configurations.
• Wrong move: Stating that first ionisation energy increases continuously across a period with no exceptions. Why: Students memorise the general trend but ignore subshell energy differences and electron repulsion effects. Correct move: Memorise the two Period 3 exceptions (Mg > Al, P > S) and be prepared to explain them using subshell structure.
• Wrong move: Using the mass of the most abundant isotope as the relative atomic mass of an element. Why: Students confuse the mass number of a single isotope with the weighted average for the element. Correct move: Always use the weighted sum formula to calculate relative atomic mass from mass spectrometry data.
• Wrong move: Omitting gaseous state symbols in ionisation energy equations. Why: Students forget that ionisation energies are defined for gaseous species only. Correct move: Always include (g) state symbols for all reactants and products in ionisation energy equations, as examiners deduct marks for missing state symbols here.
• Wrong move: Assigning visible emission lines to the Lyman series. Why: Students mix up the series names and their associated regions of the electromagnetic spectrum. Correct move: Remember the mnemonic: Lyman = UV = Lowest final level ($n=1$), Balmer = Visible = $n=2$, Paschen = IR = $n=3$.

8. Practice Questions (IB Chemistry HL Style)

Question 1

(a) Write the full electron configuration for a Cr²⁺ ion ($Z=24$). Note that neutral Cr has an exception electron configuration with a half-full 3d subshell for extra stability. (2 marks) (b) Explain why the first ionisation energy of aluminium is lower than that of magnesium, even though aluminium has a higher nuclear charge. (3 marks)

Solution 1

(a) Neutral Cr: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^5$ (half-full 3d subshell is more stable). Remove 4s electron first for ion formation: Cr²⁺ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^4$. 1 mark for removing 4s electrons first, 1 mark for correct full configuration. (b) Magnesium has an outer electron configuration of $3s^2$, with its highest energy electron in the lower-energy 3s subshell. Aluminium has an outer electron configuration of $3s^{2}3p^1$, with its highest energy electron in the higher-energy 3p subshell, which is shielded from the nucleus by the 3s electrons. Less energy is required to remove the higher-energy 3p electron from Al, outweighing the small increase in nuclear charge between Mg and Al. 1 mark for identifying the difference in outer subshell, 1 mark for noting 3p is higher energy than 3s, 1 mark for linking to lower ionisation energy.

Question 2

The mass spectrum of zirconium shows 5 peaks with the following data: m/z 90 (51.5% abundance), 91 (11.2%), 92 (17.1%), 94 (17.4%), 96 (2.8%). Calculate the relative atomic mass of zirconium to 1 decimal place. (2 marks)

Solution 2

$$A_r=\frac{(90\times 51.5)+(91\times 11.2)+(92\times 17.1)+(94\times 17.4)+(96\times 2.8)}{100}$$ $$=\frac{4635+1019.2+1573.2+1635.6+268.8}{100}=\frac{9131.8}{100}=91.3$$ 1 mark for correct substitution into the weighted average formula, 1 mark for correct final answer to 1 decimal place.

Question 3

A transition in the hydrogen emission spectrum releases 183 kJ mol⁻¹ of energy. (a) Calculate the wavelength of the emitted radiation in nanometres. (2 marks) (b) State which series this transition belongs to, and justify your answer. (2 marks)

Solution 3

(a) First convert energy per mole to energy per atom: $\frac{183000{\text{ J mol}}^{-1}}{6.022\times 10^{23}{\text{ mol}}^{-1}}=3.04\times 10^{-19}\text{ J per atom}$. Rearrange $\Delta E=hc/\lambda$: $$\lambda =\frac{6.626\times 10^{-34}\times 3\times 10^8}{3.04\times 10^{-19}}=6.54\times 10^{-7}\text{ m}=654\text{ nm}$$ 1 mark for correct conversion to per-atom energy and calculation, 1 mark for correct conversion to nanometres. (b) 654 nm falls in the visible region of the electromagnetic spectrum, so it belongs to the Balmer series, where all transitions end at the $n=2$ energy level. 1 mark for identifying the Balmer series, 1 mark for linking to visible radiation.

9. Quick Reference Cheatsheet

10. What's Next

Atomic structure is the foundational topic for all subsequent IB Chemistry HL content. A strong grasp of electron configuration directly informs your understanding of chemical bonding (Topic 4), periodicity (Topic 3), and transition metal chemistry (Topic 13), while mass spectrometry is used extensively in organic chemistry (Topics 10 and 20) to identify unknown compounds. Ionisation energy trends also link directly to reactivity trends across groups and periods, which you will apply when studying group 1, 7, and p-block element chemistry. Even if you only plan to revise atomic structure for short-answer questions, mastering this content will save you significant time when learning more complex later topics.

If you have any questions about specific concepts, worked examples, or exam technique for this topic, you can ask Ollie for personalised explanations and extra practice questions at any time on the homepage. For your next study session, we recommend moving to Periodicity (Topic 3) to apply the foundational knowledge of atomic structure you have just revised.