Atomic Structure Revision Notes
Subject: Chemistry | Level: A-Level | Exam Board: WJEC
Master the core of A-Level Chemistry with our guide to Atomic Structure. We'll break down quantum orbitals, ionisation energy trends, and mass spectrometry to help you secure top marks in your WJEC exams.
Revision Notes & Key Concepts
Revision Podcast Transcript
Welcome to the A-Level Chemistry Study Podcast. I'm your host, and today we're diving deep into one of the most fundamental — and most examined — topics in WJEC A-Level Chemistry: Atomic Structure, Topic 1.1. Whether you're just starting out or doing a final revision sweep, this episode is going to give you everything you need to walk into that exam with confidence. So grab a pen, maybe a cup of tea, and let's get into it. By the end of this episode, you'll be able to define relative atomic mass precisely enough to earn that mark, write electronic configurations for transition metals without falling into the classic traps, explain ionisation energy trends using the three key factors, calculate relative atomic masses from mass spectrometry data, and interpret emission spectra to find ionisation energy. That's a lot — but we'll take it step by step. Let's start with the big picture. An atom is made up of a tiny, dense nucleus containing protons and neutrons, surrounded by electrons arranged in shells and subshells. Now, at GCSE, you probably thought of electrons as sitting in neat circular orbits. At A-Level, we upgrade that model significantly. We use the quantum mechanical model, which tells us that electrons don't travel in fixed paths — instead, they exist in regions of space called orbitals, where there's a high probability of finding them. There are four types of orbital you need to know: s, p, d, and f. The s-orbital is spherical — just a ball of electron density around the nucleus. Each shell has one s-orbital, and it holds a maximum of two electrons. The p-orbitals are dumbbell-shaped — like two lobes pointing in opposite directions. There are three p-orbitals in each p-subshell, oriented along the x, y, and z axes. Each holds two electrons, so the p-subshell holds six electrons in total. Then we have the d-orbitals — five of them, with more complex shapes, holding up to ten electrons. And finally f-orbitals — seven of them, holding up to fourteen electrons. The order in which these fill up is critical. You need to know the Aufbau principle: electrons fill the lowest energy orbitals first. The order is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, and so on. Notice something interesting there — 4s fills before 3d. This trips up so many candidates. The 4s orbital is slightly lower in energy than the 3d, so it fills first. Here's a memory trick: think of it as "4 before 3 in the d-block". When you're writing configurations for elements in the first transition series — that's scandium through zinc — you write 4s first, then 3d. Let's also talk about Hund's rule. When electrons are filling orbitals of equal energy — called degenerate orbitals — they fill each orbital singly before any pairing occurs. Think of it like seats on a bus: everyone takes their own seat before anyone has to share. This minimises electron-electron repulsion and gives the lowest energy arrangement. Now, here are the two most important exceptions you absolutely must memorise for WJEC: Chromium and Copper. Chromium, atomic number 24, has the configuration: argon, 3d to the power of 5, 4s to the power of 1. NOT 3d4, 4s2 as you might expect. Why? Because a half-filled d-subshell — all five d-orbitals each containing one electron — is particularly stable due to the symmetrical distribution of electron density and minimised repulsion. Copper, atomic number 29, has the configuration: argon, 3d to the power of 10, 4s to the power of 1. NOT 3d9, 4s2. A fully filled d-subshell is also especially stable. Remember: Chromium and Copper both "steal" one electron from the 4s to achieve these stable arrangements. Now, what about ions? This is another classic exam trap. When transition metals form ions, they lose the 4s electrons first — not the 3d. So iron, which is argon 3d6 4s2, becomes Fe2+ by losing both 4s electrons: argon 3d6. And Fe3+ loses one more 3d electron: argon 3d5. Many candidates incorrectly remove 3d electrons first. Don't be one of them. Let's move on to relative atomic mass and mass spectrometry. This is a calculation topic that WJEC loves to test, and the good news is it's very mechanical once you know the formula. The relative atomic mass, Ar, is defined as: the weighted mean mass of an atom of an element relative to one-twelfth of the mass of an atom of carbon-12. Every single word in that definition can earn you a mark, so learn it precisely. The key phrases are "weighted mean mass", "relative to", and "one-twelfth of the mass of carbon-12". Omitting any of these is a common error. In a mass spectrometer, a sample is vaporised, ionised, accelerated, deflected by a magnetic field, and detected. The output is a mass spectrum — a graph of relative abundance against mass-to-charge ratio, or m/z. For singly-charged ions, m/z equals the relative isotopic mass. To calculate relative atomic mass from a mass spectrum, you use this formula: Ar equals the sum of (isotopic mass multiplied by percentage abundance) divided by 100. Or if you're given actual abundances rather than percentages, divide by the total abundance instead. Let's do a quick example. Chlorine has two isotopes: chlorine-35 with 75% abundance, and chlorine-37 with 25% abundance. Ar equals (35 times 75 plus 37 times 25) divided by 100. That's (2625 plus 925) divided by 100, which equals 3550 divided by 100, which equals 35.5. Always show full working — WJEC often awards marks for the method even if you make an arithmetic error. Now let's talk about ionisation energies — this is where atomic structure gets really beautiful, because the patterns in ionisation energy data tell us so much about electronic structure. The first ionisation energy is defined as the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of singly-charged gaseous cations. The equation is: X gas gives X-plus gas plus one electron. The key words here are "gaseous" — the atoms must be in the gas phase — and "one mole". Three factors determine the magnitude of ionisation energy: nuclear charge, atomic radius, and electron shielding. Nuclear charge: more protons means a stronger pull on the outer electrons, so higher ionisation energy. Atomic radius: the further the electron is from the nucleus, the weaker the attraction, so lower ionisation energy. Shielding: inner-shell electrons repel outer electrons and reduce the effective nuclear charge felt by the outer electron. More shielding means lower ionisation energy. Now, here's a critical point about shielding that candidates frequently get wrong: shielding remains approximately constant across a period. The electrons being added across a period go into the same shell, and electrons in the same shell don't shield each other effectively. So across Period 3, the shielding stays roughly the same, but nuclear charge increases — hence the general increase in ionisation energy across a period. But there are two dips in Period 3 that you must be able to explain. The first dip is from Magnesium to Aluminium. Magnesium has configuration 3s2 — its outer electron is in a 3s orbital. Aluminium has configuration 3s2 3p1 — its outer electron is in a 3p orbital. The 3p orbital is higher in energy and further from the nucleus than the 3s, so it's easier to remove. Hence aluminium has a lower first ionisation energy than magnesium, despite having a higher nuclear charge. The second dip is from Phosphorus to Sulphur. Phosphorus has configuration 3p3 — three electrons, one in each p-orbital, all unpaired. Sulphur has configuration 3p4 — four electrons, so one p-orbital must contain a pair. That paired electron experiences extra repulsion from its partner in the same orbital, making it easier to remove. Hence sulphur has a lower first ionisation energy than phosphorus. These two explanations are worth practising until you can write them fluently. WJEC mark schemes are very specific: for the Mg-Al dip, you must mention the higher energy of the p-orbital compared to the s-orbital. For the P-S dip, you must mention electron-pair repulsion within the same orbital. Now let's look at successive ionisation energies — the energies required to remove the first, second, third electron, and so on from the same atom. As you remove more electrons, each successive ionisation energy is generally larger, because you're removing from an increasingly positive ion. But there are sharp jumps when you move from one shell to the next — removing an electron from a shell closer to the nucleus requires much more energy. These jumps allow us to deduce the group of an element. For example, if successive ionisation energies show a large jump between the second and third ionisation energies, the element is in Group 2 — it has two outer-shell electrons that are relatively easy to remove, then a third electron in a shell much closer to the nucleus. Finally, let's connect emission spectra to ionisation energy. When electrons in atoms absorb energy, they jump to higher energy levels. When they fall back down, they release that energy as photons of light. The emission spectrum of hydrogen shows distinct coloured lines — each line corresponds to electrons falling to a particular energy level. The Balmer series, which falls in the visible region, corresponds to transitions down to n equals 2. Here's the key exam point: as you look at the emission spectrum from low frequency to high frequency, the lines get closer and closer together. This convergence happens because the energy levels get closer together as n increases. At the convergence limit — the point where the lines merge — the electron has just enough energy to escape the atom entirely. This convergence limit corresponds to the ionisation energy of hydrogen. WJEC specifically tests whether candidates can identify the convergence limit and state what it represents. One mark is awarded for identifying it as the point where the electron leaves the atom, or where ionisation occurs. Now let's go through the top exam tips and common mistakes. I want you to really listen to these, because they come up year after year. Number one: when defining relative atomic mass, always include "one-twelfth of the mass of carbon-12". Many candidates write "relative to carbon-12" and lose the mark. The reference point is specifically one-twelfth of a carbon-12 atom. Number two: Copper's configuration. It is argon, 3d10, 4s1. Not 3d9, 4s2. Write it out on a flashcard tonight and test yourself tomorrow. Number three: when writing configurations for transition metal ions, remove 4s electrons first. Fe2+ is argon 3d6. Not argon 3d4 4s2. The 4s electrons are always lost first during ionisation. Number four: shielding does NOT increase across a period. It stays approximately constant. If you write "shielding increases across Period 3" in an exam, you will lose marks. The increase in ionisation energy across a period is due to increasing nuclear charge with approximately constant shielding. Number five: for the Mg-Al ionisation energy dip, you must explicitly say the 3p orbital is higher in energy than the 3s orbital. Just saying "aluminium has a p-electron" is not enough. Number six: for mass spectrometry calculations, always show full working. Write out the formula, substitute the values, and show the arithmetic. Error carried forward is often available if your method is clear. Number seven: orbital sketches. If asked to sketch an s-orbital, draw a circle or sphere. For a p-orbital, draw two lobes — a dumbbell shape — and label the axis. WJEC may ask you to show the nodal plane, which is the plane where the probability of finding an electron is zero, passing through the nucleus between the two lobes. Now, let's do a quick-fire recall quiz. I'll ask a question, give you a few seconds to think, then give the answer. Ready? Question one: What is the electronic configuration of Copper? Think... The answer is: argon, 3d10, 4s1. Question two: What are the three factors that affect ionisation energy? Think... Nuclear charge, atomic radius or distance from nucleus, and electron shielding. Question three: Why is the first ionisation energy of aluminium lower than that of magnesium? Think... Because aluminium's outer electron is in a 3p orbital, which is higher in energy and further from the nucleus than magnesium's 3s orbital. Question four: What does the convergence limit in an emission spectrum represent? Think... It represents the ionisation energy — the point at which the electron has enough energy to escape the atom. Question five: When calculating relative atomic mass, what is the formula? Think... Ar equals the sum of (isotopic mass times percentage abundance) divided by 100. How did you do? If you got all five, brilliant — you're in great shape. If you missed any, go back and re-read that section of your notes. Let's wrap up with a quick summary of the key points from today. First: electrons occupy orbitals — s, p, d, f — filling in order of increasing energy following the Aufbau principle, with Hund's rule applying to degenerate orbitals. Second: Chromium is argon 3d5 4s1, and Copper is argon 3d10 4s1. These are the two exceptions you must memorise. Third: when forming transition metal ions, 4s electrons are always removed before 3d electrons. Fourth: relative atomic mass is the weighted mean mass of an atom relative to one-twelfth of the mass of a carbon-12 atom. Calculate it as the sum of (isotopic mass times abundance) divided by total abundance. Fifth: ionisation energy is affected by nuclear charge, atomic radius, and shielding. Shielding stays approximately constant across a period. Sixth: the Mg-Al dip in ionisation energy is due to the higher energy of the 3p orbital. The P-S dip is due to electron-pair repulsion in a doubly-occupied 3p orbital. Seventh: the convergence limit in an emission spectrum corresponds to the ionisation energy. That's everything for today's episode on Atomic Structure. You've covered the quantum mechanical model, electronic configurations including the tricky exceptions, mass spectrometry calculations, ionisation energy trends, and emission spectra. That is a substantial chunk of Unit 1, and you should feel really good about what you've just revised. Keep testing yourself — use flashcards for the definitions, practise writing configurations from memory, and work through past paper questions on ionisation energy calculations. The more you practise retrieving this information, the more firmly it will stick. Good luck with your revision, and I'll see you in the next episode. You've got this!
Key Terms & Definitions
- Relative Atomic Mass (Ar)
- The weighted mean mass of an atom of an element relative to 1/12th of the mass of an atom of carbon-12.
- First Ionisation Energy
- The energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.
- Orbital
- A region within an atom that can hold up to two electrons with opposite spins.
- Isotopes
- Atoms of the same element (same number of protons) with different numbers of neutrons.
- Shielding
- The repulsion between electrons in inner shells and electrons in the outer shell, which reduces the effective nuclear charge experienced by the outer electrons.
- Aufbau Principle
- The rule that electrons fill atomic orbitals from the lowest energy level upwards.
Worked Examples
Worked Example
Question: The mass spectrum of a sample of zirconium shows five isotopes. Use the data below to calculate the relative atomic mass of zirconium. Give your answer to one decimal place. (4 marks) | m/z | 90 | 91 | 92 | 94 | 96 | |-----------|-------|-------|-------|-------|-------| | Abundance | 51.5 | 11.2 | 17.1 | 17.4 | 2.8 |
Solution: Step 1: State the formula for relative atomic mass. `Ar = (Σ (isotope mass × abundance)) / (Σ abundance)` Step 2: Calculate the sum of (mass × abundance). `(90 × 51.5) + (91 × 11.2) + (92 × 17.1) + (94 × 17.4) + (96 × 2.8)` `= 4635 + 1019.2 + 1573.2 + 1635.6 + 268.8 = 9131.8` Step 3: Calculate the sum of the abundances. `51.5 + 11.2 + 17.1 + 17.4 + 2.8 = 100` Step 4: Calculate Ar and give the answer to one decimal place. `Ar = 9131.8 / 100 = 91.318` Final answer: **91.3**
Worked Example
Question: Explain the trend in first ionisation energies for the Period 3 elements from Sodium to Argon. Your answer should refer to the factors that influence ionisation energy and explain the drops between Group 2 and 3, and Group 5 and 6. (6 marks)
Solution: Step 1: Describe the general trend. Across Period 3 from Na to Ar, the first ionisation energy generally increases. Step 2: Explain the general trend using the three key factors. This is because the nuclear charge increases as more protons are added to the nucleus. The electrons are added to the same principal energy level (n=3), so the shielding from inner shells remains approximately constant. The increasing nuclear attraction pulls the electron shells closer, decreasing the atomic radius. The stronger effective nuclear charge requires more energy to remove the outermost electron. Step 3: Explain the drop between Group 2 (Mg) and Group 3 (Al). There is a drop in ionisation energy from Mg to Al. The outer electron in Al (1s²2s²2p⁶3s²3p¹) is in a 3p orbital, whereas in Mg (1s²2s²2p⁶3s²) it is in a 3s orbital. The 3p orbital is at a higher energy level and is further from the nucleus than the 3s orbital. This outweighs the increase in nuclear charge, so less energy is needed to remove the electron. Step 4: Explain the drop between Group 5 (P) and Group 6 (S). There is a drop in ionisation energy from P to S. The outer electron configuration of P is 3p³, with one electron in each p-orbital. The configuration of S is 3p⁴, which has one doubly-occupied p-orbital. The paired electrons in this orbital experience mutual repulsion, making it easier to remove one of them. This repulsion effect requires less energy to overcome than the stability of the half-filled subshell in phosphorus.
Worked Example
Question: Write the full electronic configurations for a copper atom (Cu) and a Fe³⁺ ion. (3 marks)
Solution: Step 1: Determine the number of electrons for the copper atom (Z=29). Cu atom has 29 electrons. Step 2: Write the configuration for Cu, remembering the exception. **Cu: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰** (or [Ar] 4s¹ 3d¹⁰) Step 3: Determine the configuration for the Fe atom (Z=26) and then the Fe³⁺ ion. Fe atom: [Ar] 4s² 3d⁶ To form Fe³⁺, remove 3 electrons. The 4s electrons are removed first, then one 3d electron. **Fe³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵** (or [Ar] 3d⁵)
Practice Questions
Question: The element with the atomic number 24 has an anomalous electronic configuration. State and explain this configuration. (3 marks)
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Question: Explain why the second ionisation energy of sodium is significantly higher than its first ionisation energy. (3 marks)
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Question: A sample of magnesium contains three isotopes: ²⁴Mg (78.6%), ²⁵Mg (10.1%), and ²⁶Mg (11.3%). Calculate the relative atomic mass of magnesium to three significant figures. (3 marks)
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Question: Describe the shape of a p-orbital and state how many p-orbitals are found in a p-subshell. (2 marks)
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Question: The successive ionisation energies of an element X are shown below (in kJ/mol): 738, 1451, 7733, 10543. In which group of the periodic table is element X found? Explain your reasoning. (3 marks)
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