Atomic Structure and the Periodic Table (Edexcel IAL Chemistry Unit 1)
Edexcel International A-Level Chemistry· Spec points 2.1–2.18· 45 min read
1. Subatomic Particles, Isotopes and Mass Spectrometry★★☆☆☆⏱ 10 min
Atoms contain three subatomic particles: protons (charge +1, relative mass 1), neutrons (charge 0, relative mass 1), and electrons (charge -1, relative mass 1/1840). The atomic number (Z) equals the number of protons, and mass number (A) equals protons + neutrons. Ions have more or fewer electrons than protons.
Isotope
Atoms of the same element with identical proton count but different neutron count, sharing chemical properties but differing in physical properties like density and boiling point.
Calculate the number of protons, neutrons and electrons in a ³⁷Cl⁻ ion.
- 1
Atomic number of Cl is 17, so proton count = 17
- 2
Mass number = 37, so neutron count = 37 - 17 = 20
- 3
1- charge means 1 extra electron, so electron count = 17 + 1 = 18
Mass spectrometers measure isotopic abundances to calculate relative atomic mass (Ar). For diatomic molecules like Cl₂, you will see 3 mass peaks corresponding to combinations of isotopes.
Chlorine has two isotopes: ³⁵Cl (75% abundance) and ³⁷Cl (25% abundance). Calculate Ar of chlorine.
- 1
Use the formula:
- 2
2. Ionisation Energies and Shell/Sub-shell Evidence★★★☆☆⏱ 12 min
First Ionisation Energy
The energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous 1+ ions, always endothermic (positive value).
Successive ionisation energies show large jumps when electrons are removed from a shell closer to the nucleus, indicating the number of outer shell electrons and group of the element.
Successive ionisation energies (kJ mol⁻¹) of an element are: 738, 1451, 7733, 10540. Identify its group.
- 1
Locate the first large jump in IE values, between 2nd (1451) and 3rd (7733) IE
- 2
This means 2 electrons are in the outer shell, so the element is in Group 2
Across a period, first IE generally increases (higher nuclear charge, same shielding, smaller atomic radius). Dips occur between Group 2 and 13 (electron removed from higher energy p orbital) and Group 15 and 16 (electron removed from paired p orbital with repulsion). Down a group, IE decreases (more shielding, weaker nuclear attraction).
3. Orbitals and Electron Configuration Rules★★★☆☆⏱ 10 min
Orbital
Region of space around the nucleus holding up to 2 electrons with opposite spin. s orbitals are spherical, p orbitals are dumbbell shaped.
Electrons fill orbitals following three rules: Aufbau principle (lowest energy first: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p), Pauli exclusion principle (max 2 electrons per orbital, opposite spin), Hund's rule (fill orbitals singly with parallel spin before pairing). Exceptions: Cr = [Ar]4s¹3d⁵, Cu = [Ar]4s¹3d¹⁰ (stable half/full d subshells). For ions, 4s electrons are lost before 3d.
Write the full electron configuration of the Cu²⁺ ion.
- 1
Neutral Cu has 29 electrons: [Ar]4s¹3d¹⁰
- 2
Remove 2 electrons, first from 4s then 3d: remove 1 from 4s, 1 from 3d
- 3
Final configuration: 1s²2s²2p⁶3s²3p⁶3d⁹
4. Periodic Table Blocks and Periodic Trends★★★★☆⏱ 8 min
The periodic table is divided into blocks based on the highest energy subshell containing electrons: s block (Groups 1-2, outer electrons in s orbital), p block (Groups 13-18, outer electrons in p orbital), d block (transition metals, highest energy electrons in d orbital).
Identify the block of an element with electron configuration 1s²2s²2p⁶3s²3p⁶3d⁵4s².
- 1
Locate the highest energy subshell with electrons, which is 3d
- 2
Element is in the d block
Melting/boiling points across Periods 2 and 3 increase from Group 1 to 14 (giant metallic to giant covalent structures with strong bonds), then drop sharply from Group 15 to 18 (simple molecular structures with weak intermolecular forces).
5. Interpreting Ionisation Energy Graphs★★★★☆⏱ 5 min
Log first ionisation energy graphs for elements 1–36 show a periodic repeating pattern, providing evidence for shell and sub-shell structure. You will be required to explain trends and dips in these graphs in exams.
State and explain the general trend in first ionisation energy across Period 3 from Na to Ar.
- 1
General trend: increases across the period
- 2
Explanation: Nuclear charge increases across the period, shielding remains similar as electrons are added to the same shell
- 3
Atomic radius decreases, so attraction between nucleus and outer electron increases, requiring more energy to remove the electron
6. Common Pitfalls
Wrong move:
Calculating Ar using an unweighted average of isotope masses
Why:
Isotopes are not present in equal amounts, so unweighted averages are inaccurate
Correct move:
Multiply each isotope mass by its percentage abundance, sum values and divide by 100
Wrong move:
Removing 3d electrons before 4s when writing transition metal ion configurations
Why:
Once 3d orbitals are filled, their energy is lower than 4s, so 4s electrons are lost first
Correct move:
Always remove 4s electrons before 3d for transition metal ions
Wrong move:
Writing Cr and Cu configurations with 4s² instead of 4s¹
Why:
Half-full and full d subshells have extra stability, making these valid exceptions to the Aufbau rule
Correct move:
Recall Cr = [Ar]4s¹3d⁵ and Cu = [Ar]4s¹3d¹⁰
Wrong move:
Stating ionisation energies are exothermic
Why:
Energy is required to overcome attraction between the negative electron and positive nucleus
Correct move:
All ionisation energies are endothermic, with positive enthalpy values
Wrong move:
Attributing the Group 15-16 IE dip to increased shielding
Why:
Shielding is identical across the same period, the dip arises from electron-electron repulsion in paired p orbitals
Correct move:
Explain the dip as removal of an electron from a paired p orbital, where repulsion reduces the energy required for removal
7. Quick Reference Cheatsheet
Concept | Key Rule/Value |
|---|---|
Subatomic particles | Proton: +1, 1 mass; Neutron: 0, 1 mass; Electron: -1, 1/1840 mass |
Ar calculation | Sum (isotope mass × % abundance) ÷ 100 |
Electron filling order | 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p |
Transition metal ions | Remove 4s electrons before 3d |
Cr/Cu exceptions | Cr = [Ar]4s¹3d⁵; Cu = [Ar]4s¹3d¹⁰ |
IE trend down group | Decreases: more shielding, weaker nuclear attraction |
IE trend across period | General increase: higher nuclear charge, same shielding |
8. Frequently Asked
Why do 4s electrons fill before 3d but remove first?
4s orbitals have lower energy than 3d when empty, so they fill first. Once 3d orbitals are occupied, their energy drops below 4s, so 4s electrons are lost first during ionisation.
Are Cr and Cu electron configuration exceptions required?
Yes, you must recall Cr = [Ar]4s¹3d⁵ and Cu = [Ar]4s¹3d¹⁰, as well as configurations for their ions.
Is mass spectrometry fragmentation in organic molecules covered here?
No, organic mass spectrometry fragmentation is covered in Topics 10 and 15 of the Edexcel IAL Chemistry specification.
What's Next
Now that you have mastered atomic structure and the periodic table, you are prepared to move on to bonding and structure topics in Edexcel IAL Chemistry Unit 1. Understanding electron configurations will help you explain ionic, covalent and metallic bonding, molecular shapes and intermolecular forces. This topic is also the foundation for later content including Group 2 and Group 7 periodicity, and transition metal chemistry in Unit 2. Practice past paper questions on electron configurations and ionisation energy explanations, as these are common high-mark questions in the WCH11 exam. Use the cheatsheet for quick last-minute revision before your test.
