An atom consists of three fundamental sub‑atomic particles. Their charge, relative mass and typical location are summarised below.
| Particle | Charge | Relative mass (u) | Location |
|---|---|---|---|
| Proton (p⁺) | +1 | ≈ 1.007 u | In the nucleus |
| Neutron (n⁰) | 0 (neutral) | ≈ 1.009 u | In the nucleus |
| Electron (e⁻) | –1 | ≈ 0.00055 u | In electron shells (cloud) surrounding the nucleus |
Atomic number (Z) = number of protons. In a neutral atom Z also equals the number of electrons.
Mass number (A) = protons + neutrons.
⁽ᴬ⁾₍ᶻ₎X (e.g. 126C).| Isotope | Mass number (A) | Neutron number (N) | Natural abundance (%) |
|---|---|---|---|
| 12C | 12 | 6 | 98.9 |
| 13C | 13 | 7 | 1.1 |
| 14C | 14 | 8 | trace |
For a mono‑isotopic element the spectrum shows a single peak; for elements with several stable isotopes (e.g. Cl) two or more peaks appear.
Orbitals are filled in order of increasing n + ℓ; if two orbitals have the same n + ℓ, the one with the lower n fills first.
Sequence (first 20 orbitals):
$$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s \dots$$
Each box = one orbital; an arrow = one electron (↑ or ↓). Example – ground‑state carbon (Z = 6):
1s : ↑↓ 2s : ↑↓ 2p : ↑ ↑ (two electrons occupy two of the three 2p boxes singly – Hund’s rule)
Compact form: 1s² 2s² 2p².
A species with one or more unpaired electrons. Example – methyl radical:
·CH₃ Electronic configuration: 1s² 2s² 2p¹
X(g) → X⁺(g) + e⁻Xⁿ⁺(g) → X⁽ⁿ⁺¹⁾⁺(g) + e⁻X(g) + e⁻ → X⁻(g)| Trend | Reason |
|---|---|
| Across a period (left → right) | Increasing nuclear charge with little change in shielding → stronger attraction → IE rises. |
| Down a group | Additional shells increase radius and shielding → IE falls. |
| Between successive IE’s | Large jump when a new, more tightly bound sub‑shell is reached (e.g. after removing the ns electron of an alkali metal). |
| Half‑filled or fully‑filled sub‑shells | Extra stability causes a slight increase in IE. |
| Element | IE₁ |
|---|---|
| Li | 520 |
| Be | 900 |
| B | 801 |
| C | 1086 |
| N | 1402 |
| O | 1314 |
| F | 1681 |
| Ne | 2081 |
Given IE₁ = 1314 kJ mol⁻¹ and IE₂ = 3388 kJ mol⁻¹, the values match oxygen. Hence the ground‑state configuration is 1s² 2s² 2p⁴.
mass (g) = moles × M
Create an account or Login to take a Quiz
Log in to suggest improvements to this note.
Your generous donation helps us continue providing free Cambridge IGCSE & A-Level resources, past papers, syllabus notes, revision questions, and high-quality online tutoring to students across Kenya.