2.1Lesson 1: Early atomic models — Thomson and Rutherford2.21.1 Thomson's atomic model (1897)2.31.2 Rutherford's gold foil experiment2.41.3 Postulates of Rutherford's atomic model (1911)2.52.1 Atomic emission (line) spectrum2.62.2 Bohr's atomic model (1913)2.72.3 The hydrogen spectrum explained2.82.4 Success of Bohr's model2.92.5 Shortcomings of Bohr's model2.10Lesson 3: The modern atomic theory2.11Lesson 4: Quantum numbers2.124.1 Principal quantum number (n)2.134.2 Subsidiary quantum number (ℓ)2.144.3 Magnetic quantum number (mℓ)2.154.4 Spin quantum number (ms)2.164.5 Relations between n, sublevels, orbitals and electrons2.174.6 Worked examples2.18Lesson 5: Distribution of electrons in the atom| Electron | n | ℓ | mℓ | ms |
|---|---|---|---|---|
| first | 3 | 0 | 0 | +½ |
| second | 3 | 0 | 0 | −½ |
| Comparison aspect | Main idea | Keywords |
|---|---|---|
| • | Worked example 5.1 (book examples) | Workedexamplebookexamples |
| ₁₁Na | 1s2 2s2 2p6 3s1 | |
| ₂₀Ca | 1s2 2s2 2p6 3s2 3p6 4s2 | |
| ₃₀Zn | 1s2 2s2 2p6 3s2 3p6 4s2 3d10 | 3d10 |
| • | Worked example 5.2 — new atoms | Workedexampleatoms |
| ₁₇Cl | 1s2 2s2 2p6 3s2 3p5 (2 + 2 + 6 + 2 + 5 = 17) | |
| ₂₆Fe | 1s2 2s2 2p6 3s2 3p6 4s2 3d6 (total 26) | total |
| ₃₅Br | 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 (total 35) | 3d10total |
| ₁₀Ne |
| ₁₈Ar |
| Worked example 5.3 — number of occupied levels and valence electrons. |
| PhosphorusThreemainshell |
| Notes on special cases. |
| stabilityhalffilledcompletely |
| Comparison aspect | Main idea | Keywords |
|---|---|---|
| • | Worked example 5.4 — unpaired electrons | Workedexampleunpairedelectrons |
| Carbon ₆C | 1s2 2s2 2px1 2py1: two unpaired electrons | 2px12py1unpairedelectrons |
| Oxygen |
| unpaired |
| Fluorine | one unpaired | unpaired |
| Neon | none | none |
| Phosphorus ₁₅P (3p3) | three unpaired electrons | threeunpairedelectrons |
| • |
| Workedexamplediagramcorrect |
| • | 2p must show one pair and two single electrons with parallel spins: [↑↓][↑][↑] | mustshowpairsingle |
| • |
| diagramshowingpairsempty |
| Worked example 5.6 — find the atomic number. | An atom has the configuration 1s2 2s2 2p6 3s2 3p6 4s1. The sum of the electrons is 19, so Z = 19 (potassium). In its ground state, its last electron has n = 4, ℓ = 0, mℓ = 0, ms = +½. | configurationelectronspotassiumground |
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