Notes
Moles and mass
- Avogadro's number: 1 mol $= 6.022 \times 10^{23}$ particles.
- Moles $\displaystyle = \frac{\text{mass}}{\text{molar mass}}$. Molar mass comes from the periodic table (water: 18.02 g/mol).
- Mass spectrometry shows isotopes and their relative abundance; the average atomic mass is the weighted average.
- Empirical formula: the simplest whole-number ratio of atoms. Convert percent by mass to grams, then moles, then divide by the smallest.
- A pure substance has a fixed composition; a mixture can vary.
Electron configuration
- Fill orbitals in order: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p \dots$ Oxygen is $1s^2 2s^2 2p^4$.
- Photoelectron spectroscopy (PES) shows peaks for each subshell. Higher binding energy means electrons closer to the nucleus. Peak height shows the number of electrons.
- Coulomb's law: force $\displaystyle \propto \frac{q_1q_2}{r^2}$. Attraction is stronger with more charge and shorter distance.
Periodic trends
| Trend | Across a period (left to right) | Down a group |
|---|---|---|
| Atomic radius | Decreases | Increases |
| Ionization energy | Increases | Decreases |
| Electronegativity | Increases | Decreases |
- Across a period, protons increase but shielding stays about the same, so effective nuclear charge rises and electrons are held tighter. Down a group, electrons are in higher shells, farther away and more shielded.
- Cations are smaller than their atoms; anions are larger.
Valence electrons and ions
Valence electrons determine reactivity. Main-group elements tend to form ions with noble-gas configurations: Na⁺, Mg²⁺, Cl⁻, O²⁻.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle \frac{36.0}{18.0} = 2.00$ mol.
16 electrons: 2 + 2 + 6 + 2 + 4.
K is lower in the same group as Na, so its valence electrons are in a higher shell.
Higher effective nuclear charge at a similar distance means stronger attraction.
Per 100 g: 3.33 mol C, 6.7 mol H, 3.33 mol O. Divide by 3.33: 1 : 2 : 1.
Electrons closest to the nucleus need the most energy to remove.
$0.758(35) + 0.242(37) \approx 35.5$.
Losing the valence shell makes a cation smaller.
Free response
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Free-response question
Consider the elements Mg and Al.
(a) Write the electron configurations of Mg and Al.
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Mg: $1s^2 2s^2 2p^6 3s^2$. Al: $1s^2 2s^2 2p^6 3s^2 3p^1$.
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(b) Al has a lower first ionization energy than Mg, even though Al has more protons. Explain.
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Al's first electron removed is a $3p$ electron, which is higher in energy and slightly farther from the nucleus and more shielded by the $3s$ electrons than Mg's $3s$ electron. So it's easier to remove.
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(c) Predict which has the larger second ionization energy, Mg or Al. Explain.
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Al. For Mg, the second electron is still a $3s$ valence electron. For Al, the second electron comes from the $3s$ subshell with a greater nuclear charge (13 protons vs 12) at a similar distance, so it's harder to remove.
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