Structure 1. Models of the particulate nature of matter

Diagnostic check: a particle box containing two unbonded particle types represents a mixture even if it looks uniform. For an unknown mixture, complete three columns—chosen method, physical-property reason, and observable recovered fraction. Reject any proposal that changes the substances chemically or cannot state where each component goes.

Worked example — convert before interpreting

For $25.0\,^\circ\mathrm{C}$, substitute into the conversion: $T=25.0+273.15=298.15\,\mathrm{K}$, reported as $298.2\,\mathrm{K}$ to one decimal place. The Kelvin value is the absolute temperature used for kinetic-energy comparisons; a rise of $10\,\mathrm{K}$ is the same temperature interval as a rise of $10\,^\circ\mathrm{C}$.

Apply the symbols in a fixed order. For ³⁵₁₇Cl⁻, Z = 17 gives 17 protons, A − Z gives 18 neutrons, and the 1− charge means one more electron than protons, so there are 18 electrons. The ion charge changes electron count, never the element identity.

For a two-isotope sample containing 75% mass 35 and 25% mass 37, the weighted mean is (35 × 0.75) + (37 × 0.25) = 35.5. An answer between the isotope masses is a useful check; simply averaging 35 and 37 would ignore abundance.

Read isotope identity from the pair of nuclear symbols: ³⁵₁₇Cl and ³⁷₁₇Cl are both chlorine because Z = 17 in each, but they contain 18 and 20 neutrons respectively because A differs. Mass number belongs to one nuclide; relative atomic mass is the abundance-weighted mean for a sample.

If singly charged isotope peaks occur at m/z 35 and 37 in a 3:1 height ratio, use fractional abundances 0.75 and 0.25 to obtain Aᵣ = 35.5. First confirm the charge state: m/z is a ratio, so a multiply charged ion cannot be read as mass alone.

Read every spectral transition in two directions: absorption raises an electron by ΔE, while a downward transition emits a photon with ΔE = hf = hc/λ. A shorter-wavelength line therefore represents a larger energy gap, not a higher line intensity.

Orient the spectrum before comparing lines: radio → microwave → infrared → visible → ultraviolet → X-ray → gamma is increasing frequency and photon energy, and decreasing wavelength. Within visible light, red has longer wavelength and lower photon energy than violet.

Compare lines by their energy gaps. Transitions ending at n = 2 form the visible series, and lines crowd together as the starting level rises because adjacent high-n levels are closer in energy. The convergence limit represents removal of the electron, not one more bound-state transition.

For n = 3, the theoretical capacity is 2(3²) = 18 electrons. This is a capacity, not a claim that every third shell is full: the actual occupancy depends on the atom and the relative energies of available sublevels.

Use the hierarchy as a classification test: a p sublevel contains three orbitals, and each orbital can hold two electrons. An orbital describes a probability region with a characteristic shape; it is not a circular route travelled by an electron.

Build an orbital diagram by checking electron total, energy order, single occupation of equal-energy orbitals, then opposite-spin pairing. For transition-metal ions remove 4s electrons before 3d, and verify Cr and Cu against the stated exceptions rather than forcing the simple filling pattern.

Worked ion check: Fe has 26 electrons and condensed configuration [Ar] 4s² 3d⁶. To form Fe³⁺, remove the two electrons from the highest principal level, 4s, before removing one 3d electron, giving [Ar] 3d⁵. The final superscripts total 23 electrons, matching 26 − 3.

Worked example — hydrogen convergence limit

The local course book gives $\lambda=9.12\times10^{-8}\,\mathrm{m}$. First, $f=c/\lambda=(3.00\times10^8\,\mathrm{m\,s^{-1}})/(9.12\times10^{-8}\,\mathrm{m})=3.29\times10^{15}\,\mathrm{s^{-1}}$. Then $E_{\text{photon}}=hf=(6.63\times10^{-34}\,\mathrm{J\,s})(3.29\times10^{15}\,\mathrm{s^{-1}})=2.18\times10^{-18}\,\mathrm{J}$. Convert one-photon energy to one mole and joules to kilojoules: $IE=(2.18\times10^{-18})(6.02\times10^{23})/1000=1.31\times10^3\,\mathrm{kJ\,mol^{-1}}$. This is the molar energy for the first ionization process $\mathrm{H(g)\rightarrow H^+(g)+e^-}$.

At the convergence limit, convert wavelength or frequency to energy per photon with E = hf, then multiply by the Avogadro constant and convert J mol⁻¹ to kJ mol⁻¹. Across-period trends are general patterns; subshell energy and electron pairing explain the named dips.

Locate the first order-of-magnitude jump before naming the group. A large jump after the second electron is removed shows two outer-shell electrons; the third electron would come from a lower shell, supporting a group 2 assignment.

Specify the entity before calculating. One mole of H₂O contains one mole of molecules, two moles of H atoms and one mole of O atoms; multiplying by Nₐ without the formula-unit multiplier answers a different particle question.

For Ca(OH)₂, include both bracketed groups: Mᵣ = Aᵣ(Ca) + 2[Aᵣ(O) + Aᵣ(H)]. Keep Mᵣ dimensionless; attach g mol⁻¹ only when the same numerical total is used as a molar mass.

Use units to choose the direction: 9.0 g of H₂O divided by 18.0 g mol⁻¹ gives 0.50 mol. To find molecules, continue from moles to nNₐ; do not multiply mass directly by the Avogadro constant.

A composition of 40.0% C, 6.7% H and 53.3% O gives the simplest ratio CH₂O after division by atomic masses. If the molar mass is 180 g mol⁻¹, compare it with the empirical-formula mass 30 to obtain the multiplier 6 and molecular formula C₆H₁₂O₆.

Do not round a ratio such as 1 : 1.50 : 1 to 1 : 2 : 1. Preserve the calculated values and multiply every ratio by the same small integer: ×2 converts halves, while ×3 can resolve values close to thirds such as 1.33 or 1.67. Round only after the common multiplier produces values consistent with the data precision.

Convert volume before substitution: 500 cm³ = 0.500 dm³, so 0.250 mol in that solution gives C = 0.500 mol dm⁻³. The denominator is the final solution volume, not the volume or mass of solute alone.

Adding solvent increases the final solution volume while the amount of solute stays constant, so concentration decreases. Use the final solution volume—not the solvent volume added—and do not apply c₁V₁ = c₂V₂ when solute reacts or is removed.

For N₂ + 3H₂ → 2NH₃ at one temperature and pressure, one gas volume of N₂ requires three equal gas volumes of H₂ and forms two of NH₃. This direct volume ratio works because the gases share conditions; otherwise convert through moles.

Use the assumptions to make predictions: compressing a gas until particle volume is no longer negligible weakens the model, while raising temperature usually reduces the relative importance of attractions. Ideal particles still move and collide; only the collisions are treated as elastic.

Gas pressure arises from particle collisions with the container walls and the associated momentum transfer. At a higher Kelvin temperature, the greater average kinetic energy established in Structure 1.1.3 changes the collision behaviour; this does not replace the separate assumptions that particle volume and intermolecular attractions are negligible.

Diagnose the cause from the condition. Cooling makes attractive forces more important because particle kinetic energy is lower; strong compression exposes both attractions and finite particle volume. Name the failed ideal assumption rather than stating only that the gas is 'non-ideal'.

Attach conditions to every molar volume. Once Vₘ is valid for the stated temperature and pressure, convert gas volume to moles, apply the balanced-equation ratio, then convert back if needed. Do not carry one tabulated Vₘ into a different set of conditions.

Worked example — amount and molar mass from gas data

The local course book gives $P=101.3\,\mathrm{kPa}$, $V=1.91\,\mathrm{dm^3}$, $m=3.30\,\mathrm{g}$ and $T=150\,^\circ\mathrm{C}=423.15\,\mathrm{K}$. Because $\mathrm{kPa\,dm^3=J}$, use $R=8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$: $n=PV/(RT)=(101.3\times1.91)/(8.31\times423.15)=0.0550\,\mathrm{mol}$. Then $M=m/n=3.30\,\mathrm{g}/0.0550\,\mathrm{mol}=60.0\,\mathrm{g\,mol^{-1}}$. The final value is the mass of one mole of the vaporized compound under the stated ideal-gas model.

Before solving, write a unit line beside P, V and T. With R = 8.31 J mol⁻¹ K⁻¹, use pressure in Pa, volume in m³ and temperature in K; using kPa with dm³ is also consistent because kPa·dm³ equals J. Judge model suitability before trusting the numerical result.