HSC Exam Practice

Sample response. Isotopes are atoms of the same element that have the same number of protons (and therefore the same atomic number Z) but different numbers of neutrons, giving them different mass numbers (A). Because they have the same Z, all isotopes of an element have the same number of electrons and the same electron configuration.

Marking notes. 1 mark for correctly stating what is the same (protons / atomic number / electrons); 1 mark for correctly stating what differs (neutrons / mass number). Both criteria must be explicitly named.

Sample response. Chlorine occurs naturally as a mixture of two isotopes: ³⁵Cl (75.77%) and ³⁷Cl (24.23%). The relative atomic mass is the weighted average of these isotopic masses, weighted by their fractional abundances. Using fractional abundances: A_r(Cl) = (35 × 0.7577) + (37 × 0.2423) = 26.52 + 8.97 = 35.49 ≈ 35.5. The result is not a whole number because the two isotopes are present in unequal amounts, so the weighted average falls between 35 and 37 rather than at either whole-number value.

Marking notes. 1 mark for identifying that A_r is a weighted average of the two isotopes; 1 mark for using correct fractional abundances (not raw percentages) in a calculation or correctly applying the formula; 1 mark for reaching A_r ≈ 35.45–35.5 with a clear explanation that the non-integer result reflects unequal isotopic proportions.

Sample response. A mass spectrometer ionises atoms and accelerates them into a magnetic field. Heavier ions are deflected less than lighter ions, separating them by their mass-to-charge ratio (m/z). A detector records how many ions arrive at each m/z value, producing a mass spectrum with the m/z ratio on the x-axis and relative abundance (%) on the y-axis. Each peak corresponds to one isotope: the peak’s position (x-value) gives the mass number of that isotope, and the peak’s height gives its relative abundance in the sample. This data allows A_r to be calculated.

Marking notes. 1 mark for correctly describing separation by m/z (heavier = less deflection); 1 mark for correctly identifying the axes (x = m/z; y = relative abundance %); 1 mark for correctly stating what each peak gives (position = mass number; height = abundance).

Sample response. The mass number (A) of a nuclide is always a whole number; it equals the total number of protons plus neutrons in the nucleus of that specific isotope. For example, the mass number of ³⁵Cl is 35 (17 protons + 18 neutrons). The relative atomic mass (A_r) of an element is a decimal (usually); it is the weighted average mass of all naturally occurring isotopes, relative to 1/12 the mass of ¹²C. For example, A_r(Cl) = 35.45. The two must not be confused: mass number refers to one specific nuclide and is a count; relative atomic mass refers to all natural isotopes of the element and is a statistical average.

Marking notes. 1 mark for correctly stating mass number is a whole number and giving its definition (protons + neutrons for one nuclide); 1 mark for a correct example of mass number (e.g. ³⁵Cl = 35); 1 mark for correctly defining A_r as a weighted average that is not necessarily a whole number; 1 mark for a correct example of A_r for an element with isotopes (e.g. Cl = 35.45, B = 10.81).

Sample response. ¹²C and ¹⁴C are both carbon atoms: each has 6 protons and therefore 6 electrons in a neutral atom. Chemical behaviour — including bonding, valence, and reactivity — is determined solely by the electron configuration. Both ¹²C and ¹⁴C have the same electron configuration (2, 4), so they form the same two C=O bonds with two oxygen atoms to produce CO₂. The different number of neutrons (6 in ¹²C vs 8 in ¹⁴C) does not affect electron configuration and therefore has no effect on chemical properties.

Marking notes. 1 mark for identifying that both isotopes have 6 protons and 6 electrons, giving the same electron configuration; 1 mark for stating that chemical behaviour is determined by electron configuration (not neutrons); 1 mark for explicitly explaining that the different neutron count in ¹⁴C has no effect on bonding and therefore both form CO₂ identically.

Sample response. The student used raw percentage values (19.9 and 80.1) instead of fractional abundances. The correct formula requires fractional abundances: fractional abundance = % ÷ 100. Fractional abundances: ¹⁰B = 19.9 ÷ 100 = 0.199; ¹¹B = 80.1 ÷ 100 = 0.801. Correct calculation: A_r(B) = (10 × 0.199) + (11 × 0.801) = 1.990 + 8.811 = 10.80 (to 4 s.f.). The student’s answer of 1079 is clearly unreasonable — A_r must lie between the two isotopic masses (i.e. between 10 and 11).

Marking notes. 1 mark for identifying the error (used raw % instead of fractional abundance); 1 mark for correctly converting to fractional abundances and applying the formula; 1 mark for arriving at the correct A_r(B) = 10.80–10.81 and noting the answer must lie between 10 and 11 as a reasonableness check.

Sample response (a). Fractional abundances: ⁷⁹Br = 50.69 ÷ 100 = 0.5069; ⁸¹Br = 49.31 ÷ 100 = 0.4931. Check: 0.5069 + 0.4931 = 1.0000 ✓ [1 mark]. A_r(Br) = (79 × 0.5069) + (81 × 0.4931) = 40.045 + 39.941 = 79.99 ≈ 80.0 [1 mark]. This agrees with the IUPAC value of 79.90 (accept 79.9–80.0 depending on significant figures used). [1 mark for correct final value with units-free dimensionless result and reasonable s.f.]

Sample response (b). The two peaks being of almost equal height means that ⁷⁹Br and ⁸¹Br are present in nearly equal proportions (~50% each) [1 mark]. When two isotopes are present in equal amounts, the weighted average falls exactly halfway between their masses. Since ⁷⁹ and ⁸¹ are the two isotope masses and each is ~50% abundant, A_r will be closest to 80 (the midpoint between 79 and 81) [1 mark]. The slight excess of ⁷⁹Br (~50.7% vs ~49.3%) pulls A_r fractionally below 80 [1 mark].

Sample response (c). A_r = (79 × 50.69 + 81 × 49.31) ÷ 100 = (4004.5 + 3994.1) ÷ 100 = 7998.6 ÷ 100 = 79.99 [1 mark]. This is algebraically identical to using fractional abundances because multiplying by the percentage and then dividing the whole expression by 100 is the same as dividing each percentage by 100 individually before multiplying. In both cases you are computing the weighted sum Σ(mass × fractional abundance) [1 mark].

Marking notes (8 marks): Part (a): 1 = fractional abundances correct with sum check; 1 = correct products; 1 = correct A_r 79.9–80.0. Part (b): 1 = interprets equal peak heights as roughly equal abundance; 1 = predicts A_r closest to 80 with justification; 1 = identifies the slight ⁷⁹Br excess pulling A_r below 80. Part (c): 1 = reaches same numerical answer using the ÷100 approach; 1 = explains the algebraic equivalence.

Sample response. The claim that “isotopes of the same element are chemically identical but physically different” is largely valid but requires qualification. The chemical identity of isotopes arises directly from atomic structure: all isotopes of an element have the same atomic number Z and therefore the same number of protons, giving neutral atoms an identical number of electrons. Since chemical behaviour — bonding, valence, and reactivity — is determined by electron configuration, isotopes of the same element exhibit the same chemical properties under most conditions. For example, ¹⁶O, ¹⁷O, and ¹⁸O all react with hydrogen to form water (H₂O) under identical conditions; and ¹²C and ¹⁴C both bond with oxygen to form CO₂ identically, which is why carbon-14 can be metabolised by all living organisms in the same chemical pathways as carbon-12 — making radiocarbon dating possible (medical/forensic application). In geochemistry, ²³⁸U decays to ²⁰⁶Pb over billions of years (a physical process), and this radiogenic ²⁰⁶Pb is chemically indistinguishable from primordial ²⁰⁶Pb, meaning geologists exploit the difference in mass number — a physical property — rather than any chemical difference, to date rocks via uranium–lead mass spectrometry. However, the claim that isotopes are completely “chemically identical” is an oversimplification. The kinetic isotope effect means that heavier isotopes react slightly more slowly than lighter ones because the vibrational frequency of bonds involving heavier nuclei is lower (from quantum mechanics: zero-point energy scales with 1/√m). For example, the carbon-14 kinetic isotope effect means that metabolic enzymes process ¹²CO₂ fractionally faster than ¹⁴CO₂, creating isotopic fractionation in biological tissues. Similarly, deuterium (²H) reacts measurably more slowly than protium (¹H) in C–H bond-breaking reactions. This fractionation effect is exploited in isotope-ratio mass spectrometry to trace food authenticity and climate records in ice cores. Therefore, while the claim captures the essential principle, a more precise statement is: isotopes of the same element have the same chemical properties in the absence of kinetic isotope effects, but may show small but measurable rate differences in reactions involving bond making or breaking, particularly for light elements such as hydrogen, carbon, oxygen, and nitrogen.

Marking criteria (7 marks). 1 = correctly explains the structural basis for chemical identity (same protons → same electrons → same electron configuration). 1 = correctly identifies at least one physical difference (mass, density, nuclear stability, melting/boiling point). 1 = gives a correctly described medical or forensic isotope application (radiocarbon dating; PET scanning using ¹⁸F; ²H-labelling in pharmacokinetics; stable isotope forensics). 1 = gives a correctly described geochemical or industrial isotope application (uranium–lead dating; heavy water reactors using ²H; oxygen isotope analysis in climate science). 1 = identifies a limitation of the claim: kinetic isotope effect or isotopic fractionation means isotopes are not always perfectly chemically identical. 1 = provides a correct example of where isotopic rates differ (C–H bond breaking; enzyme discrimination between ¹²C and ¹⁴C; ¹H vs ²H reaction rates). 1 = reaches a nuanced evaluative judgement: the claim is valid as a first approximation but must be qualified by the kinetic isotope effect, especially for light elements.