How to use this review
Use this page after the Unit 1 diagnostic, once you have revisited the topics it flagged. Answer all seven questions on paper before you open any answer. Each one combines two or more topics.
These are original Marlbridge practice questions, not past exam questions. The elements in Questions 1 and 4 are fictional, and photoelectron binding energies are illustrative values, not measured data. Each rubric is a suggested Marlbridge rubric, not official scoring, and your total gives no predicted score.
Big ideas of the unit
- You count particles by weighing them. The molar mass links a measured mass to a number of particles (Topic 1.1).
- Molar masses come from mass spectra. A periodic-table mass is a weighted average that assumes the natural isotope mix (Topic 1.2).
- Mass data become formulas only through moles. Compare moles, never grams (Topic 1.3).
- A pure compound has a fixed composition; a mixture does not. So elemental analysis can test purity and find a mixture’s make-up (Topic 1.4).
- Electrons sit in shells and subshells, filled in Aufbau order. Cations lose electrons from the highest shell first (Topic 1.5).
- Photoelectron spectra are the evidence for that model. Peak heights give the configuration; peak positions show how strongly each subshell is held (Topic 1.6).
- One law explains every trend. Coulomb’s law, shells and shielding explain radius, ionization energy, electron affinity and electronegativity, and support estimates (Topic 1.7).
- Valence electrons decide chemistry. They set the typical ion charge, so a group forms analogous compounds (Topic 1.8).
Key relationships and methods
| Idea | Relationship or method | Topic |
|---|---|---|
| Moles and particles | n = m / M; N = n × N_A | 1.1 |
| Average atomic mass | Σ (isotope mass × fractional abundance); heights scaled to 100 must be divided by their total | 1.2 |
| Percent by mass | (atoms in formula × atomic mass) ÷ molar mass × 100 | 1.3 |
| Empirical formula | grams → moles → divide by the smallest → whole numbers (× 2, 3 or 4 if needed) | 1.3 |
| Purity | mass of compound = mass of element ÷ its mass fraction; ÷ sample mass × 100 | 1.4 |
| Two-component mixture | one equation for total mass, one for the analysed element | 1.4 |
| Coulomb’s law | F ∝ q₁q₂ / r²; core charge ≈ protons − core electrons | 1.5 |
| PES | one peak per subshell; height ∝ electrons; position = binding energy | 1.6 |
| Trends | across a period: Z_eff rises, same shell; down a group: new shell, farther out | 1.7 |
| Ionic formulas | charge from valence electrons; smallest ratio with total charge zero | 1.8 |
Question 1 (multiple choice · mixed)
The mass spectrum of a fictional element shows two peaks: m/z 58.0 with 70.0% abundance and m/z 61.0 with 30.0% abundance. How many atoms are in a 2.945 g sample of the element?
- (A) 3.01 × 10²² atoms
- (B) 3.06 × 10²² atoms
- (C) 2.98 × 10²² atoms
- (D) 1.20 × 10²⁵ atoms
Answer and explanation
Answer: (A). Molar mass from the spectrum (Topic 1.2): (58.0 × 0.700) + (61.0 × 0.300) = 58.90 g mol⁻¹. Then count atoms (Topic 1.1): n = 2.945 ÷ 58.90 = 0.05000 mol, and N = 0.05000 × 6.022 × 10²³ = 3.01 × 10²² atoms.
- (B) uses 58.0, the most abundant isotope’s mass. A real sample contains both isotopes.
- (C) uses the simple mean, 59.5, ignoring abundance.
- (D) flips the fraction; the units do not cancel to “atoms”.
Question 2 (multiple choice · mixed)
The photoelectron spectrum of a species Y has three peaks with relative heights 2 : 2 : 6. Its 1s peak is at a lower binding energy than the 1s peak of Na⁺. Which could Y be, and how does its radius compare with that of Na⁺?
- (A) Mg²⁺, which is smaller than Na⁺
- (B) F⁻, which is larger than Na⁺
- (C) F⁻, which is smaller than Na⁺
- (D) Mg²⁺, which is larger than Na⁺
Answer and explanation
Answer: (B). Heights 2 : 2 : 6 mean 1s²2s²2p⁶, the same 10 electrons as Na⁺ (Topic 1.6). The 1s position then depends on nuclear charge: a lower binding energy means fewer than 11 protons, so Y could be F⁻ (9 protons). Fewer protons pulling the same 10 electrons make F⁻ larger than Na⁺ (Topic 1.7).
- (A) and (D) Mg²⁺ has 12 protons, so its 1s peak would be at a higher binding energy.
- (C) reverses the size argument: for isoelectronic species, fewer protons means a larger particle.
Question 3 (multiple choice · mixed)
An element in group 13 forms an oxide that is 52.9% of the element by mass. Which formula and element fit these data?
- (A) Al₂O₃, aluminium
- (B) AlO, aluminium
- (C) Ga₂O₃, gallium
- (D) B₂O₃, boron
Answer and explanation
Answer: (A). A group 13 element forms M³⁺ and oxygen forms O²⁻, so the oxide is M₂O₃ (Topic 1.8). Then use percent by mass (Topic 1.3): 2M ÷ (2M + 48.00) = 0.529 gives M = 27.0 g mol⁻¹, which is aluminium. Check: Al₂O₃ is 53.96 ÷ 101.96 = 52.92% Al.
- (B) is not neutral: Al³⁺ and O²⁻ cannot combine 1 : 1. AlO would be 62.77% Al.
- (C) Ga₂O₃ is 74.39% Ga: heavier metal atoms take a larger share of the mass.
- (D) B₂O₃ is only 31.05% B.
Question 4 (constructed response · mixed)
A fictional element Zq gives a mass spectrum with three peaks. The tallest peak is scaled to 100.
| m/z | Relative height |
|---|---|
| 88.0 | 100 |
| 90.0 | 60 |
| 91.0 | 40 |
(a) Calculate the average atomic mass of Zq. (b) 2.500 g of Zq reacts completely with oxygen to give 3.173 g of a pure oxide. Find its empirical formula. (c) Calculate the number of Zq atoms in the 3.173 g sample of oxide. (d) A second student uses 88.0 g mol⁻¹, the mass of the most common isotope, in part (b). Explain whether this changes the empirical formula. (e) Zq forms ions with the same charge in its chloride as in its oxide. Predict the formula of the chloride.
Worked solution and suggested Marlbridge rubric
(a) The heights add up to 100 + 60 + 40 = 200, so the fractions are 0.500, 0.300 and 0.200. Average = (88.0 × 0.500) + (90.0 × 0.300) + (91.0 × 0.200) = 44.0 + 27.0 + 18.2 = 89.20 amu. (Dividing by 100 instead gives 178.40, which is impossible.)
(b) Oxygen: 3.173 − 2.500 = 0.673 g. Moles: Zq 2.500 ÷ 89.20 = 0.02803 mol; O 0.673 ÷ 16.00 = 0.04206 mol. O : Zq = 1.50, so multiply by 2: Zq₂O₃.
(c) All the Zq atoms came from the 2.500 g of metal: 0.02803 mol × 6.022 × 10²³ mol⁻¹ = 1.69 × 10²² atoms.
(d) With 88.0 g mol⁻¹, n(Zq) = 0.02841 mol and O : Zq = 1.48, which still rounds to 1.5, so the formula does not change here. The method is still wrong: the sample contains all three isotopes, so the molar mass must be the weighted average. The 1.4% error in moles could give the wrong formula with less clear-cut data.
(e) In Zq₂O₃ the charges must balance: 2x + 3(−2) = 0, so Zq forms Zq³⁺. With Cl⁻ the chloride is ZqCl₃.
| Point | What earns it |
|---|---|
| 1 | (a) Divides each height by the total (200) to get fractions |
| 1 | (a) 89.20 amu |
| 1 | (b) Oxygen mass by subtraction and both amounts in moles |
| 1 | (b) Ratio 1 : 1.50 turned into Zq₂O₃ (not rounded to ZqO or ZqO₂) |
| 1 | (c) 1.69 × 10²² atoms (carry forward from (b)) |
| 1 | (d) Formula unchanged here, and explains that the weighted average is the correct molar mass |
| 1 | (e) ZqCl₃, with the 3+ charge taken from the oxide formula |
Total: 7 points.
Question 5 (constructed response · mixed)
An element gives this illustrative photoelectron spectrum (peaks from highest to lowest binding energy).
| Peak | Binding energy (MJ mol⁻¹) | Relative height |
|---|---|---|
| 1 | 348 | 2 |
| 2 | 36.6 | 2 |
| 3 | 28.7 | 6 |
| 4 | 3.36 | 2 |
| 5 | 1.77 | 6 |
| 6 | 0.42 | 1 |
(a) Write the electron configuration of the element and identify it. (b) Which peak gives an estimate of the first ionization energy? Give the estimate in kJ mol⁻¹. (c) Describe the spectrum of the element’s 1+ ion. A student predicts that the ion’s 1s peak will be far from peak 1 “because the ion has lost an electron”. Evaluate the prediction. (d) For the next element in the periodic table, the lowest-energy peak is at a higher binding energy than peak 6. Give one other difference in that peak, and explain the higher energy. (e) Explain why the 1+ ion is smaller than the atom.
Worked solution and suggested Marlbridge rubric
(a) Assign subshells in order of decreasing binding energy, with heights as superscripts: 1s²2s²2p⁶3s²3p⁶4s¹. 19 electrons, so potassium, K.
(b) Peak 6, the most loosely held electron: 0.42 MJ mol⁻¹ = 420 kJ mol⁻¹.
(c) K⁺ is 1s²2s²2p⁶3s²3p⁶: five peaks, heights 2, 2, 6, 2, 6, no 4s peak. The prediction is not supported. K⁺ keeps all 19 protons, and the removed 4s electron, in the outermost shell, did almost nothing to shield the innermost 1s electrons. Their attraction to the nucleus hardly changes, so the 1s peak stays almost where it was (at most slightly higher).
(d) Calcium, [Ar]4s², has a 4s peak twice as tall (2 electrons). Both atoms have 18 core electrons shielding the 4s electrons, but calcium has 20 protons against 19, so the core charge is about +2 rather than +1. At a similar distance, Coulomb’s law gives a stronger attraction.
(e) Forming K⁺ removes the only n = 4 electron, so the outermost electrons are now in n = 3, much closer to the nucleus, and 19 protons pull on only 18 electrons.
| Point | What earns it |
|---|---|
| 1 | (a) Correct configuration and potassium |
| 1 | (b) Peak 6 and 420 kJ mol⁻¹ |
| 1 | (c) Five peaks, heights 2, 2, 6, 2, 6, no 4s peak |
| 1 | (c) Rejects the prediction: same 19 protons, and the removed outer 4s electron hardly shielded the 1s electrons |
| 1 | (d) Peak twice as tall (two 4s electrons) |
| 1 | (d) More protons, same core shielding, so larger effective charge at similar distance |
| 1 | (e) Loss of the n = 4 shell, so the outer electrons are closer to the nucleus |
Total: 7 points. In (c), agreeing that the 1s peak moves a long way does not earn the point: only an outer electron was removed, and the nuclear charge is unchanged.
Question 6 (constructed response · mixed)
A 3.000 g powder is a mixture of CaCl₂ and MgCl₂ only. Elemental analysis shows it contains 2.044 g of chlorine.
(a) Calculate the percent by mass of chlorine in the sample, in pure CaCl₂ and in pure MgCl₂. Explain how these values support the claim that the powder is a mixture. (b) Calculate the mass of each salt in the sample. (c) Calculate the ratio of calcium atoms to magnesium atoms in the sample. Comment on whether a non-whole-number ratio is a problem. (d) Calculate the number of chloride ions in the sample. (e) Explain why calcium and magnesium form chlorides with the same formula type.
Worked solution and suggested Marlbridge rubric
(a) Sample: 2.044 ÷ 3.000 × 100 = 68.13% Cl. CaCl₂: M = 40.08 + 70.90 = 110.98 g mol⁻¹, so 70.90 ÷ 110.98 × 100 = 63.89% Cl. MgCl₂: M = 24.31 + 70.90 = 95.21 g mol⁻¹, so 70.90 ÷ 95.21 × 100 = 74.47% Cl. 68.13% matches neither fixed value but lies between them, as a mixture of the two must.
(b) Let a = mol CaCl₂ and b = mol MgCl₂. Each formula unit holds two Cl.
- Chlorine: 2(a + b) = 2.044 ÷ 35.45 = 0.05766 mol, so a + b = 0.02883 mol.
- Mass: 110.98a + 95.21b = 3.000.
- Substitute b = 0.02883 − a: 2.745 + 15.77a = 3.000, so a = 0.01618 mol and b = 0.01265 mol.
- Masses: CaCl₂ 0.01618 × 110.98 = 1.80 g; MgCl₂ 0.01265 × 95.21 = 1.20 g (total 3.00 g).
(c) Ca : Mg = 0.01618 : 0.01265 = 1.28 : 1. Not a problem: a mixture can have any proportions, so its atom ratios need not be whole numbers.
(d) 0.05766 mol × 6.022 × 10²³ mol⁻¹ = 3.47 × 10²² chloride ions.
(e) Both are in group 2 with two valence electrons (ns²). Each loses both to form M²⁺, which needs two Cl⁻ to balance: elements in one group form analogous compounds.
| Point | What earns it |
|---|---|
| 1 | (a) All three percentages correct |
| 1 | (a) Sample value between the pure values, matching neither |
| 1 | (b) Two correct equations (chlorine and total mass) |
| 1 | (b) 1.80 g CaCl₂ and 1.20 g MgCl₂ |
| 1 | (c) 1.28 : 1, with the comment that mixtures need not give whole-number ratios |
| 1 | (d) 3.47 × 10²² ions |
| 1 | (e) Same group, two valence electrons, so both form M²⁺ and analogous MCl₂ |
Total: 7 points. Accept carried-forward values in (c) and (d).
Question 7 (constructed response · mixed)
Selenium, Se (Z = 34), is below sulfur in group 16.
(a) Write the noble-gas configurations of a selenium atom and of the selenide ion, Se²⁻. (b) Predict the formula of the ionic compound of aluminium and selenium. Justify the charge of each ion. (c) Rank S²⁻, Se²⁻ and Br⁻ in order of increasing radius. Justify your ranking. (d) The first ionization energies of sulfur and tellurium are 1000 and 869 kJ mol⁻¹. Tellurium is below selenium. Estimate the first ionization energy of selenium and justify your estimate.
Worked solution and suggested Marlbridge rubric
(a) Se: [Ar]4s²3d¹⁰4p⁴ (34 electrons). Se²⁻ gains two electrons into 4p: [Ar]4s²3d¹⁰4p⁶ = [Kr].
(b) Aluminium (group 13, 3s²3p¹) loses its 3 valence electrons to form Al³⁺; a fourth would come from the tightly held core. Selenium (4s²4p⁴) gains two electrons to fill 4p, forming Se²⁻; a third would have to start a new, more distant shell. 2(+3) + 3(−2) = 0, so Al₂Se₃, analogous to Al₂O₃.
(c) S²⁻ < Br⁻ < Se²⁻. S²⁻ has 18 electrons in three shells; Se²⁻ and Br⁻ have 36, with an occupied fourth shell, so S²⁻ is smallest. Se²⁻ and Br⁻ are isoelectronic, and Br⁻ has one more proton (35 against 34), so it pulls the same electron cloud in more tightly.
(d) Selenium’s valence electrons are in n = 4: farther out than sulfur’s (n = 3), closer than tellurium’s (n = 5), with similar effective nuclear charge. So the value lies between: (1000 + 869) ÷ 2 = 934.5, about 935 kJ mol⁻¹. (The measured value is 941 kJ mol⁻¹.)
| Point | What earns it |
|---|---|
| 1 | (a) Both configurations correct |
| 1 | (b) Al³⁺ and Se²⁻ with a valence-electron reason for each |
| 1 | (b) Al₂Se₃ with a charge check |
| 1 | (c) Correct order S²⁻ < Br⁻ < Se²⁻ |
| 1 | (c) Extra shell for Se²⁻ and Br⁻, and more protons in Br⁻ for the same electrons |
| 1 | (d) Estimate between 869 and 1000 kJ mol⁻¹, justified by the intermediate shell |
Total: 6 points.
How did you do?
Total: 30 points (3 multiple choice, 27 constructed response). Where you lost points matters more.
- Moles, spectra or formulas (Questions 1, 3, 4, 6): revisit Topics 1.1 to 1.4.
- Configurations, PES or trends (Questions 2, 5, 7): revisit Topics 1.5 to 1.7.
- Ion charges or analogous formulas (3, 4(e), 6(e), 7(b)): revisit Topic 1.8.
- Explanations marked down: name the cause (charge, distance, shielding or repulsion) and which way it pushes.
If several topics were weak, return to the Unit 1 diagnostic. Then tick off each topic checklist: