These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: binding energies are in MJ mol⁻¹ (1 MJ mol⁻¹ = 1000 kJ mol⁻¹); every spectrum is illustrative, with values chosen to follow the real pattern rather than measured data; peaks are listed from highest to lowest binding energy; use a periodic table for atomic numbers.
Question 1 (multiple choice · foundation)
The photoelectron spectrum of a neutral atom has five peaks. From highest to lowest binding energy, their relative heights are 2, 2, 6, 2 and 1. Which element is it?
- (A) Sodium
- (B) Magnesium
- (C) Aluminium
- (D) Silicon
Answer and explanation
Answer: (C). The heights give the configuration 1s² 2s² 2p⁶ 3s² 3p¹. The total is 2 + 2 + 6 + 2 + 1 = 13 electrons, so the atom has 13 protons: aluminium.
- (A) Sodium (1s² 2s² 2p⁶ 3s¹) would give only four peaks, the last with height 1.
- (B) Magnesium (12 electrons) is what you get if you ignore the small final peak. A small peak is still a subshell.
- (D) Silicon (1s² 2s² 2p⁶ 3s² 3p²) would need a 3p peak as tall as the 3s peak.
Question 2 (multiple choice · core)
In their photoelectron spectra, the 1s peak of oxygen is at a higher binding energy than the 1s peak of nitrogen. Which statement best explains this?
- (A) Oxygen has one more proton, so its 1s electrons, at a similar distance from the nucleus, are attracted more strongly.
- (B) Oxygen has more electrons in its 1s subshell than nitrogen does.
- (C) Oxygen’s extra 2p electron shields its 1s electrons from the nucleus.
- (D) Oxygen’s 1s electrons are farther from the nucleus than nitrogen’s.
Answer and explanation
Answer: (A). By Coulomb’s law, attraction increases with the charges involved. Oxygen’s nucleus is +8 against nitrogen’s +7, and the 1s electrons in both atoms are innermost, so oxygen holds its 1s electrons more tightly.
- (B) Both atoms have exactly 2 electrons in 1s, so the 1s peaks have the same height.
- (C) An electron in an outer subshell does not shield an inner one. Even if it did, more shielding would lower the binding energy, not raise it.
- (D) Greater distance would lower the binding energy. A larger nuclear charge actually pulls the 1s electrons slightly closer.
Question 3 (multiple choice · core)
How does the photoelectron spectrum of Mg²⁺ differ from that of a magnesium atom, Mg?
- (A) The Mg²⁺ spectrum has no 3s peak, and its remaining peaks have relative heights 2 : 2 : 6.
- (B) The Mg²⁺ spectrum has no 1s peak, because the electrons closest to the nucleus are removed first.
- (C) Both spectra have four peaks, but the 3s peak of Mg²⁺ is half as tall.
- (D) The Mg²⁺ spectrum has an extra 3p peak.
Answer and explanation
Answer: (A). Mg is 1s² 2s² 2p⁶ 3s². Forming Mg²⁺ removes the two 3s electrons, the most loosely held, leaving 1s² 2s² 2p⁶. So there is no 3s peak, and the three remaining peaks have heights 2, 2 and 6.
- (B) The 1s electrons are the hardest to remove, so they are the last to go, not the first.
- (C) A 3s peak of half height (one electron) describes Mg⁺, not Mg²⁺.
- (D) A cation has fewer electrons than the atom; there is nothing to put into a new subshell.
Question 4 (calculation · core)
A spectrum of a neutral atom has the following peaks. The heights are raw readings from the graph, in arbitrary units.
| Peak | Binding energy (MJ mol⁻¹) | Height (arbitrary units) |
|---|---|---|
| 1 | 239 | 3.9 |
| 2 | 22.7 | 4.1 |
| 3 | 16.5 | 12.0 |
| 4 | 2.05 | 4.0 |
| 5 | 1.00 | 8.1 |
(a) Work out the number of electrons in each subshell. (b) Write the electron configuration and identify the element. (c) How many core electrons and how many valence electrons does the atom have? Which peak gives an estimate of the first ionization energy, and what is that estimate in kJ mol⁻¹?
Worked solution
(a) Divide each height by the smallest (3.9): 1.0, 1.05, 3.08, 1.03, 2.08. The first peak must be 1s, which holds 2 electrons, so multiply by 2: 2.0, 2.1, 6.15, 2.05, 4.15. Round to whole electrons: 2, 2, 6, 2, 4. The heights are only ideally proportional, so small deviations are expected.
(b) 1s² 2s² 2p⁶ 3s² 3p⁴. Total 16 electrons, so 16 protons: sulfur.
(c) Core electrons (shells 1 and 2): 2 + 2 + 6 = 10. Valence electrons (shell 3): 2 + 4 = 6. Peak 5, the lowest binding energy, estimates the first ionization energy: 1.00 MJ mol⁻¹ = 1000 kJ mol⁻¹.
Suggested mark points (3): 1 for the ratio 2 : 2 : 6 : 2 : 4 with working; 1 for the configuration and sulfur; 1 for 10 core and 6 valence electrons and peak 5 with 1000 kJ mol⁻¹.
Common error: rounding 12.0 ÷ 3.9 = 3.08 to “3 electrons” without scaling by 2. The 1s peak sets the scale at 2 electrons.
Question 5 (constructed response · core)
Two neighbouring elements in period 2, Y and Z, give these spectra.
| Element | Peak binding energies (MJ mol⁻¹) and relative heights |
|---|---|
| Y | 39.6 (2), 2.45 (2), 1.40 (3) |
| Z | 52.6 (2), 3.04 (2), 1.31 (4) |
(a) Write the electron configuration of each element and identify Y and Z. (b) Explain, in terms of Coulomb’s law, why the 1s peak of Z is at a higher binding energy than the 1s peak of Y. (c) Convert the lowest-energy peak of Y to kJ mol⁻¹ and state what quantity this value estimates.
Worked solution and suggested Marlbridge rubric
(a) Y: 1s² 2s² 2p³, 7 electrons, so Y is nitrogen. Z: 1s² 2s² 2p⁴, 8 electrons, so Z is oxygen.
(b) Z (oxygen) has 8 protons; Y (nitrogen) has 7. The 1s electrons in both atoms are the innermost electrons, at a similar distance from the nucleus, with no electrons between them and the nucleus. Coulomb’s law says the attraction increases with the size of the charges, so oxygen’s larger nuclear charge holds its 1s electrons more strongly, and more energy is needed to remove one.
(c) 1.40 MJ mol⁻¹ × 1000 = 1400 kJ mol⁻¹. This estimates the first ionization energy of nitrogen, because it is the energy to remove the most loosely held (2p) electron from the atom.
| Point | What earns it |
|---|---|
| 1 | Both configurations correct, with Y = N and Z = O |
| 1 | States that Z has more protons (greater nuclear charge) than Y |
| 1 | Links greater nuclear charge to a stronger attraction for 1s electrons at a similar distance, so a higher binding energy |
| 1 | 1400 kJ mol⁻¹ and identifies it as an estimate of the first ionization energy |
Do not award the third point for “Z has more electrons” or “Z has more shells”: the 1s subshells are identical in both atoms.
Looking ahead: you may notice that Z’s 2p peak (1.31) is slightly lower than Y’s (1.40), even though oxygen has more protons. Topic 1.7 explains this exception to the ionization energy trend.
Question 6 (model analysis · core)
A student uses a simple shell model in which every electron in the same shell has the same energy.
(a) Using this model, predict the number of peaks in the spectrum of silicon and the relative height of each. (b) The illustrative spectrum of silicon has peaks at 178 (2), 15.1 (2), 10.3 (6), 1.46 (2) and 0.79 (2) MJ mol⁻¹, with relative heights in brackets. Explain whether the simple shell model is consistent with these data. Refer to specific features of the spectrum. (c) Explain why the peak at 178 MJ mol⁻¹ is at a much higher binding energy than the peak at 0.79 MJ mol⁻¹.
Worked solution and suggested Marlbridge rubric
(a) Silicon has 14 electrons: 2 in shell 1, 8 in shell 2 and 4 in shell 3. The model predicts three peaks with relative heights 2, 8 and 4.
(b) The model is not consistent with the data. The spectrum has five peaks, not three. The 8 electrons of shell 2 appear as two peaks (heights 2 and 6, at 15.1 and 10.3 MJ mol⁻¹), and the 4 electrons of shell 3 appear as two peaks (heights 2 and 2, at 1.46 and 0.79 MJ mol⁻¹). So electrons in the same shell do not all have the same energy: each shell is split into s and p subshells, as in the configuration 1s² 2s² 2p⁶ 3s² 3p².
(c) The 178 MJ mol⁻¹ peak is the 1s subshell; the 0.79 MJ mol⁻¹ peak is the 3p subshell. The 1s electrons are much closer to the nucleus (smaller r) and have no electrons shielding them, so they feel almost the full +14 nuclear charge. The 3p electrons are in the outermost shell, much farther away, and are shielded by 10 core electrons. By Coulomb’s law the attraction on the 1s electrons is far stronger, so far more energy is needed to remove one.
| Point | What earns it |
|---|---|
| 1 | Three peaks with heights 2, 8, 4 from the shell model |
| 1 | States the model is not consistent and cites five peaks (or the split of a shell into two peaks) as evidence |
| 1 | Distance: 1s electrons are much closer to the nucleus than 3p electrons |
| 1 | Shielding/effective nuclear charge: 3p electrons are shielded by core electrons, linked to weaker attraction through Coulomb’s law |
Accept any one shell split described correctly for the second point. A statement like “the model is wrong” without citing data does not earn it.
Question 7 (constructed response · stretch)
S²⁻, Ar and Ca²⁺ all have 18 electrons.
(a) State two features that are the same in the photoelectron spectra of these three species. (b) Rank the three species in order of increasing binding energy of their 1s peaks. Justify your ranking. (c) A fourth species with 18 electrons has its 1s peak between those of Ar and Ca²⁺. Identify it.
Model answer and suggested Marlbridge rubric
(a) All three are 1s² 2s² 2p⁶ 3s² 3p⁶, so their spectra have the same number of peaks (five) and the same relative heights (2 : 2 : 6 : 2 : 6).
(b) S²⁻ < Ar < Ca²⁺. The species have the same electrons in the same subshells, so the distance and shielding of the 1s electrons are similar. What differs is the nuclear charge: +16 for sulfur, +18 for argon and +20 for calcium. By Coulomb’s law, a larger nuclear charge attracts the 1s electrons more strongly, so more energy is needed to remove one.
(c) The nuclear charge must lie between +18 and +20, so it is +19: the species is K⁺.
| Point | What earns it |
|---|---|
| 1 | Two correct shared features (number of peaks, relative heights, or the same configuration stated as the reason) |
| 1 | Correct order S²⁻ < Ar < Ca²⁺ |
| 1 | Justification based on increasing nuclear charge (number of protons) with the same electron arrangement |
| 1 | K⁺ |
Do not award the justification point for “Ca²⁺ has lost electrons”: all three species have the same number of electrons.
How did you do?
- Q1 or Q4 wrong: re-read “How to read a spectrum” and Worked example 1 in the study guide.
- Q2, Q5 or Q7 incomplete: your explanation needs both Coulomb’s law factors: distance and the charge the electron feels.
- Q3 wrong: write the ion’s configuration first, then draw its spectrum.
- Q6 incomplete: a model judgement must quote the data that agree or disagree with it.
Then tick off the topic checklist.