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. Use a periodic table. Ionization energies (kJ mol⁻¹, rounded) are given where a question needs them. In Question 4 the data describe an unidentified element and are rounded.
Question 1 (multiple choice · foundation)
Which list puts the atoms in order of increasing atomic radius?
- (A) F < Li < Na
- (B) Na < Li < F
- (C) Li < F < Na
- (D) Li < Na < F
Answer and explanation
Answer: (A). Li and F are both in period 2. Fluorine has 9 protons and lithium has 3, but both have the same core (1s²). Fluorine’s larger effective nuclear charge pulls its n = 2 shell in more, so F is smaller than Li. Sodium is below lithium in group 1 and has its outer electron in the n = 3 shell, so Na is larger than Li.
- (B) is the decreasing order, the right trend read backwards.
- (C) assumes that more electrons make a bigger atom across a period. Across a period, radius decreases.
- (D) gets Li < Na right but puts F as the largest atom, which ignores the decrease in radius across period 2.
Question 2 (multiple choice · core)
The first ionization energy of sulfur (1000 kJ mol⁻¹) is slightly lower than that of phosphorus (1012 kJ mol⁻¹). Which statement best explains this?
- (A) Sulfur has a smaller nuclear charge than phosphorus.
- (B) The electron removed from sulfur is in a higher shell than the electron removed from phosphorus.
- (C) The electron removed from sulfur is in a 3p orbital that already holds another electron, and the repulsion between the two makes it easier to remove.
- (D) Sulfur has more core electrons than phosphorus, so its valence electrons are more shielded.
Answer and explanation
Answer: (C). Phosphorus is [Ne] 3s² 3p³, with one electron in each 3p orbital. Sulfur is [Ne] 3s² 3p⁴, so one 3p orbital holds a pair. Repulsion between the paired electrons makes one of them easier to remove, and this outweighs sulfur’s extra proton.
- (A) is false: sulfur has 16 protons and phosphorus has 15.
- (B) is false: in both atoms the electron comes from the 3p subshell of the n = 3 shell.
- (D) is false: both atoms have the same 10 core electrons (1s² 2s² 2p⁶), so core shielding is the same.
Question 3 (multiple choice · core)
Which of these elements has the highest electronegativity: nitrogen, oxygen, sulfur or chlorine?
- (A) Nitrogen
- (B) Oxygen
- (C) Sulfur
- (D) Chlorine
Answer and explanation
Answer: (B). Pauling values: N 3.04, O 3.44, S 2.58, Cl 3.16. Oxygen is to the right of nitrogen in period 2, so it has a larger effective nuclear charge acting on the same shell. Oxygen is above sulfur in group 16, so its bonding electrons are closer to the nucleus.
- (A) Nitrogen is in the same period as oxygen but further left, so its effective nuclear charge is smaller.
- (C) Sulfur is below oxygen; its bonding electrons are in the n = 3 shell, farther from the nucleus.
- (D) is the common trap. Chlorine is further right than oxygen, but its bonding electrons are in the n = 3 shell. The extra distance outweighs the larger nuclear charge, so chlorine is less electronegative than oxygen.
Question 4 (multiple choice · core)
The first four ionization energies of an element in period 3 are 580, 1820, 2750 and 11 600 kJ mol⁻¹. Which ion is this element most likely to form in its compounds?
- (A) X⁺
- (B) X²⁺
- (C) X³⁺
- (D) X⁴⁺
Answer and explanation
Answer: (C). The increases are 1240, 930 and 8850 kJ mol⁻¹. The huge jump is between the third and fourth ionization energies, so the fourth electron must come from an inner shell (n = 2). The element has three valence electrons (it is in group 13), so it forms X³⁺.
- (A) would need the big jump after the first ionization energy (group 1).
- (B) would need the big jump after the second (group 2). The rise from 580 to 1820 (1240) is larger than the next rise (930), but it is small compared with the 8850 kJ mol⁻¹ jump that marks a new shell.
- (D) would need four valence electrons, but removing the fourth electron costs about four times as much energy as the third, which shows it comes from the core.
Question 5 (constructed response · core)
The first ionization energies of three period 3 elements are: Na 496, Mg 738 and Al 578 kJ mol⁻¹.
(a) Explain why the first ionization energy of magnesium is greater than that of sodium. (b) Explain why the first ionization energy of aluminium is less than that of magnesium. (c) Predict whether the first ionization energy of silicon is greater or less than that of aluminium. Justify your prediction.
Worked solution and suggested Marlbridge rubric
(a) Both electrons are removed from the 3s subshell, the same distance from the nucleus. Magnesium has 12 protons and sodium has 11, with the same 10 core electrons shielding. Magnesium’s effective nuclear charge is larger, so by Coulomb’s law its 3s electron is attracted more strongly and needs more energy to remove.
(b) Aluminium ([Ne] 3s² 3p¹) loses a 3p electron, but magnesium ([Ne] 3s²) loses a 3s electron. The 3p electron is in a higher-energy subshell and is partly shielded by the 3s electrons, so it is held less strongly. This outweighs aluminium’s extra proton.
(c) Greater. Silicon ([Ne] 3s² 3p²) also loses a 3p electron from the same shell. It has one more proton than aluminium with the same core shielding, and its two 3p electrons are in separate orbitals, so there is no pairing repulsion. The larger effective nuclear charge holds the electron more strongly.
| Point | What earns it |
|---|---|
| 1 | (a) Mg has a greater nuclear charge (or Z_eff) and the electron is removed from the same shell or subshell with the same shielding |
| 1 | (b) Identifies that the Al electron is removed from 3p, while the Mg electron is removed from 3s |
| 1 | (b) Links the 3p electron to higher energy or extra shielding by the 3s electrons, so it is held less strongly |
| 1 | (c) Predicts “greater” with a reason: more protons, same shell, no new subshell or pairing |
Do not award (a) for “magnesium has a full 3s subshell” alone; the answer must name the attraction. In (c), a prediction without a reason earns no point.
Question 6 (constructed response · core)
The first ionization energies of potassium and caesium are 419 and 376 kJ mol⁻¹. Rubidium lies between them in group 1.
(a) Estimate the first ionization energy of rubidium. (b) Justify the direction of the trend from potassium to caesium using Coulomb’s law. (c) Strontium is next to rubidium in period 5. Predict whether strontium’s first ionization energy is greater or less than rubidium’s, and justify your prediction.
Worked solution and suggested Marlbridge rubric
(a) Rubidium’s value should lie between the two. Mean: (419 + 376) ÷ 2 = 397.5, so about 398 kJ mol⁻¹. Any value between about 380 and 415 kJ mol⁻¹ is a reasonable estimate. (The measured value is 403 kJ mol⁻¹, so the mean is within about 1.5%.)
(b) Each atom loses one ns¹ electron, and the effective nuclear charge acting on it is roughly the same (about +1 by the simple estimate). Going down the group the outer electron is in a higher shell: 4s for K, 5s for Rb, 6s for Cs. It is farther from the nucleus, so by Coulomb’s law the attraction is weaker and less energy is needed to remove it.
(c) Greater. Rubidium is [Kr] 5s¹ and strontium is [Kr] 5s². Both lose a 5s electron from the same shell, with the same core shielding, but strontium has one more proton. Its larger effective nuclear charge holds the 5s electron more strongly.
| Point | What earns it |
|---|---|
| 1 | (a) An estimate between the two given values, with a method (such as the mean) |
| 1 | (b) Outer electron in a higher shell, farther from the nucleus, with similar Z_eff |
| 1 | (b) Links greater distance to weaker attraction (Coulomb’s law), so less energy needed |
| 1 | (c) “Greater”, because Sr has more protons, the same shell and the same core shielding |
An answer to (b) that says only “more shielding” earns at most one point: it must connect distance (or shielding) to a weaker attraction.
Question 7 (constructed response · stretch)
A student writes: “A chloride ion, Cl⁻, is smaller than a chlorine atom, Cl. The extra electron is attracted to the nucleus, so it pulls the electron cloud inward.”
(a) Evaluate the student’s claim. (b) The ions K⁺ and Cl⁻ both have 18 electrons. Identify the larger ion and explain your answer. (c) Less energy is released when an electron is added to a gaseous bromine atom than when one is added to a gaseous chlorine atom. Explain this difference.
Worked solution and suggested Marlbridge rubric
(a) The claim is incorrect. Cl⁻ is larger than Cl. The nuclear charge (17 protons) does not change, but the ion has 18 electrons instead of 17. The extra electron adds electron–electron repulsion in the outer shell, and the same nuclear charge is now spread over more electrons, so the electron cloud expands. An added electron does not pull other electrons inward; electrons repel each other.
(b) Cl⁻ is larger. Both ions have the configuration 1s² 2s² 2p⁶ 3s² 3p⁶. K⁺ has 19 protons and Cl⁻ has 17. With the same number of electrons in the same shells, the larger nuclear charge of K⁺ pulls the electron cloud closer, so K⁺ is smaller.
(c) The added electron enters the 4p subshell in bromine but the 3p subshell in chlorine. In bromine it is farther from the nucleus, with a similar effective nuclear charge, so by Coulomb’s law it is attracted less strongly and less energy is released.
| Point | What earns it |
|---|---|
| 1 | (a) States that Cl⁻ is larger, so the claim is wrong |
| 1 | (a) Same number of protons with more electrons, so more repulsion (or less pull per electron) |
| 1 | (b) Cl⁻ is larger because it has fewer protons (17 vs 19) for the same 18 electrons |
| 1 | (c) Added electron is in a higher shell (n = 4 vs n = 3), farther from the nucleus, so weaker attraction |
In (a), “the claim is wrong” without a reason does not earn the second point. In (b), stating the larger ion without comparing proton numbers earns no point.
How did you do?
- Q1 or Q3 wrong: re-read “The model behind every trend” and Figure 1 in the study guide.
- Q2, Q4 or Q5 wrong: work through Figure 2 and Worked example 2 (the dips and successive ionization energies).
- Q6 or Q7 incomplete: your explanation needs the cause (nuclear charge, distance, shielding or repulsion), not only the trend. See Worked examples 1 and 3.
Then tick off the topic checklist.