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: atomic numbers Li 3, Be 4, Na 11, Mg 12, S 16, Cl 17, Ar 18, K 19, Ti 22, Mn 25, Fe 26, As 33. Coulomb’s law: F ∝ q₁q₂ / r².
Question 1 (multiple choice · foundation)
Which is the ground-state electron configuration of an arsenic atom, As?
- (A) [Ar]4s²4p³
- (B) [Ar]4s²3d¹⁰4p³
- (C) [Ar]4s²3d¹⁰4p¹
- (D) [Ne]3s²3p⁶3d¹⁰4p⁵
Answer and explanation
Answer: (B). Arsenic has 33 electrons. Walking period 4 of the table: 4s², then all ten 3d, then three 4p. Check: 18 + 2 + 10 + 3 = 33.
- (A) skips the d block. It holds only 23 electrons, which is vanadium’s count, not arsenic’s.
- (C) has 31 electrons. That is gallium, the first p-block element in period 4.
- (D) has the right count (33) but leaves 4s empty and puts five electrons in 4p instead of three. In the ground state 4s fills before 3d and 4p.
Question 2 (multiple choice · core)
Which is the ground-state electron configuration of the Mn²⁺ ion?
- (A) [Ar]3d⁵
- (B) [Ar]4s²3d³
- (C) [Ar]4s²3d⁷
- (D) [Ar]4s¹3d⁴
Answer and explanation
Answer: (A). A manganese atom is [Ar]4s²3d⁵ (25 electrons). Mn²⁺ has 23 electrons. Electrons are removed from the highest shell first, n = 4, so both 4s electrons go: [Ar]3d⁵.
- (B) removes two 3d electrons, the “last in, first out” error. It has the right count (23) but the wrong subshells.
- (C) adds two electrons instead of removing them (27 electrons), which would describe a negative ion.
- (D) removes one 4s and one 3d electron. The 4s subshell must empty first.
Question 3 (multiple choice · core)
In arrangement X, a +2 charge and a −1 charge are a distance d apart. In arrangement Y, a +4 charge and a −1 charge are a distance 2d apart. How does the attractive force in Y compare with the force in X?
- (A) F(Y) = ¼ F(X)
- (B) F(Y) = ½ F(X)
- (C) F(Y) = F(X)
- (D) F(Y) = 2 F(X)
Answer and explanation
Answer: (B). Write each force as a ratio: F(X) ∝ (2 × 1) / d² = 2/d². F(Y) ∝ (4 × 1) / (2d)² = 4/(4d²) = 1/d². So F(Y) / F(X) = 1/2.
- (A) applies the distance effect (¼) but forgets that the charge doubled.
- (C) uses 1/r instead of 1/r²: doubling the charge and doubling the distance would then cancel.
- (D) doubles for the charge and ignores the distance completely.
Question 4 (configurations · core)
Write the ground-state electron configuration for each species. Use noble-gas shorthand for (b) to (d).
(a) S (full configuration) (b) S²⁻ (c) Ti (d) Ti²⁺ (e) How many core electrons and how many valence electrons does a sulfur atom have?
Worked solution
(a) 1s²2s²2p⁶3s²3p⁴ (2 + 2 + 6 + 2 + 4 = 16). (b) S²⁻ has 18 electrons: two more go into 3p, giving 1s²2s²2p⁶3s²3p⁶ = [Ar]. (c) Ti has 22 electrons: [Ar]4s²3d². (d) Ti²⁺ has 20 electrons. Remove from n = 4 first: [Ar]3d². (e) 10 core electrons (the [Ne] core) and 6 valence electrons (3s²3p⁴).
Suggested mark points (5): 1 for each of (a) to (d); 1 for both counts in (e). Accept [Ar]3d²4s² for Ti (shell-number order). Do not accept [Ar]4s² for Ti²⁺.
Question 5 (calculation · core)
A student uses a simple Coulomb model for four electrons, P, Q, R and S. Each electron feels a core charge q at an average distance r from the nucleus. The values below are rounded model values made up for this question.
| Electron | Core charge q | Distance r (pm) |
|---|---|---|
| P | +1 | 180 |
| Q | +2 | 150 |
| R | +6 | 100 |
| S | +1 | 240 |
(a) Calculate the force on each electron as a multiple of the force on P. (b) Rank the electrons from hardest to easiest to remove, according to this model.
Worked solution
(a) Use F ∝ q / r² (the electron’s charge is the same each time, so it cancels in the ratio).
- Q: (2 / 150²) ÷ (1 / 180²) = 2 × (180/150)² = 2 × 1.44 = 2.88
- R: (6 / 100²) ÷ (1 / 180²) = 6 × (180/100)² = 6 × 3.24 = 19.4
- S: (1 / 240²) ÷ (1 / 180²) = (180/240)² = 0.75² = 0.56
- P: 1 (by definition)
(b) Larger force means harder to remove: R > Q > P > S.
Suggested mark points (3): 1 for setting up the ratio with r squared; 1 for all three ratios correct (to 2 or 3 significant figures); 1 for a ranking consistent with the ratios. Carry forward an error from (a) into (b).
Common error: forgetting to square r gives Q = 2.4, R = 10.8 and S = 0.75. The ranking happens to be the same here, but the numbers are wrong, so do not rely on the ranking to check your method.
Question 6 (constructed response · core)
An ion M³⁺ has a mass number of 56 and contains 23 electrons.
(a) How many protons and neutrons does the ion have? Identify element M. (b) Write the noble-gas configurations of the M atom and of the M³⁺ ion. (c) A student says: “The 3d subshell fills after 4s, so when M forms M³⁺ the three electrons come from 3d.” Evaluate this claim. (d) Name a 2+ ion that has the same electron configuration as M³⁺.
Worked solution and suggested Marlbridge rubric
(a) A 3+ ion has 3 fewer electrons than protons: protons = 23 + 3 = 26, so M is iron, Fe. Neutrons = 56 − 26 = 30.
(b) Fe atom: [Ar]4s²3d⁶. Fe³⁺ (23 electrons): remove both 4s electrons, then one 3d electron: [Ar]3d⁵.
(c) The claim is incorrect. The order of filling does not decide the order of removal. Once 3d holds electrons, the 4s electrons are in the highest shell (n = 4), farthest from the nucleus on average, so they are lost first. Only the third electron comes from 3d. Removing three 3d electrons would give [Ar]4s²3d³, which has the right count but is not the ground state.
(d) Mn²⁺: manganese has 25 protons, so Mn²⁺ also has 23 electrons, [Ar]3d⁵.
| Point | What earns it |
|---|---|
| 1 | 26 protons, 30 neutrons and M = Fe |
| 1 | Both configurations correct: [Ar]4s²3d⁶ and [Ar]3d⁵ |
| 1 | States the claim is wrong and explains that 4s electrons (largest n) are removed first |
| 1 | Mn²⁺ |
Accept [Ar]3d⁶4s² for the atom. Do not award the point in (c) for “4s goes first” with no reason.
Question 7 (explanation · stretch)
Use Coulomb’s law to explain each observation.
(a) Removing an electron from the 1s subshell of a lithium atom needs far more energy than removing the 2s electron. (b) Removing a 2s electron from a beryllium atom needs more energy than removing the 2s electron from a lithium atom.
Model answer and suggested Marlbridge rubric
(a) Lithium (1s²2s¹) has 3 protons. The 2s electron is in shell 2, so it is farther from the nucleus, and the two 1s electrons shield it: it feels a core charge of only about 3 − 2 = +1. A 1s electron is in shell 1, much closer, and has no inner shell shielding it (only the other 1s electron shields it a little), so it feels most of the +3 charge. By F ∝ q₁q₂ / r², a larger charge at a smaller distance gives a much stronger attraction, so more energy is needed to remove it.
(b) Both electrons are in the 2s subshell, at a similar distance and shielded by the same two 1s electrons. Beryllium has 4 protons, so its 2s electrons feel a core charge of about +2 against +1 for lithium. A larger effective charge gives a stronger attraction, so more energy is needed.
| Point | What earns it |
|---|---|
| 1 | (a) The 1s electron is closer to the nucleus (smaller r) |
| 1 | (a) The 1s electron is not shielded by inner electrons / feels a larger effective charge than the 2s electron |
| 1 | (b) Same subshell, so similar distance and the same shielding by 1s² |
| 1 | (b) Be has more protons, so a larger effective nuclear charge and a stronger attraction |
Accept “effective nuclear charge” or “core charge”. In (a), an answer that gives only one factor (distance or shielding) earns at most 1 point. In (b), an answer that says only “Be has more protons”, without noting the similar distance and equal shielding, earns at most 1 point.
How did you do?
- Q1 or Q4 wrong: walk the periodic table block by block, as in Figure 1 of the study guide.
- Q2 or Q6 wrong: re-read “Configurations of ions” and Worked example 1. Remove electrons from the largest n first.
- Q3 or Q5 wrong: practise the ratio method; remember the r².
- Q7 incomplete: name both factors, distance and effective charge, as in Worked example 2.
Then tick off the topic checklist.