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: ionic radii (pm) Li⁺ 76, Mg²⁺ 72, Al³⁺ 53.5, K⁺ 138, Ca²⁺ 100, F⁻ 133, Cl⁻ 181, Br⁻ 196, O²⁻ 140, S²⁻ 184; N_A = 6.022 × 10²³ mol⁻¹. The molecule in Questions 1 and 4 is fictional, and its data are invented for practice.
Question 1 (multiple choice · foundation)
The potential energy curve for a diatomic molecule Q₂ falls from high positive values, crosses zero at about 125 pm, reaches a minimum of −250 kJ mol⁻¹ at 160 pm, then rises towards zero at large distances. Which statement is correct?
- (A) The bond length is 160 pm and 250 kJ is needed to break one mole of Q–Q bonds.
- (B) The bond length is 125 pm and 250 kJ is needed to break one mole of Q–Q bonds.
- (C) The bond length is 160 pm and 250 kJ is released when one mole of Q–Q bonds is broken.
- (D) The bond length is 125 pm and the bond energy is zero.
Answer and explanation
Answer: (A). The bond length is the distance at the minimum (160 pm). The bond energy is the depth of the well, 250 kJ mol⁻¹, and it must be supplied to separate the atoms.
- (B) reads the bond length where the curve crosses zero. At 125 pm the atoms are closer than their equilibrium distance and repulsion is raising the energy.
- (C) has the right distance but the wrong direction: breaking a bond absorbs energy. 250 kJ is released when the bonds form.
- (D) combines the zero-crossing error with the idea that the energy at that point is the bond energy.
Question 2 (multiple choice · core)
Which bond is the longest?
- (A) H–F
- (B) H–Cl
- (C) H–Br
- (D) H–I
Answer and explanation
Answer: (D). All four are single bonds to hydrogen, so bond order is the same; only the size of the halogen changes. Iodine is lowest in group 17, with the most occupied shells and the largest core, so its nucleus stays furthest from the hydrogen nucleus.
- (A) H–F is the shortest: fluorine is the smallest halogen. Choosing it confuses “most electronegative” with “longest bond”.
- (B) and (C) have halogens smaller than iodine, so their bonds are shorter than H–I.
Question 3 (multiple choice · core)
Which pair of ions has the strongest attraction when the ions are in contact?
- (A) Ca²⁺ and O²⁻
- (B) Ca²⁺ and S²⁻
- (C) K⁺ and Cl⁻
- (D) K⁺ and Br⁻
Answer and explanation
Answer: (A). By Coulomb’s law the attraction is proportional to q₁q₂ / r². Ca²⁺O²⁻ has a charge product of 4 and the shortest distance (100 + 140 = 240 pm), so q₁q₂ / r² = 6.94 × 10⁻⁵ pm⁻².
- (B) Same charges as (A), but S²⁻ is larger (r = 284 pm), so 4.96 × 10⁻⁵ pm⁻²: weaker.
- (C) and (D) have a charge product of only 1 and larger distances (319 pm and 334 pm). Both are several times weaker; (D) is the weakest because Br⁻ is larger than Cl⁻.
Question 4 (graphing · core)
The table gives model potential energies for the fictional molecule X₂.
| Internuclear distance (pm) | 100 | 110 | 120 | 130 | 140 | 150 | 160 | 180 | 200 | 250 | 300 | 400 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Potential energy (kJ mol⁻¹) | +151 | −97 | −227 | −285 | −300 | −290 | −267 | −209 | −154 | −63 | −24 | −3 |
(a) Plot the data and draw a smooth curve. Choose and label your axes, with units and a suitable scale. (b) Use your graph to state the bond length and the bond energy of X₂. (c) Explain why the potential energy is positive at 100 pm. (d) Explain why the potential energy is close to zero at 400 pm.
Worked solution and suggested Marlbridge rubric
(a) Distance on the horizontal axis (for example 0 to 400 pm, 1 cm = 25 pm). Potential energy on the vertical axis (for example −350 to +200 kJ mol⁻¹), with zero clearly marked. Plot all twelve points and draw a smooth curve through them: steep on the left, a rounded minimum, and a long, gentle rise towards zero.
(b) The minimum is at 140 pm and −300 kJ mol⁻¹. The bond length is 140 pm; the bond energy is 300 kJ mol⁻¹ (energy needed to separate the atoms).
(c) At 100 pm the nuclei are much closer than the bond length. Repulsion between the two positive nuclei (and between the electron clouds) outweighs the attractions, so the energy is above that of the separated atoms.
(d) At 400 pm the atoms are so far apart that they barely interact: the attractions and repulsions are both negligible, so the energy approaches the zero chosen for separated atoms.
| Point | What earns it |
|---|---|
| 1 | Axes labelled with quantity and unit (pm; kJ mol⁻¹) |
| 1 | Linear scales that use most of the grid and include all data, including the positive value |
| 1 | Points plotted accurately and joined by a smooth curve with one minimum |
| 1 | Bond length 140 pm and bond energy 300 kJ mol⁻¹ (positive) |
| 1 | (c) repulsion between nuclei dominates at short distance |
| 1 | (d) atoms too far apart to interact, so energy approaches zero |
Question 5 (data analysis · core)
Three nitrogen–nitrogen bonds have these average values:
| Bond | Length (pm) | Bond energy (kJ mol⁻¹) |
|---|---|---|
| P | 125 | 418 |
| R | 145 | 160 |
| T | 110 | 941 |
(a) Identify which of P, R and T is the single, double and triple bond. (b) Explain the relationship between bond order, bond length and bond energy shown by the data. (c) Describe how the potential energy curve for bond T would differ from the curve for bond R if both were drawn on the same axes.
Worked solution and suggested Marlbridge rubric
(a) R (longest, weakest) is the single bond, N–N. P is the double bond, N=N. T (shortest, strongest) is the triple bond, N≡N.
(b) As bond order increases from 1 to 3, more electron pairs are shared between the same two nuclei. The nuclei are pulled closer (145 → 125 → 110 pm) and held more strongly (160 → 418 → 941 kJ mol⁻¹). Higher bond order: shorter and stronger.
(c) T’s minimum is further left (110 pm instead of 145 pm) and much deeper (−941 instead of −160 kJ mol⁻¹). Both curves rise steeply at short distance and level off at zero far apart.
| Point | What earns it |
|---|---|
| 1 | All three assigned correctly |
| 1 | Higher bond order linked to more shared pairs (or more electron density) between the nuclei |
| 1 | Correct trend: shorter and larger bond energy with higher order, using the data |
| 1 | T’s minimum at shorter distance |
| 1 | T’s well deeper, and both curves approach zero at large distance |
Question 6 (constructed response · core)
Each of the cations Li⁺, Mg²⁺ and Al³⁺ is placed in contact with a fluoride ion, F⁻.
(a) Calculate q₁q₂ / r² for each ion pair, where r is the sum of the ionic radii in pm. (b) Express each result relative to the Li⁺–F⁻ pair. (c) Explain the ranking. Which factor, charge or size, has the larger effect here?
Worked solution and suggested Marlbridge rubric
(a)
| Pair | q₁q₂ | r (pm) | q₁q₂ / r² (pm⁻²) |
|---|---|---|---|
| Li⁺–F⁻ | 1 | 76 + 133 = 209 | 2.29 × 10⁻⁵ |
| Mg²⁺–F⁻ | 2 | 72 + 133 = 205 | 4.76 × 10⁻⁵ |
| Al³⁺–F⁻ | 3 | 53.5 + 133 = 186.5 | 8.63 × 10⁻⁵ |
(b) Relative to Li⁺–F⁻: 1.00 : 2.08 : 3.77.
(c) Order of attraction: Al³⁺–F⁻ > Mg²⁺–F⁻ > Li⁺–F⁻. The cation charge rises from +1 to +3, which on its own multiplies the attraction by 2 and then 3. The cations also get slightly smaller, which shortens r and adds a little more. Charge has the larger effect: the radius changes by only a few percent from Li⁺ to Mg²⁺.
| Point | What earns it |
|---|---|
| 1 | Correct r values for all three pairs |
| 1 | Correct q₁q₂ / r² values (3 significant figures) |
| 1 | Correct relative values 2.08 and 3.77 (allow 2.1 and 3.8) |
| 1 | Correct ranking, linked to Coulomb’s law |
| 1 | Identifies charge as the dominant factor, with size as a smaller, reinforcing effect |
Using the energy form q₁q₂ / r gives relative values 1.00 : 2.04 : 3.36 and the same ranking; accept it if stated.
Question 7 (explanation · stretch)
A student writes: “To make a bond, energy has to be put in to push the two atoms together. The most stable arrangement is when the nuclei are as close as possible.”
Evaluate both sentences, using a potential energy curve in your answer.
Model answer and suggested Marlbridge rubric
Model answer. Both sentences are wrong.
First sentence. As two atoms approach from far apart, the attraction of each nucleus for the other atom’s electrons lowers the potential energy. The curve goes down towards the minimum, so energy is released when the bond forms. Energy has to be put in only to break the bond.
Second sentence. The most stable arrangement is at the minimum of the curve, the equilibrium bond length. If the nuclei are pushed closer than this, repulsion between the positive nuclei (and between the electrons) dominates and the potential energy rises steeply, so the arrangement becomes less stable, not more.
| Point | What earns it |
|---|---|
| 1 | Energy decreases as the atoms approach (attraction), so bond formation releases energy |
| 1 | Energy is needed to break, not to form, a bond |
| 1 | Most stable at the minimum (bond length), not at the closest distance |
| 1 | Nucleus–nucleus repulsion explains the steep rise at short distance |
How did you do?
- Q1 or Q4 wrong: re-read “Reading a potential energy graph” and Worked example 1 in the study guide.
- Q2 or Q5 wrong: look again at the two factors for covalent bonds and Figure 2.
- Q3 or Q6 wrong: practise Worked example 3; check both charge and distance every time.
- Q7 incomplete: your answer needs both attraction and repulsion, and the direction of energy change.
Then tick off the topic checklist.