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: valence electrons H 1, C 4, N 5, O 6, Cl 7. Electronegativity order used here: O > Cl > N > C > H. Formal charge FC = valence electrons − nonbonding electrons − ½ × bonding electrons.
Question 1 (multiple choice · foundation)
In the Lewis diagram of the chlorate ion, ClO₃⁻, in which every atom has an octet, chlorine has three single bonds to oxygen and one lone pair. What is the formal charge on the chlorine atom?
- (A) −1
- (B) 0
- (C) +2
- (D) +4
Answer and explanation
Answer: (C). Cl has 7 valence electrons, 2 nonbonding electrons (one lone pair) and 6 bonding electrons (three bonds): FC = 7 − 2 − ½(6) = +2. Check: each O has one bond and three lone pairs, FC = 6 − 6 − 1 = −1, and +2 + 3(−1) = −1, the charge of the ion. ✓
- (A) subtracts all 6 bonding electrons instead of half of them (7 − 2 − 6).
- (B) is the formal charge of Cl in a different diagram, with two Cl=O double bonds, where Cl has more than an octet. The question specifies the octet diagram.
- (D) forgets the lone pair (7 − 0 − 3).
Question 2 (multiple choice · core)
The nitrate ion, NO₃⁻, has nitrogen bonded to three oxygen atoms. What is the average N–O bond order?
- (A) 1
- (B) 1.33
- (C) 1.5
- (D) 2
Answer and explanation
Answer: (B). Nitrate has 5 + 3(6) + 1 = 24 electrons. Each octet structure has one N=O and two N–O bonds: 4 bonds over 3 positions. There are three equivalent resonance structures (the double bond can be to any O), so every N–O bond has order 4 ÷ 3 = 1.33.
- (A) ignores the double bond.
- (C) divides by 2, as if there were only two N–O positions (as in nitrite).
- (D) treats every bond as double, which would give nitrogen 12 electrons, impossible for a period 2 atom.
Question 3 (multiple choice · core)
Using Lewis diagrams, including resonance where needed, which list puts the carbon–oxygen bonds in CO, CO₂ and CO₃²⁻ in order from shortest to longest?
- (A) CO < CO₂ < CO₃²⁻
- (B) CO₃²⁻ < CO₂ < CO
- (C) CO₂ < CO₃²⁻ < CO
- (D) CO₂ < CO < CO₃²⁻
Answer and explanation
Answer: (A). Bond orders: CO has a triple bond (3); CO₂ has two double bonds (2 each); CO₃²⁻ has three equivalent resonance structures with 4 bonds over 3 positions (1.33 each). A higher bond order means a shorter bond, so CO (3) < CO₂ (2) < CO₃²⁻ (1.33).
- (B) is the order from longest to shortest.
- (C) treats CO as a single bond.
- (D) puts CO after CO₂, as if CO had a lower bond order than CO₂. This happens if you draw C=O and leave carbon with only 6 electrons; the octet diagram of CO has a triple bond.
Question 4 (constructed response · core)
The azide ion, N₃⁻, is linear, with one nitrogen in the centre.
(a) Calculate the number of valence electrons. (b) Draw (or describe) three resonance structures in which every atom has an octet, and give the formal charge on each atom in each. (c) Identify the structure that contributes most, and justify your choice. (d) Use your answer to (c) to predict how the two N–N bond lengths compare.
Worked solution and suggested Marlbridge rubric
(a) 3(5) + 1 = 16 electrons. Needed for three octets: 24, so 8 shared electrons = 4 bonds.
(b)
| Structure | Lone pairs (left N, centre N, right N) | Formal charges | Sum |
|---|---|---|---|
| [N=N=N]⁻ | 2, 0, 2 | 5 − 4 − 2 = −1; 5 − 0 − 4 = +1; −1 | −1 ✓ |
| [N≡N–N]⁻ | 1, 0, 3 | 5 − 2 − 3 = 0; +1; 5 − 6 − 1 = −2 | −1 ✓ |
| [N–N≡N]⁻ | 3, 0, 1 | −2; +1; 0 | −1 ✓ |
(c) [N=N=N]⁻ contributes most. Its formal charges (−1, +1, −1) are all of size 1, whereas the other two structures each contain a −2 charge. All three atoms are the same element, so electronegativity cannot be used to favour one of the −2 structures.
(d) The main structure is symmetrical, and the two minor structures are mirror images of each other and contribute equally. So the model predicts the two N–N bonds are equal in length.
| Point | What earns it |
|---|---|
| 1 | 16 electrons |
| 1 | Three valid octet structures drawn or described |
| 1 | Correct formal charges for all three structures, summing to −1 |
| 1 | Chooses [N=N=N]⁻, justified by smaller formal charges (no −2) |
| 1 | Equal bond lengths, linked to the symmetry of the structures |
Question 5 (constructed response · core)
Ozone, O₃, is a bent molecule with one central oxygen atom. A student draws one Lewis diagram, O=O–O, and claims: “Ozone has one short O=O bond and one longer O–O bond.”
(a) Show that the student’s diagram has the correct number of electrons, and calculate the formal charge on each atom. (b) Evaluate the student’s claim using resonance. (c) Predict how an O–O bond in ozone compares in length with the O=O bond in O₂ and the O–O single bond in hydrogen peroxide, H₂O₂.
Worked solution and suggested Marlbridge rubric
(a) Total = 3(6) = 18. In O=O–O: the double-bonded end O has 2 lone pairs, the central O has 1 lone pair, the single-bonded end O has 3 lone pairs. Count: 3 bonds (6) + 6 lone pairs (12) = 18. ✓ Formal charges: double-bonded O: 6 − 4 − 2 = 0; central O: 6 − 2 − 3 = +1; single-bonded O: 6 − 6 − 1 = −1. Sum = 0 ✓.
(b) The claim is incorrect. An equivalent structure, O–O=O, has the double bond on the other side. Both structures contribute equally, so the real molecule is a resonance hybrid in which the two O–O bonds are identical, each with bond order 3 ÷ 2 = 1.5. The student’s single diagram is not wrong as a structure, but on its own it gives a wrong prediction.
(c) Bond order 1.5 lies between 1 (H₂O₂) and 2 (O₂), so each ozone bond is shorter than the O–O bond in H₂O₂ and longer than the O=O bond in O₂.
| Point | What earns it |
|---|---|
| 1 | 18 electrons confirmed and formal charges 0, +1, −1 |
| 1 | Draws or describes the second resonance structure |
| 1 | Rejects the claim: the bonds are identical in the hybrid |
| 1 | Bond order 1.5 used to place the length between single and double |
Question 6 (constructed response · core)
Hypochlorous acid has the formula HOCl. Two skeletons are proposed: H–O–Cl and H–Cl–O. Both can be drawn with an octet on O and Cl.
(a) For each skeleton, describe the Lewis diagram and calculate the formal charge on every atom. (b) Use formal charge to decide which skeleton is more likely. Explain.
Worked solution and suggested Marlbridge rubric
Total = 1 + 6 + 7 = 14 electrons for both.
(a)
- H–O–Cl: O has 2 bonds and 2 lone pairs; Cl has 1 bond and 3 lone pairs. FC: H = 1 − 0 − 1 = 0; O = 6 − 4 − 2 = 0; Cl = 7 − 6 − 1 = 0. Count: 4 + 4 + 6 = 14 ✓.
- H–Cl–O: Cl has 2 bonds and 2 lone pairs; O has 1 bond and 3 lone pairs. FC: H = 0; Cl = 7 − 4 − 2 = +1; O = 6 − 6 − 1 = −1. Count: 4 + 4 + 6 = 14 ✓.
(b) H–O–Cl is more likely. Every atom has a formal charge of zero, while H–Cl–O needs a +1 and a −1 charge. Lower formal charges give the better model.
| Point | What earns it |
|---|---|
| 1 | Correct diagram and formal charges (all 0) for H–O–Cl |
| 1 | Correct diagram and formal charges (0, +1, −1) for H–Cl–O |
| 1 | Chooses H–O–Cl because its formal charges are all zero (closer to zero) |
Question 7 (constructed response · stretch)
Chlorine dioxide, ClO₂, has chlorine bonded to two oxygen atoms. A student claims: “With enough trial and error, I can draw a Lewis diagram of ClO₂ in which every atom has exactly eight electrons around it.”
(a) Calculate the number of valence electrons in ClO₂. (b) Evaluate the student’s claim. (c) ClO₂ is attracted into a magnetic field, a property shown by substances that contain unpaired electrons. Explain how this observation supports your answer to (b), and state what it shows about the Lewis model.
Model answer and suggested Marlbridge rubric
(a) 7 + 2(6) = 19 electrons.
(b) The claim is incorrect. Lewis diagrams place electrons as bonding pairs and lone pairs. With an odd total, one electron must be left unpaired, so at least one atom is surrounded by an odd number of electrons and cannot have exactly eight. No amount of rearranging changes the total.
(c) Attraction into a magnetic field is evidence of an unpaired electron, which is exactly what an odd electron count requires. The observation confirms that ClO₂ cannot be described by an all-paired, all-octet diagram. It shows a limitation of the Lewis model: the model works with electron pairs and octets, and it cannot fully describe species with an odd number of electrons, including where the unpaired electron is.
| Point | What earns it |
|---|---|
| 1 | 19 electrons |
| 1 | Rejects the claim: an odd number of electrons means one is unpaired, so not every atom can have 8 |
| 1 | Links the magnetic behaviour to the unpaired electron |
| 1 | States that this is a limitation of the Lewis (paired-electron, octet) model |
How did you do?
- Q1 or Q6 wrong: practise the formal charge formula in the study guide and always check the sum.
- Q2, Q3 or Q5 wrong: re-read “The resonance hybrid” and “Average bond order”, then Worked example 1.
- Q4 wrong: compare your work with Worked example 2 (N₂O), where structures are ranked by formal charge.
- Q7 incomplete: see Worked example 3 on odd-electron species.
Then tick off the topic checklist.