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, B 3, C 4, N 5, O 6, P 5, F 7, Cl 7, I 7. In written answers, describe a Lewis diagram in words if you cannot draw it (for example “N with three single bonds to F and one lone pair”).
Question 1 (multiple choice · foundation)
How many valence electrons must be shown in the Lewis diagram of the phosphate ion, PO₄³⁻?
- (A) 26
- (B) 29
- (C) 32
- (D) 40
Answer and explanation
Answer: (C). P has 5 and each O has 6: 5 + 4(6) = 29. The 3− charge means 3 extra electrons: 29 + 3 = 32.
- (A) subtracts the charge instead of adding it (29 − 3). A negative ion has gained electrons.
- (B) ignores the charge.
- (D) is 8 × 5 atoms, the number of electrons the atoms would need for separate octets, not the number available.
Question 2 (multiple choice · core)
In which molecule does the central atom have fewer than eight electrons around it in the best Lewis diagram?
- (A) NH₃
- (B) BCl₃
- (C) CCl₄
- (D) PCl₅
Answer and explanation
Answer: (B). BCl₃ has 3 + 3(7) = 24 valence electrons. Three B–Cl bonds use 6, and three lone pairs on each Cl use 18. Nothing is left, so boron has only 6 electrons around it. Boron, with 3 valence electrons, commonly ends with an incomplete octet.
- (A) N has 3 bonds and 1 lone pair: 8 electrons.
- (C) C has 4 bonds: 8 electrons.
- (D) P has 5 bonds: 10 electrons. This is an expanded octet (more than 8), not fewer.
Question 3 (multiple choice · core)
How many lone pairs are on the iodine atom in the Lewis diagram of ICl₄⁻, in which iodine is bonded to all four chlorine atoms?
- (A) 0
- (B) 1
- (C) 2
- (D) 3
Answer and explanation
Answer: (C). Total = 7 + 4(7) + 1 = 36 electrons. Four I–Cl bonds use 8; three lone pairs on each Cl use 24; 36 − 8 − 24 = 4 electrons left, which go on iodine as 2 lone pairs. Iodine ends with 12 electrons (4 bonds + 2 lone pairs), which is allowed for a period 5 atom.
- (A) stops at the octet rule and leaves 4 electrons out of the diagram.
- (B) ignores the 1− charge (35 electrons), which leaves 3 electrons and cannot make whole pairs.
- (D) would need 38 electrons.
Question 4 (constructed response · core)
Hydrazine (N₂H₄), diazene (N₂H₂) and nitrogen (N₂) all contain a bond between two nitrogen atoms. In N₂H₄ each N is bonded to two H atoms; in N₂H₂ each N is bonded to one H atom.
(a) Calculate the total number of valence electrons in each species. (b) Draw (or describe) the Lewis diagram of each. (c) State the number of electrons shared between the two nitrogen atoms in each species.
Worked solution and suggested Marlbridge rubric
(a) N₂H₄: 2(5) + 4(1) = 14. N₂H₂: 2(5) + 2(1) = 12. N₂: 2(5) = 10.
(b)
- N₂H₄: H₂N–NH₂. A single N–N bond; each N has two N–H bonds and one lone pair. Count: 5 bonds × 2 = 10, plus 2 lone pairs = 4; total 14. ✓
- N₂H₂: H–N=N–H. A double N=N bond; each N has one N–H bond and one lone pair. Count: N–H bonds 4 + N=N 4 + lone pairs 4 = 12. ✓
- N₂: N≡N with one lone pair on each N. Count: 6 + 4 = 10. ✓
Check with the shortcut: N₂H₄ needs 8(2) + 2(4) = 24, has 14, so 5 bonds (4 N–H + 1 N–N). N₂H₂ needs 20, has 12, so 4 bonds (2 N–H + 2 in N=N). N₂ needs 16, has 10, so 3 bonds (N≡N).
(c) N₂H₄: 2 electrons (single bond). N₂H₂: 4 (double bond). N₂: 6 (triple bond).
| Point | What earns it |
|---|---|
| 1 | All three electron totals correct |
| 1 | N₂H₄ diagram: N–N single bond, two H on each N, one lone pair on each N |
| 1 | N₂H₂ diagram: N=N double bond, one H on each N, one lone pair on each N |
| 1 | N₂ diagram: triple bond with one lone pair on each N |
| 1 | Shared electrons 2, 4 and 6, consistent with the diagrams drawn |
Common error: drawing N₂H₂ with a single N–N bond and two lone pairs on one nitrogen. That gives the right total but leaves one N with only 6 electrons.
Question 5 (constructed response · core)
In the hydronium ion, H₃O⁺, the central atom is oxygen. In the amide ion, NH₂⁻, the central atom is nitrogen.
(a) Calculate the number of valence electrons in each ion. (b) Draw (or describe) the Lewis diagram of each ion, showing how the charge is indicated. (c) Explain why the two ions have different numbers of lone pairs on the central atom even though the totals in (a) are the same.
Worked solution and suggested Marlbridge rubric
(a) H₃O⁺: 6 + 3(1) − 1 = 8. NH₂⁻: 5 + 2(1) + 1 = 8.
(b)
- H₃O⁺: O with three O–H single bonds and one lone pair (6 + 2 = 8), inside square brackets with + outside.
- NH₂⁻: N with two N–H single bonds and two lone pairs (4 + 4 = 8), inside square brackets with − outside.
(c) Both central atoms are surrounded by 8 electrons. Each bond to H uses 2 of them. H₃O⁺ has three bonds (6 electrons), leaving one lone pair. NH₂⁻ has only two bonds (4 electrons), leaving two lone pairs. The number of lone pairs depends on how many of the 8 electrons are used in bonds, not on the total alone.
| Point | What earns it |
|---|---|
| 1 | Both totals correct, with the charge subtracted for H₃O⁺ and added for NH₂⁻ |
| 1 | H₃O⁺: three bonds, one lone pair, brackets and + charge |
| 1 | NH₂⁻: two bonds, two lone pairs, brackets and − charge |
| 1 | Explanation linking the number of bonds to the number of electrons left as lone pairs |
Question 6 (constructed response · core)
A student draws this Lewis diagram for nitrogen trifluoride, NF₃: nitrogen in the centre with single bonds to three fluorine atoms; each fluorine has three lone pairs; nitrogen has no lone pairs.
(a) Calculate the number of electrons in the student’s diagram and the number there should be. (b) Identify the error and describe the correct Lewis diagram. (c) The student says, “My diagram must be right because every fluorine has an octet.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) Student’s diagram: 3 bonds × 2 = 6, plus 9 lone pairs × 2 = 18, total 24. Correct total: 5 + 3(7) = 26.
(b) The student has left out 2 electrons. After the bonds and the fluorine lone pairs, 26 − 6 − 18 = 2 electrons remain and belong on nitrogen as one lone pair. Correct diagram: N with three N–F single bonds and one lone pair; each F with three lone pairs. Nitrogen now has 8 electrons (in the student’s version it had only 6).
(c) The claim is not valid. Outer-atom octets are only part of the check. A correct diagram must also (i) contain exactly the number of valence electrons available, 26, and (ii) give nitrogen, a period 2 atom, its octet. The student’s diagram fails both checks.
| Point | What earns it |
|---|---|
| 1 | 24 in the student’s diagram and 26 correct |
| 1 | Adds one lone pair on N (describes the correct diagram) |
| 1 | Rejects the claim, citing the wrong total or nitrogen having only 6 electrons |
| 1 | Both reasons given in (c) |
Question 7 (constructed response · stretch)
Chlorine trifluoride, ClF₃, is a molecule in which chlorine is bonded to three fluorine atoms.
(a) Draw (or describe) the Lewis diagram of ClF₃ and state how many electrons surround the chlorine atom. (b) A student proposes a molecule NCl₅, with nitrogen bonded to five chlorine atoms, “by analogy with PCl₅”. Explain why no valid Lewis diagram can be drawn for this proposed molecule.
Worked solution and suggested Marlbridge rubric
(a) Total = 7 + 3(7) = 28. Three Cl–F bonds use 6; three lone pairs on each F use 18; 28 − 6 − 18 = 4 electrons left, which go on Cl as 2 lone pairs. Chlorine is surrounded by 3 bonds (6) + 2 lone pairs (4) = 10 electrons, an expanded octet. This is allowed because chlorine is in period 3.
(b) Five N–Cl bonds would put 10 electrons around nitrogen. Nitrogen is in period 2: its valence shell (n = 2) can hold at most 8 electrons, so it cannot form more than four bonds, and with 5 valence electrons a neutral N normally forms three. Phosphorus is in the same group but in period 3, and larger period 3 atoms can be surrounded by more than 8 electrons, so PCl₅ has a valid Lewis diagram and NCl₅ does not.
| Point | What earns it |
|---|---|
| 1 | Correct total (28) and two lone pairs on Cl |
| 1 | States Cl is surrounded by 10 electrons and links this to Cl being in period 3 |
| 1 | NCl₅ would need 10 electrons around N, more than a period 2 atom can hold |
| 1 | Contrasts N (period 2) with P (period 3) in the same group |
How did you do?
- Q1, Q3 or Q5(a) wrong: practise step 1 of the method in the study guide, especially adjusting for charge.
- Q4 wrong: re-read Worked examples 1 and 2 on making double and triple bonds, and use the needed-minus-available shortcut.
- Q2, Q3 or Q7 wrong: review the section “When the central atom does not end with eight”.
- Q6 incomplete: a full check means recounting the total and every atom’s electrons.
Then tick off the topic checklist.