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, F 7, P 5, Cl 7, Br 7, I 7, Te 6, Xe 8. Electronegativity order: F > O > Cl > N > Br > C > H.
Question 1 (multiple choice · foundation)
Which species has a trigonal pyramidal molecular geometry?
- (A) BCl₃
- (B) PF₃
- (C) ClF₃
- (D) NO₃⁻
Answer and explanation
Answer: (B). PF₃ has 5 + 3(7) = 26 electrons. Three P–F bonds (6) and three lone pairs on each F (18) leave 2 electrons: one lone pair on P. So P has 3 bonded atoms + 1 lone pair = 4 domains. Tetrahedral electron-domain geometry with one lone pair gives a trigonal pyramid.
- (A) BCl₃ has 24 electrons and no lone pair on B: 3 domains, trigonal planar.
- (C) ClF₃ has 28 electrons and two lone pairs on Cl: 5 domains, T-shaped. The formula “AX₃” alone does not fix the shape.
- (D) NO₃⁻ has 3 domains on N (resonance does not add domains) and no lone pair: trigonal planar.
Question 2 (multiple choice · core)
Propyne has the structure CH₃–C≡CH. What is the hybridization of the three carbon atoms, from left to right?
- (A) sp³, sp, sp
- (B) sp³, sp², sp
- (C) sp³, sp³, sp
- (D) sp², sp, sp
Answer and explanation
Answer: (A). Count domains on each carbon. The CH₃ carbon has 4 bonded atoms (3 H and 1 C): 4 domains, sp³. The middle carbon has 2 bonded atoms (CH₃ carbon and the end carbon): 2 domains, sp. The end carbon has 2 bonded atoms (C and H): 2 domains, sp.
- (B) treats the middle carbon as if it had 3 domains, perhaps counting the triple bond as two. A triple bond is one domain.
- (C) gives the middle carbon 4 domains, as if each of its four shared pairs were a separate domain. A triple bond is one domain, so that carbon has only 2.
- (D) mistakes the CH₃ carbon for sp²; it has four single bonds and so four domains.
Question 3 (multiple choice · core)
Which molecule has no dipole moment, even though its bonds are polar?
- (A) NH₃
- (B) CH₂F₂
- (C) PCl₅
- (D) BrF₅
Answer and explanation
Answer: (C). PCl₅ has 40 electrons and no lone pair on P: 5 domains, trigonal bipyramidal. The two axial P–Cl dipoles point in opposite directions and cancel; the three equatorial dipoles are 120° apart and cancel. The molecule has no dipole moment.
- (A) NH₃ is trigonal pyramidal: the lone pair makes the shape unsymmetrical, so the N–H dipoles add up.
- (B) CH₂F₂ is tetrahedral, but the C–F and C–H bonds are different, so the bond dipoles do not cancel.
- (D) BrF₅ has 42 electrons and one lone pair on Br: square pyramidal, so the dipoles do not cancel.
Question 4 (constructed response · core)
Propenal has the structure CH₂=CH–CH=O, with the O atom bonded only to the end carbon.
(a) Show that a Lewis diagram with the bonds shown and two lone pairs on oxygen uses every valence electron. (b) State the number of σ bonds and π bonds in the molecule. (c) Give the hybridization of each carbon atom and of the oxygen atom. (d) Predict the approximate H–C=O bond angle at the end carbon. (e) Compare the C=O bond with the C–C single bond in propenal in terms of length and bond energy. Justify your answer.
Worked solution and suggested Marlbridge rubric
(a) Total: 3(4) + 4(1) + 6 = 22 electrons. Drawn: C=C (4) + C–C (2) + C=O (4) + four C–H (8) + two lone pairs on O (4) = 22. ✓
(b) σ bonds: C=C, C–C, C=O and four C–H = 7 σ. π bonds: one in C=C and one in C=O = 2 π.
(c) Each carbon has 3 bonded atoms and no lone pair: 3 domains, sp² (all three). Oxygen has 1 bonded atom + 2 lone pairs = 3 domains: sp².
(d) The end carbon is sp² with 3 domains: H–C=O about 120°. (Slightly more than 120° is also reasonable, because the C=O double bond pushes harder than the single bonds.)
(e) The C=O bond is shorter and stronger than the C–C bond. It has a higher bond order (2 instead of 1): two shared pairs pull the nuclei closer and hold them more tightly. Oxygen is also a smaller atom than carbon, which shortens the bond further.
| Point | What earns it |
|---|---|
| 1 | 22 electrons, with a count of the drawn diagram |
| 1 | 7 σ and 2 π |
| 1 | All three carbons sp² and oxygen sp², from domain counts |
| 1 | About 120°, linked to 3 domains |
| 1 | C=O shorter and stronger, justified by bond order (atomic radius accepted as an extra reason) |
Do not award (c) for oxygen if the lone pairs are ignored (calling it “sp” because it has one bonded atom).
Question 5 (constructed response · core)
Dinitrogen difluoride, N₂F₂, has the skeleton F–N–N–F and exists as two different compounds with the same formula.
(a) Draw (or describe) the Lewis diagram in which every atom has an octet. (b) Give the hybridization of each nitrogen atom and predict the F–N=N bond angle. (c) Explain why N₂F₂ exists as two geometric isomers. (d) One isomer has a dipole moment and the other does not. Describe the two isomers and identify which one is polar.
Worked solution and suggested Marlbridge rubric
(a) Total: 2(5) + 2(7) = 24 electrons. Diagram: F–N=N–F. Each N has one lone pair; each F has three lone pairs. Count: 4 bonding pairs (8) + 2 lone pairs on N (4) + 6 lone pairs on F (12) = 24. ✓ Each N: 1 single + 1 double + 1 lone pair = 8 electrons.
(b) Each N has 2 bonded atoms + 1 lone pair = 3 domains: sp². VSEPR predicts an F–N=N angle of about 120°, a little less because the lone pair on N pushes the bonds together. (The measured angle is smaller still: VSEPR gives the direction of the change, not its exact size.)
(c) The N=N double bond contains a π bond. Its side-on overlap would be broken by twisting, so there is no rotation about the N=N bond. The F atoms are fixed either on the same side or on opposite sides, giving two different compounds.
(d) In the cis isomer both F atoms are on the same side of the N=N bond. In the trans isomer they are on opposite sides. The N–F bonds are polar. In trans, the two halves point in opposite directions, so their dipoles cancel: no dipole moment. In cis, both N–F dipoles point partly towards the same side and do not cancel, so the cis isomer is polar.
| Point | What earns it |
|---|---|
| 1 | Correct diagram with 24 electrons and octets |
| 1 | sp² for each N and an angle of about 120° or a little less |
| 1 | π bond in N=N prevents rotation |
| 1 | cis and trans correctly described |
| 1 | cis polar, trans not, with a reason based on whether the bond dipoles cancel |
Question 6 (constructed response · core)
Copy and complete the table for tellurium tetrachloride, TeCl₄, the dichloroiodate ion, ICl₂⁻, and xenon tetrafluoride, XeF₄. Then answer (b) and (c).
| Species | Valence electrons | Lone pairs on central atom | Electron domains | Electron-domain geometry | Molecular geometry |
|---|---|---|---|---|---|
| TeCl₄ | |||||
| ICl₂⁻ | |||||
| XeF₄ |
(b) Explain why the lone pairs in ICl₂⁻ are placed where they are. (c) Which of TeCl₄ and XeF₄ has a dipole moment? Justify your answer.
Worked solution and suggested Marlbridge rubric
(a)
| Species | Valence electrons | Lone pairs on central atom | Electron domains | Electron-domain geometry | Molecular geometry |
|---|---|---|---|---|---|
| TeCl₄ | 6 + 4(7) = 34 | 1 | 5 | trigonal bipyramidal | seesaw |
| ICl₂⁻ | 7 + 2(7) + 1 = 22 | 3 | 5 | trigonal bipyramidal | linear |
| XeF₄ | 8 + 4(7) = 36 | 2 | 6 | octahedral | square planar |
Working for lone pairs: subtract the bonding electrons and 6 per outer halogen. TeCl₄: 34 − 8 − 24 = 2 (one pair). ICl₂⁻: 22 − 4 − 12 = 6 (three pairs). XeF₄: 36 − 8 − 24 = 4 (two pairs).
(b) In a trigonal bipyramid, an equatorial position has two neighbours at 90° but an axial position has three. Lone pairs repel more strongly than bonding pairs, so all three lone pairs take the equatorial positions. The two Cl atoms are left on the axis, 180° apart: a linear ion.
(c) TeCl₄ has a dipole moment. Its lone pair sits in one equatorial position, so the seesaw is unsymmetrical and the Te–Cl dipoles do not cancel. In XeF₄ the two lone pairs are opposite each other and the four Xe–F bonds form a square: opposite dipoles cancel, so XeF₄ has no dipole moment.
| Point | What earns it |
|---|---|
| 1 | All three electron counts correct |
| 1 | Lone pairs and domains correct for all three |
| 1 | Both geometries correct for all three species |
| 1 | Lone pairs equatorial, explained by fewer 90° neighbours (less repulsion) |
| 1 | TeCl₄ polar, XeF₄ not, each with a symmetry reason |
Award the geometry point for answers carried forward correctly from wrong domain counts.
Question 7 (constructed response · stretch)
A student writes: “Xenon difluoride, XeF₂, and water, H₂O, both have two atoms bonded to the central atom. So both molecules must be bent, and both must have a dipole moment.”
(a) Determine the number of electron domains on the central atom of each molecule. (b) Evaluate the student’s claim about shape for each molecule. (c) Evaluate the student’s claim about dipole moments.
Model answer and suggested Marlbridge rubric
(a) XeF₂: 8 + 2(7) = 22 electrons. Two Xe–F bonds (4) and three lone pairs on each F (12) leave 6 electrons: three lone pairs on Xe. Domains: 2 + 3 = 5. H₂O: 8 electrons; O has 2 bonds and two lone pairs: 4 domains.
(b) For water the claim is correct: 4 domains with 2 lone pairs give a bent molecule (H–O–H 104.5°). For XeF₂ it is incorrect. The five domains form a trigonal bipyramid. The three lone pairs take the three equatorial positions, where repulsion is least, so the two F atoms sit on the axis at 180°: XeF₂ is linear. The number of bonded atoms alone does not decide the shape; the lone pairs do.
(c) Water has a dipole moment: the O–H bonds are polar and the bent shape means their dipoles add. XeF₂ has no dipole moment: the two Xe–F bond dipoles are equal and point in exactly opposite directions, so they cancel. The claim is only half right.
| Point | What earns it |
|---|---|
| 1 | 5 domains (3 lone pairs) on Xe and 4 domains (2 lone pairs) on O |
| 1 | Water bent, with reason |
| 1 | XeF₂ linear, because lone pairs go equatorial and the F atoms are axial |
| 1 | Water polar, XeF₂ not, with dipoles cancelling (or not) as the reason |
How did you do?
- Q1, Q6 or Q7 wrong: count the lone pairs on the central atom carefully, then use the shapes table in the study guide.
- Q2 or Q4 wrong: re-read “Hybridization: naming the atom” and Worked example 3. Count domains, not bonds.
- Q3, Q5(d), Q6(c) or Q7(c) wrong: a dipole moment needs polar bonds and a shape where they do not cancel.
Then tick off the topic checklist.