IB DP Chemistry Structure 2.2: The covalent model: formal charge, sigma and pi bonds, and hybridization – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions test Structure 2.2, The covalent model, understandings 2.2.11 to 2.2.16, as listed in the IB Chemistry guide that took effect for Diploma Programme candidates examined from 2025 (issued February 2023). All of this content is additional higher level, so every question is HL only.

Read the study guide and revision notes before attempting these. Questions 9 and 10 combine several understandings. For the wider course, use the hub for this subject and its tick-off checklist.

Questions

1. In the nitrite ion, NO₂⁻, nitrogen is bonded to two oxygen atoms.

(a) Draw the two resonance structures of NO₂⁻, showing all lone pairs. [2] (b) State and explain how the lengths of the two N–O bonds compare. [2]

2. Use these values. Cyclohexene + H₂ → cyclohexane, ΔH = −121 kJ mol⁻¹. Benzene + 3H₂ → cyclohexane, ΔH = −210 kJ mol⁻¹.

(a) Calculate the resonance energy of benzene from these data. [2] (b) Give one piece of physical evidence, other than enthalpy data, and one piece of chemical evidence for a delocalized structure in benzene. [2] (c) Suggest why benzene tends to undergo substitution rather than addition reactions. [2]

3. For each species, deduce the number of electron domains round the central atom, the electron domain geometry and molecular geometry.

(a) BrF₃ [3] (b) I₃⁻ (the central iodine is bonded to the other two) [3]

4. Dinitrogen monoxide, N₂O, has the atom order N–N–O.

(a) Calculate the formal charge on each atom in the Lewis formulas N=N=O and N≡N–O. [2] (b) Deduce which of these two Lewis formulas is preferred, giving a reason. [1] (c) Use formal charge to explain why the atom order N=O=N is not favoured. [2]

5. Two Lewis formulas for the phosphate ion, PO₄³⁻, have either four P–O single bonds or one P=O and three P–O bonds.

(a) Calculate the formal charge on the phosphorus atom in each Lewis formula. [2] (b) Deduce which Lewis formula is preferred. [1] (c) State the oxidation state of phosphorus in PO₄³⁻ and explain why it differs from its formal charge in the preferred formula. [2]

6.

(a) State the number of σ bonds and π bonds in propenoic acid, CH₂=CHCOOH. [2] (b) State the number of σ bonds and π bonds in cyanogen, N≡C–C≡N. [2] (c) State the hybridization of each carbon atom in cyanogen and deduce the shape of the molecule. [2]

7. Aminoethanenitrile has the structure H₂N–CH₂–C≡N.

(a) State the hybridization of the amine nitrogen atom, the CH₂ carbon atom and the nitrile carbon atom. [3] (b) State the C–C≡N bond angle. [1] (c) Explain, in terms of orbital overlap, how the C≡N bond forms. [3]

8.

(a) The NO₂⁺ ion is linear. Draw its Lewis formula, showing the formal charge on each atom, and deduce the hybridization of nitrogen. [3] (b) An atom in a molecule is sp² hybridized and carries one lone pair. State its electron domain geometry and the molecular geometry around it. [2]

9. Ozone, O₃, and oxygen, O₂, are both split by ultraviolet light. Use these values: bond enthalpy of O=O in O₂ = 498 kJ mol⁻¹; speed of light = 3.00 × 10⁸ m s⁻¹; Planck constant = 6.63 × 10⁻³⁴ J s; Avogadro constant = 6.02 × 10²³ mol⁻¹.

(a) Draw the two resonance structures of O₃ and calculate the formal charge on each atom in one of them. [3] (b) State the hybridization of the central oxygen atom and predict the O–O–O bond angle. [2] (c) Calculate the longest wavelength of light, in nm, that can break the bond in one O₂ molecule. [3] (d) Without further calculation, explain why ozone can be broken apart by light of longer wavelength than oxygen. [2]

10. In XeO₂F₂, xenon is bonded to two O atoms and two F atoms.

(a) Calculate the total number of valence electrons in XeO₂F₂. [1] (b) Lewis formula A has two Xe–O single bonds; B has two Xe=O double bonds. Both have two Xe–F bonds and one lone pair on Xe. Calculate the formal charges on Xe and each O in A and in B. [3] (c) Deduce which Lewis formula is preferred. [1] (d) Deduce the number of electron domains round xenon, the electron domain geometry and molecular geometry. [3] (e) State the number of σ bonds and π bonds in Lewis formula B. [2]

Answers

1. (a) 18 electrons: N with one lone pair, one N=O (two lone pairs on O) and one N–O (three lone pairs on O), in brackets with the charge [1]. Second structure with the double bond to the other O, joined by ↔ [1]. [2] (b) Equal in length [1]: the π electrons are delocalized over both positions, so each bond is between single and double [1]. [2] Examiner insight: Resonance structures with missing lone pairs or no charge lose credit; count to 18 before you finish.

2. (a) Expected for three isolated C=C bonds = 3 × (−121) = −363 kJ mol⁻¹ [1]. Resonance energy = −210 − (−363) = +153, so benzene is 153 kJ mol⁻¹ more stable than the Kekulé structure [1]. [2] (b) Physical: all six C–C bonds are equal in length, between single and double [1]. Chemical: only one 1,2-disubstituted isomer exists (or no ready decolourization of bromine water) [1]. [2] (c) Addition would break up the delocalized π system and lose the resonance energy [1]; substitution keeps the ring intact, so it is favoured [1]. [2] Examiner insight: In (a), state the answer as a stabilization and keep the sign consistent with your wording; a bare “−153” alongside “more stable” contradicts itself.

3. (a) Br: 7 valence electrons; 3 in bonds, leaving 2 lone pairs, so 5 domains [1]. Electron domain geometry trigonal bipyramidal [1]. Both lone pairs are equatorial, so the molecule is T-shaped [1]. [3] (b) Central I: 7 + 1 = 8 electrons; 2 in bonds, leaving 3 lone pairs, so 5 domains [1]. Electron domain geometry trigonal bipyramidal [1]. All three lone pairs are equatorial, so the ion is linear (180°) [1]. [3] Examiner insight: “Trigonal bipyramidal” alone gives only the electron domain geometry; the molecular geometry must describe the atoms only.

4. (a) N=N=O: terminal N −1, central N +1, O 0 [1]. N≡N–O: terminal N 0, central N +1, O −1 [1]. [2] (b) N≡N–O: the charges are equally small, but the negative one sits on oxygen, the more electronegative atom [1]. [1] (c) In N=O=N each N is −1 and O is +2 [1]. The charges are larger, and the most electronegative atom is positive, so it is not favoured [1]. [2] Examiner insight: Show the formal charge on every atom; a conclusion with no charges to support it is hard to credit.

5. (a) Four single bonds: P = 5 − (0 + 4) = +1 [1]. One P=O: P = 5 − (0 + 5) = 0 [1]. [2] (b) The formula with one P=O double bond, as its charges are closer to zero [1]. [1] (c) +5 [1]. Oxidation state gives all bonding electrons to the more electronegative O; formal charge assumes equal sharing [1]. [2] Examiner insight: Check that the formal charges in your chosen structure add to −3; a total that does not match the ion’s charge shows a counting error.

6. (a) Bonds: three C–H, C=C, C–C, C=O, C–O and O–H. 8 σ [1] and 2 π [1]. [2] (b) Two C≡N and one C–C: 3 σ [1] and 4 π [1]. [2] (c) Each carbon has two electron domains: sp [1]. Both carbons are linear with no lone pairs, so the molecule is linear [1]. [2] Examiner insight: In (a), the O–H bond is easily missed; list every bond before totalling.

7. (a) Amine N (three bonds and one lone pair): sp³ [1]. CH₂ carbon (four bonds): sp³ [1]. Nitrile carbon (two domains): sp [1]. [3] (b) 180° [1]. [1] (c) Head-on overlap of an sp orbital on C with an sp orbital on N gives one σ bond, with density along the axis [1]. Each atom keeps two unhybridized p orbitals at right angles to the axis [1]. These overlap sideways in pairs to give two π bonds, with density on opposite sides of the axis [1]. [3] Examiner insight: “Three overlapping orbitals” earns little; name which orbitals form the σ bond and which form the π bonds.

8. (a) [O=N=O]⁺ with two lone pairs on each O and none on N [1]; formal charges: N +1, each O 0 [1]. N has two electron domains, so it is sp hybridized [1]. [3] (b) Trigonal planar electron domain geometry [1]; bent (V-shaped), slightly below 120° [1]. [2] Examiner insight: In (a), the formal charges must add up to +1; checking this catches a missing or extra lone pair.

9. (a) O=O–O ↔ O–O=O, with one lone pair on the central O, two on the double-bonded O and three on the single-bonded O [1]. Central O: 6 − (2 + 3) = +1; double-bonded O: 6 − (4 + 2) = 0 [1]; single-bonded O: 6 − (6 + 1) = −1 [1]. [3] (b) Central O has three domains (two bonding, one lone pair): sp² [1]; bond angle slightly less than 120° because the lone pair repels more strongly [1]. [2] (c) Energy per bond = 498 000 ÷ (6.02 × 10²³) = 8.27 × 10⁻¹⁹ J [1]. λ = hc ÷ E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (8.27 × 10⁻¹⁹) [1] = 2.40 × 10⁻⁷ m = 240 nm [1]. [3] (d) Delocalization makes each O–O bond in ozone between single and double, so weaker than O=O [1]. Less energy is needed, so lower-frequency, longer-wavelength photons can break it [1]. [2] Examiner insight: In (c), converting kJ to J and per mole to per bond are separate steps; show each on its own line so the working can be followed.

10. (a) 8 + (2 × 6) + (2 × 7) = 34 [1]. [1] (b) A: Xe = 8 − (2 + 4) = +2 [1]; each O = 6 − (6 + 1) = −1 [1]. B: Xe = 8 − (2 + 6) = 0 and each O = 6 − (4 + 2) = 0 [1]. [3] (c) B, because every atom has a formal charge of zero [1]. [1] (d) Four bonding domains plus one lone pair = 5 domains [1]; electron domain geometry trigonal bipyramidal [1]; with the lone pair equatorial the molecule is seesaw [1]. [3] (e) 4 σ bonds [1] and 2 π bonds [1]. [2] Examiner insight: In (d), each Xe=O double bond is one domain; counting it as two gives seven domains and a wrong geometry.

Where marks are usually lost

• Leaving lone pairs, brackets or the charge off resonance structures of ions.
• Confusing electron domain geometry with molecular geometry.
• Placing lone pairs axially in a trigonal bipyramid.
• Using all the bonding electrons, instead of half, in the formal charge formula.
• Choosing a structure with a negative formal charge on the less electronegative atom.
• Counting two or three σ bonds in a multiple bond.
• Deducing hybridization from attached atoms instead of electron domains.

Next steps

• All free 10-minute diagnostics
• Revision notes for a quick recap
• Study guide for full explanations
• Hub page for the whole course
• Unit-by-unit checklist to track progress
• Book a free trial class if you want help with this unit

Official syllabus

Source: the IB Chemistry guide for the Diploma Programme (International Baccalaureate Organization), issued in February 2023 for first assessment in 2025. This page covers Structure 2.2, The covalent model, understandings 2.2.11 to 2.2.16, all of them additional higher level.