2.2 Exam Practice
Exam-style practice questions on The Covalent Model
Section B: Data Analysis (Paper 1B Style)
Calculator and Data Booklet permitted. Show all working clearly.
Question 1: Boiling Points and Intermolecular Forces Explain
5 marksThe table below shows the boiling points of the Group 14 hydrides.
| Compound | Molar mass / g mol⁻¹ | Boiling point / K |
|---|---|---|
| CH₄ | 16 | 112 |
| SiH₄ | 32 | 161 |
| GeH₄ | 77 | 185 |
| SnH₄ | 123 | 221 |
(a) Identify the type of intermolecular force present between these molecules. [1]
(b) Explain the trend in boiling points from CH₄ to SnH₄. [2]
(c) Water (H₂O, molar mass 18 g mol⁻¹) has a boiling point of 373 K. Explain why water does not fit the trend seen in Group 14 hydrides. [2]
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(a) London (dispersion) forces only. All molecules are non-polar (symmetrical tetrahedral). [1]
(b) As molar mass increases from CH₄ to SnH₄, the number of electrons increases [1]
More electrons means larger, more polarisable electron clouds, producing stronger London forces, so more energy is needed to separate molecules [1]
(c) Water has hydrogen bonding between molecules (O-H...O), which is much stronger than London forces [1]
This is possible because oxygen is highly electronegative and has lone pairs, so the boiling point is anomalously high relative to its molar mass [1]
Section C: Structured Questions (Paper 2 Style)
Show all working. State answers with appropriate significant figures and units.
Question 2: VSEPR and Molecular Geometry Predict
5 marksConsider the molecules: BF₃, NF₃, and XeF₂.
(a) Draw Lewis structures for BF₃ and NF₃. [2]
(b) Predict the molecular geometry and bond angle for each molecule. [2]
(c) Explain why BF₃ is non-polar but NF₃ is polar. [1]
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(a) BF₃: B in centre with 3 single bonds to F, no lone pairs on B (6 electrons around B) [1]
NF₃: N in centre with 3 single bonds to F and 1 lone pair on N [1]
(b) BF₃: trigonal planar, 120° [1]
NF₃: trigonal pyramidal, approximately 102° (less than 109.5° due to the lone pair compressing bond angles) [1]
(c) BF₃ is symmetrical (trigonal planar) so the individual bond dipoles cancel out. NF₃ is asymmetrical (trigonal pyramidal due to the lone pair), so the bond dipoles do not cancel, giving a net dipole moment. [1]
Question 3: Covalent Structures and Properties Compare
4 marksDiamond, graphite, and C₆₀ (buckminsterfullerene) are all allotropes of carbon.
(a) Compare the structures of diamond and graphite in terms of bonding and arrangement of carbon atoms. [2]
(b) Explain why graphite conducts electricity but diamond does not. [1]
(c) Explain why both diamond and SiO₂ have very high melting points. [1]
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(a) Diamond: each C is bonded to 4 others in a tetrahedral arrangement, forming a 3D covalent network with all single bonds (sp³) [1]
Graphite: each C is bonded to 3 others in a trigonal planar arrangement, forming flat hexagonal layers. The fourth electron is delocalised between the layers [1]
(b) Graphite has delocalised electrons between its layers that are free to move and carry charge. In diamond, all four valence electrons are used in bonds and none are free to move [1]
(c) Both are covalent network solids with strong covalent bonds throughout the entire structure. To melt them, many strong covalent bonds must be broken, requiring a very large amount of energy [1]
Question 4: Resonance and Delocalization HL Deduce
5 marksThe carbonate ion, CO₃²⁻, has three equivalent C-O bonds, each with a bond length of 129 pm. A typical C-O single bond is 143 pm and a C=O double bond is 123 pm.
(a) Draw two resonance structures for the carbonate ion. [2]
(b) Explain why all three C-O bond lengths are identical and intermediate between a single and double bond. [2]
(c) Calculate the formal charge on each oxygen atom in the resonance structure where one C=O double bond and two C-O single bonds are drawn. [1]
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(a) Any two valid resonance structures showing the double bond on a different oxygen atom in each, with the remaining oxygens having single bonds and a negative charge. Total charge 2- must be shown. [1] per correct structure [1]
(b) The true structure is a resonance hybrid; the pi electrons are delocalised across all three C-O bonds [1]
This gives each bond a bond order of 1⅓, which is between a single and double bond, explaining the intermediate bond length [1]
(c) For the double-bonded O: FC = 6 - 4 - ½(4) = 0. For each single-bonded O: FC = 6 - 6 - ½(2) = -1 [1]
Question 5: Hybridization and Sigma/Pi Bonding HL Analyse
5 marksEthanoic acid (CH₃COOH) contains different types of covalent bonds.
(a) State the hybridization of each carbon atom in ethanoic acid. [1]
(b) State the number of sigma (σ) bonds and pi (π) bonds in one molecule of ethanoic acid. [2]
(c) Explain how σ and π bonds differ in terms of orbital overlap and electron density distribution. [2]
Show Mark Scheme
(a) CH₃ carbon: sp³; COOH carbon: sp² [1]
(b) Sigma bonds: 7 (3 C-H + C-C + C=O has 1σ + C-O + O-H) [1]
Pi bonds: 1 (the second bond in C=O) [1]
(c) Sigma bonds are formed by head-on overlap of orbitals, with electron density concentrated along the internuclear axis (the bond axis) [1]
Pi bonds are formed by lateral (sideways) overlap of unhybridised p-orbitals, with electron density above and below the bond axis [1]