Edexcel International A Level Chemistry: Shapes of molecules and metallic bonding (YCH11) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set covers Topic 3 parts 3C Shapes of molecules and 3D Metallic bonding (outcomes 3.16 to 3.22) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017. All of the content is in Unit 1: Structure, Bonding and Introduction to Organic Chemistry, the IAS compulsory unit.

Related: the study guide or the revision notes. For earlier parts of Topic 3, use the Unit 1 bonding practice questions. Course hub: Edexcel International A Level Chemistry. Checklist: printable checklist. Diagnostics: free 10-minute diagnostics.

Questions

1. Define each term.

(a) Bond length [1] (b) Bond angle [1]

2. State three principles of electron-pair repulsion theory that are used to predict the shape of a simple molecule. [3]

3. This question is about gaseous BeCl₂ and BCl₃.

(a) Explain why a molecule of gaseous BeCl₂ is linear. [2] (b) State the shape of a BCl₃ molecule and its Cl-B-Cl bond angle. [2]

4. For each species, predict the shape and bond angle using electron-pair repulsion theory.

(a) The amide ion, NH₂⁻ [1] (b) The tetrafluoroborate ion, BF₄⁻ [1] (c) The hexafluorophosphate ion, PF₆⁻ [1] (d) Boron triiodide, BI₃ [1]

5. Ammonia reacts with hydrogen ions:

NH₃ + H⁺ → NH₄⁺

Explain why the H-N-H bond angle increases from about 107° in NH₃ to 109.5° in NH₄⁺. [4]

6. This question is about ethene, C₂H₄.

(a) State the shape around each carbon atom and the H-C-H bond angle. [2] (b) Explain why the C=C bond is treated as a single region of electron density, and state whether all six atoms lie in one plane. [2]

7. Phosphorus(V) chloride has different structures as a gas and as a solid.

(a) State the shape of a gaseous PCl₅ molecule and give its two bond angles. [2] (b) State how many Cl-P-Cl angles of 90° there are in one gaseous PCl₅ molecule. [1] (c) Solid phosphorus(V) chloride contains PCl₄⁺ and PCl₆⁻ ions. Predict the shape and bond angle of each ion. Show how you counted the electron pairs. [4]

8. Describe the structure of a metal and state what is meant by metallic bonding. [3]

9. Aluminium is used in overhead power cables.

(a) Explain why aluminium conducts electricity. [2] (b) Each aluminium atom contributes three electrons to the delocalised electron sea. A 1.00 m length of one cable’s aluminium core has a mass of 1.08 kg. Calculate the number of delocalised electrons in this length of core. Use Ar(Al) = 27.0 and the Avogadro constant L = 6.02 × 10²³ mol⁻¹. Give your answer to 3 significant figures. [3]

10. Use these melting temperatures.

Metal	Melting temperature / °C
lithium	181
sodium	98
potassium	63
rubidium	39
calcium	842

(a) Describe and explain the trend in melting temperature from lithium to rubidium. [3] (b) The Ca²⁺ ion is larger than the Li⁺ ion. Explain why calcium still has a much higher melting temperature than lithium. [3] (c) A student says: “When potassium melts, all of its metallic bonding is broken.” Explain why this statement is wrong. [2]

11. This question is about aluminium chloride.

(a) At high temperatures, aluminium chloride vapour contains AlCl₃ molecules. Predict and explain the shape of an AlCl₃ molecule and its Cl-Al-Cl bond angle. [3] (b) At lower temperatures the vapour contains Al₂Cl₆, in which each aluminium atom accepts a lone pair from a chlorine atom of the other AlCl₃ unit. State the number of bonding pairs and lone pairs around each aluminium atom in Al₂Cl₆, and the arrangement of chlorine atoms around each aluminium. [3] (c) Aluminium chloride reacts with chloride ions to form AlCl₄⁻. State the shape of AlCl₄⁻ and its Cl-Al-Cl bond angle, and name one other species with the same shape. [2]

12. This question is about molecules that contain multiple bonds.

(a) Hydrogen cyanide has the structure H-C≡N. Predict the shape of the molecule and the H-C-N bond angle. Give a reason. [2] (b) Methanal has the structure H₂C=O. State the shape around the carbon atom and the H-C-H bond angle you would predict. [2] (c) Use these values: N-N bond length in hydrazine, N₂H₄ = 0.145 nm; N≡N bond length in nitrogen, N₂ = 0.110 nm. Explain why the N≡N bond is shorter than the N-N bond. [2] (d) Calculate the percentage by which the N≡N bond is shorter than the N-N bond. Give your answer to 3 significant figures. [1]

Answers

1. (a) The distance between the nuclei of two atoms joined by a covalent bond [1]. (b) The angle between two bonds that meet at the same atom [1]. Examiner insight: write “nuclei” in the bond length definition; “the distance between two atoms” is vague and risks losing the mark.

2. Electron pairs in the outer shell of the central atom repel each other [1]. They take up positions as far apart as possible, so that repulsion is a minimum [1]. Lone pairs repel more strongly than bonding pairs (lone-lone > lone-bond > bond-bond) [1]. Examiner insight: three marks need three distinct ideas; “atoms repel” names the wrong particles and gains nothing.

3. (a) Beryllium has two bonding pairs and no lone pairs [1]. The pairs repel to opposite sides of the Be atom, so the molecule is linear, 180° [1]. (b) Trigonal planar [1]; 120° [1]. Examiner insight: in (a) state both the bonding pairs and the absence of lone pairs; “two bonds” alone is not an explanation.

4. (a) N: 5 - 2 + 1 = 4 electrons left, so 2 lone pairs and 2 bonding pairs: bent, about 104.5° [1]. (b) B: 3 - 4 + 1 = 0, so no lone pairs and 4 bonding pairs: tetrahedral, 109.5° [1]. (c) P: 5 - 6 + 1 = 0, so no lone pairs and 6 bonding pairs: octahedral, 90° [1]. (d) B: 3 - 3 = 0, so no lone pairs and 3 bonding pairs: trigonal planar, 120° [1]. Examiner insight: each mark needs both shape and angle; ignoring the charge gives NH₂⁻ the wrong number of lone pairs.

5. In NH₃ nitrogen has three bonding pairs and one lone pair [1]. The lone pair repels more strongly than bonding pairs, pushing the N-H bonds closer, so the angle is about 107° [1]. The lone pair forms a dative covalent bond to H⁺, so NH₄⁺ has four bonding pairs and no lone pairs [1]. These repel equally, giving a regular tetrahedron, 109.5° [1]. Examiner insight: the key link is that the lone pair becomes a bonding pair; “NH₄⁺ has four bonds” alone misses why the angle changes.

6. (a) Trigonal planar around each carbon [1]; H-C-H about 120° [1]. (b) Both pairs of the double bond lie in the same region between the two carbon atoms, so they act as one region [1]. Yes, all six atoms lie in one plane [1]. Examiner insight: counting the C=C as two regions leads to a tetrahedral answer of 109.5°, which earns nothing.

7. (a) Trigonal bipyramidal [1]; 120° and 90° [1]. (b) 6 (each of the two axial bonds is at 90° to each of the three equatorial bonds) [1]. (c) PCl₄⁺: 5 - 4 - 1 = 0, giving 4 bonding pairs with 0 lone pairs [1]; the ion is tetrahedral, 109.5° [1]. PCl₆⁻: 5 - 6 + 1 = 0, so six bonding pairs and no lone pairs [1]; octahedral, 90° [1]. Examiner insight: when asked to show your counting, a bare shape is unlikely to earn the counting marks; write the arithmetic, charge included.

8. A giant lattice of positive metal ions [1] in a sea of delocalised electrons [1]. Metallic bonding is the strong electrostatic attraction between the metal ions and the delocalised electrons [1]. Examiner insight: use “ions”, not “atoms”, and include “electrostatic”; describing metallic bonding as shared electrons uses covalent language and is not creditworthy.

9. (a) Aluminium contains delocalised electrons [1]; with a potential difference applied, these mobile electrons drift through the lattice carrying charge [1]. (b) n(Al) = 1080 g ÷ 27.0 g mol⁻¹ = 40.0 mol [1]. Moles of delocalised electrons = 3 × 40.0 = 120 mol [1]. Number = 120 × 6.02 × 10²³ = 7.22 × 10²⁵ electrons [1]. Examiner insight: in (a) the electrons carry charge, not the ions; in (b) set out each step so correct working stays visible if one line slips.

10. (a) Melting temperature decreases from lithium to rubidium [1]. All the ions are +1, but the ionic radius increases down the group [1]. The delocalised electrons are further from each ion’s centre, so attraction is weaker and less energy is needed to overcome it [1]. (b) Ca²⁺ has twice the charge of Li⁺ [1]. Calcium gives two delocalised electrons per atom compared with one for lithium [1]. These effects outweigh the larger ionic radius, so the electrostatic attraction between ions and delocalised electrons is stronger, and much more energy is needed to overcome it [1]. (c) Liquid potassium still contains K⁺ ions attracted to delocalised electrons, and it still conducts [1]. Melting only overcomes enough attraction for the ions to move past each other; separating the atoms completely, on boiling, needs much more energy [1]. Examiner insight: finish each comparison with “stronger attraction, so more energy needed”; a list of factors with no link to energy usually falls short.

11. (a) Al uses all 3 outer electrons in Al-Cl bonds: three bonding pairs, no lone pairs [1]. The pairs repel to be as far apart as possible [1]. The molecule is trigonal planar, Cl-Al-Cl 120° [1]. (b) Four bonding pairs (three covalent bonds plus the dative bond) [1]; no lone pairs [1]; the four chlorine atoms are arranged (approximately) tetrahedrally around each aluminium [1]. (c) Tetrahedral, 109.5° [1]; any one of CH₄, NH₄⁺, BF₄⁻ or another four-bonding-pair species [1]. Examiner insight: a dative bond counts as a bonding pair exactly like any other covalent bond, so do not leave it out of the count in (b).

12. (a) Carbon has two regions of electron density (the C-H bond and the C≡N bond) and no lone pairs [1]; so the molecule is linear, 180° [1]. (b) Trigonal planar [1]; about 120° [1]. (c) N≡N has three shared pairs between the nitrogen atoms, not one [1]. The greater electron density between the nuclei attracts both more strongly, pulling them closer [1]. (d) (0.145 - 0.110) ÷ 0.145 × 100 = 24.1% [1]. Examiner insight: a triple bond is still one region, so HCN is like CO₂; in (d) divide by the N-N value, the bond being compared against.

Where marks are usually lost

• Naming the shape of an ion without counting its charge into the electron total.
• Giving “tetrahedral” for a species with lone pairs, such as NH₂⁻ or NH₃.
• Treating a double or triple bond as more than one region.
• Writing “lone pairs take up more space” instead of “lone pairs repel more strongly”.
• Leaving the dative bond out of the bonding-pair count in Al₂Cl₆ or NH₄⁺.
• Explaining a melting-temperature difference by listing factors without linking them to stronger attraction and more energy.

Next steps

• Revision notes
• Study guide
• Edexcel International A Level Chemistry hub
• Printable checklist
• Find your weak spots with all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 3: Bonding and Structure, parts 3C Shapes of molecules and 3D Metallic bonding, outcomes 3.16 to 3.22.