Edexcel International A Level Chemistry: Ionic and covalent 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.

These questions cover Topic 3 parts 3A (Ionic bonding) and 3B (Covalent bonding) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017: outcomes 3.1 to 3.15. This is International AS (IAS) content for Unit 1: Structure, Bonding and Introduction to Organic Chemistry. Use the data values given in each question.

Revise first with the ionic and covalent bonding study guide or the revision notes. A free diagnostic helps you find weak spots first; the rest of the course is on the chemistry hub and in the printable checklist.

Questions

1. State what is meant by a covalent bond. [2]

2. Electron density maps can be produced for crystals of sodium chloride and for molecules of chlorine. Describe how the two maps differ, and explain what each tells you about the bonding. [3]

3. Heating sodium with phosphorus gives sodium phosphide, Na₃P, which contains P³⁻ ions.

(a) Give the electron arrangements of a sodium ion and a phosphide ion. [2] (b) Explain why the formula is Na₃P. [1] (c) Describe a dot-and-cross diagram for sodium phosphide, showing outer electrons only. [2]

4. A student places a green solution of copper(II) chromate at the bottom of a U-tube and carefully adds a colourless electrolyte solution on top in each arm. A carbon electrode dips into each arm, and a direct voltage is applied for 30 minutes.

(a) Describe what the student observes near each electrode. [2] (b) Explain why these observations are evidence for the existence of ions. [1] (c) In a second experiment, a crystal of potassium manganate(VII) is placed on moist filter paper on a microscope slide between two electrodes. State and explain the direction in which the colour moves. [2]

5. Use these ionic radii (pm): S²⁻ 184, Cl⁻ 181, K⁺ 138, Ca²⁺ 100.

(a) Show that the four ions are isoelectronic. [1] (b) Explain why the radius decreases from S²⁻ to Ca²⁺. [3]

6. Rubidium chloride and strontium sulfide have the same type of lattice. Use these ionic radii (pm): Rb⁺ 152, Cl⁻ 181, Sr²⁺ 118, S²⁻ 184.

(a) For each compound, calculate the product of the ionic charges divided by the sum of the ionic radii. Give your answers in pm⁻¹ to 3 significant figures. [2] (b) Predict which compound has the higher melting temperature, and explain your answer. [2]

7. This question is about polarisation.

(a) State what is meant by polarisation as applied to ions. [1] (b) Use these ionic radii (pm): Be²⁺ 45, Mg²⁺ 72, Ca²⁺ 100. Deduce which of beryllium chloride, magnesium chloride and calcium chloride has the most covalent character. Explain your reasoning. [3] (c) Explain why sodium iodide has more covalent character than sodium fluoride. [2]

8. Ammonia reacts with boron trifluoride to form a single product, H₃NBF₃.

(a) Explain how the bond between nitrogen and boron forms, and state the number of outer electrons around boron before and after the reaction. [3] (b) Aluminium chloride exists as Al₂Cl₆ molecules in the vapour at lower temperatures. Explain why two AlCl₃ units join together. [2]

9. For each molecule, state the number of bonding pairs and the number of lone pairs of electrons.

(a) Ethyne, C₂H₂ [2] (b) Hydroxylamine, NH₂OH [2]

10. Carbon forms several giant structures.

(a) Describe the structures of diamond and graphite, and explain why both have very high melting temperatures. [3] (b) Account for the difference in electrical conductivity between graphite and diamond. [2] (c) Describe the structure of graphene and suggest one application, linking it to a property. [2]

11. Use these Pauling electronegativity values: H 2.20, B 2.04, C 2.55, P 2.19, F 3.98.

(a) State what is meant by electronegativity. [1] (b) Calculate the electronegativity difference for the C–F, B–F, P–F and C–H bonds, and list the bonds in order of increasing polarity. [3] (c) Predict whether each molecule is polar or non-polar, giving a reason: CF₄, CH₂F₂, BF₃, PF₃. [4] (d) A charged rod is held next to a thin stream of dichloromethane, CH₂Cl₂, and then next to a stream of tetrachloromethane, CCl₄. Predict and explain what is seen. [2]

Answers

1. The strong electrostatic attraction between two nuclei [1] and the shared pair of electrons between them [1]. Examiner insight: “Sharing electrons” alone describes what happens, not what holds the atoms together; name the attraction and both things being attracted.

2. Sodium chloride: separate, nearly circular contours around each ion, with density falling almost to zero between ions [1]. Chlorine: contours enclose both nuclei, with significant density between them [1]. So sodium chloride contains separate ions, while in chlorine electrons are shared between the nuclei (covalent bonding) [1]. Examiner insight: A comparison question needs both maps described; a detailed account of one map with “the other is different” leaves the comparison incomplete.

3. (a) Na⁺ is 2.8 [1]; P³⁻ is 2.8.8 [1]. (b) Three Na⁺ (+3 in total) balance one P³⁻ (−3) [1]. (c) Three Na⁺ ions in square brackets with +, outer shell empty [1]. One P³⁻ ion in square brackets with 3−, with eight outer electrons: five of one symbol, three of the other [1]. Examiner insight: Missing brackets or charges, or the wrong number of each ion, can cost credit even when the idea is right.

4. (a) Near the negative electrode, a blue colour appears and moves up the arm [1]. Near the positive electrode, a yellow colour appears [1]. (b) Blue Cu²⁺ and yellow CrO₄²⁻ move towards opposite electrodes, so they carry opposite charges [1]. (c) The purple colour moves towards the positive electrode [1], because the coloured species is the manganate(VII) ion, MnO₄⁻, which is negatively charged [1]. Examiner insight: Link each colour to the ion responsible and to the sign of its charge; naming the electrode without the reason gives only half the answer.

5. (a) Each ion has 18 electrons: S²⁻ 16 + 2, Cl⁻ 17 + 1, K⁺ 19 − 1, Ca²⁺ 20 − 2 [1]. (b) The number of protons increases from 16 to 20 [1]. The electrons and shielding are the same in each ion [1]. So the greater nuclear charge attracts the electrons more strongly and pulls them closer, reducing the radius [1]. Examiner insight: Saying “more protons” without stating that the electron number and shielding stay the same leaves out the step that makes the comparison valid.

6. (a) RbCl: 1 ÷ (152 + 181) = 1 ÷ 333 = 0.00300 pm⁻¹ [1]. SrS: (2 × 2) ÷ (118 + 184) = 4 ÷ 302 = 0.0132 pm⁻¹ [1]. (b) Strontium sulfide [1]. Its ions have double the charge and are closer, so the attraction between ions is much stronger (about 4.4 times by this measure) and more energy is needed to separate them [1]. Examiner insight: Explain the prediction through attraction between ions; “bigger number” alone gives the conclusion without the chemistry.

7. (a) The distortion of the electron cloud of an anion by a neighbouring cation [1]. (b) All three cations have a 2+ charge, so polarising power depends on radius [1]. Be²⁺ is the smallest (45 pm), so its charge is the most concentrated and it has the greatest polarising power [1]. It distorts the chloride ion most, so beryllium chloride has the most covalent character [1]. (c) Iodide is much larger than fluoride; its outermost electrons sit well away from the nucleus and are distorted easily [1]. Iodide is more polarisable, so Na⁺ draws more electron density into the region between the ions [1]. Examiner insight: Cations have polarising power; anions have polarisability. Swapping the terms can lose credit on an otherwise correct answer.

8. (a) Nitrogen in NH₃ has a lone pair; boron in BF₃ has a vacancy in its outer shell [1]. Nitrogen donates both electrons of the lone pair to boron, forming a dative covalent bond [1]. Boron has 6 outer electrons before and 8 after [1]. (b) In AlCl₃, aluminium has only six outer electrons [1]. A chlorine atom on each unit donates a lone pair to the aluminium of the other unit, forming two dative bonds, so each aluminium has eight outer electrons [1]. Examiner insight: State which atom provides the pair. An arrow or description going from the electron-deficient atom to the donor shows the bond the wrong way round.

9. (a) 5 bonding pairs (two C–H single bonds and three pairs in the C≡C triple bond) [1]; 0 lone pairs [1]. (b) 4 bonding pairs (two N–H, one N–O, one O–H) [1]; 3 lone pairs (one on nitrogen, two on oxygen) [1]. Examiner insight: Count the outer electrons first (10 for C₂H₂, 14 for NH₂OH) and check your pairs add up; a triple bond is three bonding pairs, not one.

10. (a) Diamond: each carbon covalently bonded to four others in a tetrahedral 3D giant lattice [1]. Graphite: each carbon bonded to three others in flat hexagonal layers, with weak forces between layers [1]. Melting either breaks very many strong covalent bonds, needing a lot of energy [1]. (b) In graphite each carbon uses three outer electrons in bonds; the fourth is delocalised and moves along the layers [1]. In diamond all four are used in bonds, so there are no delocalised electrons [1]. (c) A single layer of graphite, one atom thick, of hexagonally bonded carbon atoms [1]. Any sensible application linked to a property, for example flexible electronics (conducts, very thin) or lightweight composites (very strong for its mass) [1]. Examiner insight: Melting graphite breaks covalent bonds within the layers; mentioning only the weak forces between layers explains sliding, not melting.

11. (a) The ability of an atom to attract the bonding pair of electrons in a covalent bond [1]. (b) C–F 1.43, B–F 1.94, P–F 1.79, C–H 0.35 [1] [1]. Increasing polarity: C–H < C–F < P–F < B–F [1]. (c) CF₄: non-polar; tetrahedral and symmetrical, so the C–F dipoles cancel [1]. CH₂F₂: polar; the C–H and C–F dipoles differ and do not cancel [1]. BF₃: non-polar; trigonal planar and symmetrical [1]. PF₃: polar; the lone pair makes it pyramidal, so the dipoles do not cancel [1]. (d) Dichloromethane is deflected towards the rod; tetrachloromethane is not [1]. CH₂Cl₂ is polar, so its molecules turn their oppositely charged end to the rod; CCl₄ is non-polar because its dipoles cancel [1]. Examiner insight: Give a shape or symmetry reason with each verdict; “polar” alone explains nothing.

Where marks are usually lost

• Defining ionic bonding as electron transfer instead of electrostatic attraction between ions.
• Omitting brackets or charges on ions, or drawing the wrong number of each ion in a dot-and-cross diagram.
• Explaining isoelectronic radii with “more shells”, when every ion in the set has the same electron arrangement.
• Using “polarising power” for anions or “polarisability” for cations.
• Getting the direction of a dative bond backwards in Al₂Cl₆ or an adduct such as H₃NBF₃.
• Saying that melting graphite only overcomes the forces between layers.
• Calling a molecule polar because its bonds are polar, without considering symmetry.
• Ranking polarity by the larger electronegativity value instead of the difference.

Next steps

• Ionic and covalent bonding revision notes
• Ionic and covalent bonding study guide
• More Unit 1 bonding practice
• Course hub: every Edexcel IAL Chemistry page
• Outcome checklist to print
• 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 3A Ionic bonding and 3B Covalent bonding, outcomes 3.1 to 3.15.