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A Level H2 Chemistry Atomic Structure Bonding Quiz
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Questions
A-Level Chemistry H2 Quiz - Atomic Structure Bonding
Name: ________________________
Class: ________________________
Date: ________________________
Score: ______ / 50
Duration: 45 minutes
Total Marks: 50
Instructions:
- Answer ALL questions in the spaces provided.
- Use of the Data Booklet is relevant to some questions.
- Show all working for calculation questions.
- Marks are indicated in brackets [ ].
Section A: Multiple Choice (5 × 2 marks = 10 marks)
For questions 1–5, circle the correct answer.
1. Which of the following species has the same number of electrons as the ion ³⁹K⁺?
A. ⁴⁰Ar
B. ³⁵Cl⁻
C. ⁴⁰Ca²⁺
D. ³²S²⁻
[2]
2. The first six ionisation energies (in kJ mol⁻¹) of an element are: 1090, 2350, 4620, 6220, 37800, 47300
What is the most likely identity of this element?
A. Boron
B. Carbon
C. Silicon
D. Phosphorus
[2]
3. Which molecule has a permanent dipole moment?
A. CO₂
B. BF₃
C. SO₂
D. CCl₄
[2]
4. The ion X²⁺ contains 24 electrons and 28 neutrons. What is the nucleon number of X?
A. 52
B. 54
C. 56
D. 58
[2]
5. Which statement about hydrogen bonding is correct?
A. It occurs in all molecules containing hydrogen atoms.
B. It is stronger than covalent bonding.
C. It explains why ice is less dense than liquid water.
D. It is present in gaseous hydrogen chloride.
[2]
Section B: Structured Questions (15 marks)
6. The table below shows the first seven ionisation energies of sodium.
| Ionisation | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th |
|---|---|---|---|---|---|---|---|
| IE / kJ mol⁻¹ | 496 | 4560 | 6910 | 9540 | 13400 | 16600 | 20100 |
(a) Write the full electron configuration of a sodium atom.
___________________________________________________________________________ [1]
(b) Explain why the first ionisation energy of sodium is much lower than the second ionisation energy.
___________________________________________________________________________ [3]
(c) Predict and explain the trend in the first ionisation energies across Period 3 from sodium to argon.
___________________________________________________________________________ [3]
7. Phosphorus pentachloride, PCl₅, exists as a trigonal bipyramidal molecule in the gas phase.
(a) Draw the Lewis structure of PCl₅, showing all valence electrons.
[2]
(b) State and explain the shape of the PCl₅ molecule.
___________________________________________________________________________ [2]
(c) In the solid state, PCl₅ exists as [PCl₄]⁺[PCl₆]⁻. Draw the shape of the [PCl₆]⁻ ion and state its bond angles.
Shape: ________________________
Bond angle: ________________________ [2]
8. The table below gives information about three substances: sodium chloride, diamond, and iodine.
| Substance | Melting point / °C | Electrical conductivity of solid | Electrical conductivity of liquid |
|---|---|---|---|
| Sodium chloride | 801 | Poor | Good |
| Diamond | 3550 | Poor | Poor |
| Iodine | 114 | Poor | Poor |
(a) Explain, in terms of structure and bonding, why solid sodium chloride does not conduct electricity but liquid sodium chloride does.
___________________________________________________________________________ [3]
(b) Diamond has a very high melting point. Explain this observation with reference to its structure and bonding.
___________________________________________________________________________ [2]
(c) Iodine has a relatively low melting point. Name the type of intermolecular force present in solid iodine and explain how it arises.
___________________________________________________________________________ [2]
9. The nitrate ion, NO₃⁻, can be represented by resonance structures.
(a) Draw TWO resonance structures of the nitrate ion, showing all bonds and formal charges.
[3]
(b) Explain what is meant by the term resonance and describe the actual structure of the nitrate ion.
___________________________________________________________________________ [2]
(c) State and explain the shape of the nitrate ion.
___________________________________________________________________________ [2]
10. Ammonia, NH₃, and boron trifluoride, BF₃, react to form the adduct H₃NBF₃.
(a) Draw a dot-and-cross diagram for ammonia, showing outer-shell electrons only.
[1]
(b) Explain why the H–N–H bond angle in ammonia (107°) is smaller than the ideal tetrahedral angle.
___________________________________________________________________________ [2]
(c) State the type of bond formed between NH₃ and BF₃ in the adduct H₃NBF₃, and explain how this bond is formed.
___________________________________________________________________________ [2]
(d) Predict the shape around the nitrogen atom and the boron atom in the adduct H₃NBF₃.
Around N: ________________________
Around B: ________________________ [2]
Section C: Structured Questions (15 marks)
11. The boiling points of the hydrogen halides are given below.
| Hydrogen halide | HF | HCl | HBr | HI |
|---|---|---|---|---|
| Boiling point / °C | 20 | –85 | –67 | –35 |
(a) Explain why the boiling point of HF is anomalously high compared to the other hydrogen halides.
___________________________________________________________________________ [2]
(b) Explain the trend in boiling points from HCl to HI.
___________________________________________________________________________ [2]
12. Aluminium oxide, Al₂O₃, is classified as an amphoteric oxide.
(a) What is meant by the term amphoteric?
___________________________________________________________________________ [1]
(b) Write an ionic equation for the reaction of aluminium oxide with hot aqueous sodium hydroxide.
___________________________________________________________________________ [1]
(c) Write an ionic equation for the reaction of aluminium oxide with dilute hydrochloric acid.
___________________________________________________________________________ [1]
13. The element magnesium (Mg) has three naturally occurring isotopes: ²⁴Mg, ²⁵Mg, and ²⁶Mg.
(a) Define the term isotopes.
___________________________________________________________________________ [1]
(b) The relative atomic mass of magnesium is 24.3. Explain why this value is not a whole number.
___________________________________________________________________________ [2]
(c) State the number of protons, neutrons, and electrons in a ²⁶Mg²⁺ ion.
Protons: ______ Neutrons: ______ Electrons: ______ [2]
14. The shapes of molecules and ions can be predicted using VSEPR theory.
(a) State what the acronym VSEPR stands for.
___________________________________________________________________________ [1]
(b) Predict the shape and bond angle of the SF₆ molecule.
Shape: ________________________ Bond angle: ________________________ [2]
(c) The ion I₃⁻ is linear. Explain this shape using VSEPR theory.
___________________________________________________________________________ [2]
15. The Born-Haber cycle can be used to determine the lattice energy of an ionic compound.
(a) Define the term lattice energy.
___________________________________________________________________________ [1]
(b) State whether lattice energy is an exothermic or endothermic process, and explain why.
___________________________________________________________________________ [2]
(c) The lattice energy of magnesium oxide (MgO) is much more exothermic than that of sodium chloride (NaCl). Explain this difference.
___________________________________________________________________________ [2]
Section D: Data-Based and Integrated Questions (10 marks)
16. The graph below shows the first ionisation energies of the elements from lithium (Li) to neon (Ne).
[Assume a graph is provided showing IE peaks at Be and N, and dips at B and O.]
(a) Explain why the first ionisation energy of beryllium is higher than that of boron.
___________________________________________________________________________ [2]
(b) Explain why the first ionisation energy of nitrogen is higher than that of oxygen.
___________________________________________________________________________ [2]
(c) Predict, with a reason, whether the first ionisation energy of neon is higher or lower than that of helium.
___________________________________________________________________________ [1]
17. Carbon dioxide (CO₂) and silicon dioxide (SiO₂) are both Group IV oxides, but they have very different structures and properties.
(a) State the type of structure and bonding present in solid CO₂ and solid SiO₂.
CO₂: ___________________________________________________________________________
SiO₂: ___________________________________________________________________________ [2]
(b) Explain why CO₂ sublimes at a low temperature, whereas SiO₂ has a very high melting point.
___________________________________________________________________________ [2]
(c) Write a balanced equation for the reaction of SiO₂ with hot, concentrated sodium hydroxide solution.
___________________________________________________________________________ [1]
18. The Pauling electronegativity values for some elements are: H (2.1), C (2.5), N (3.0), O (3.5), F (4.0), Cl (3.0).
(a) Define the term electronegativity.
___________________________________________________________________________ [1]
(b) Using the values above, identify the most polar bond among the following and explain your choice: C–H, N–H, O–H, F–H.
___________________________________________________________________________ [2]
(c) The molecule CH₃Cl has a dipole moment, whereas CCl₄ does not. Explain this observation.
___________________________________________________________________________ [2]
19. The table below shows the melting points of some Period 3 elements.
| Element | Na | Mg | Al | Si | P | S | Cl | Ar |
|---|---|---|---|---|---|---|---|---|
| Melting point / °C | 98 | 650 | 660 | 1410 | 44 | 119 | –101 | –189 |
(a) Explain the general trend in melting points from Na to Al.
___________________________________________________________________________ [2]
(b) Silicon has a very high melting point. Explain this observation with reference to its structure and bonding.
___________________________________________________________________________ [2]
(c) Explain why the melting point of phosphorus (P₄) is much lower than that of sulfur (S₈).
___________________________________________________________________________ [1]
20. The cyanide ion, CN⁻, and carbon monoxide, CO, are isoelectronic.
(a) State the meaning of the term isoelectronic.
___________________________________________________________________________ [1]
(b) Draw the Lewis structure of the cyanide ion, showing all bonding and non-bonding electrons, and indicate the formal charges.
[2]
(c) Explain why CO is a polar molecule, even though it is isoelectronic with the non-polar N₂ molecule.
___________________________________________________________________________ [2]
END OF QUIZ
Answers
A-Level Chemistry H2 Quiz - Atomic Structure Bonding — ANSWER KEY
Total Marks: 50
Section A: Multiple Choice (5 × 2 marks = 10 marks)
1. B. ³⁵Cl⁻
³⁹K⁺ has 19 protons, so 18 electrons (19 − 1). ³⁵Cl⁻ has 17 protons, so 18 electrons (17 + 1).
Award 2 marks for correct answer.
2. C. Silicon
Large jump between 4th and 5th IE indicates removal of electron from inner shell after 4 valence electrons removed. Silicon (Group 14) has 4 valence electrons.
Award 2 marks for correct answer.
3. C. SO₂
SO₂ is bent (V-shaped) due to lone pair on sulfur; bond dipoles do not cancel → permanent dipole. CO₂ is linear (non-polar), BF₃ is trigonal planar (non-polar), CCl₄ is tetrahedral (non-polar).
Award 2 marks for correct answer.
4. B. 54
X²⁺ has 24 electrons → neutral X has 26 electrons = 26 protons (atomic number 26, iron). Neutrons = 28. Nucleon number = 26 + 28 = 54.
Award 2 marks for correct answer.
5. C. It explains why ice is less dense than liquid water.
Hydrogen bonding in ice creates an open lattice structure, making ice less dense than liquid water.
Award 2 marks for correct answer.
Section B: Structured Questions (15 marks)
6. Ionisation energies of sodium.
(a) 1s² 2s² 2p⁶ 3s¹
Award 1 mark for correct configuration.
(b) The first electron is removed from the 3s orbital, which is further from the nucleus and experiences greater shielding from inner electrons [1]. The second electron is removed from the 2p orbital, which is closer to the nucleus and less shielded [1]. The effective nuclear charge experienced by the 2p electron is much greater, so much more energy is required to remove it [1].
Award up to 3 marks for a clear explanation referencing distance, shielding, and effective nuclear charge.
(c) Across Period 3, first ionisation energy generally increases [1]. This is because nuclear charge increases (more protons) while electrons are added to the same principal quantum shell (n = 3), so shielding remains approximately constant [1]. The increased effective nuclear charge means outer electrons are held more tightly, requiring more energy to remove [1].
Award up to 3 marks. Accept mention of small dips at Al and S due to subshell effects.
7. Phosphorus pentachloride.
(a) Lewis structure: P as central atom with 5 single bonds to Cl atoms. Each Cl has 3 lone pairs. P has no lone pairs (expanded octet, 10 electrons around P).
Award 2 marks: 1 for correct connectivity, 1 for correct lone pairs on Cl atoms.
(b) Trigonal bipyramidal [1]. There are 5 bonding pairs and 0 lone pairs around the central P atom. Electron pairs repel to positions of minimum repulsion, giving bond angles of 90° (axial–equatorial) and 120° (equatorial–equatorial) [1].
Award 2 marks.
(c) Shape: Octahedral [1]. Bond angle: 90° [1].
Award 2 marks.
8. Structure and bonding of substances.
(a) Solid NaCl has ions (Na⁺ and Cl⁻) held in fixed positions in a giant ionic lattice; ions cannot move, so no electrical conductivity [1]. When molten, the ions are free to move [1]. Mobile charge carriers (ions) allow conduction of electricity [1].
Award up to 3 marks.
(b) Diamond has a giant covalent (macromolecular) structure [1]. Each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral arrangement. A large amount of energy is required to break these strong covalent bonds throughout the lattice [1].
Award 2 marks.
(c) Instantaneous dipole–induced dipole (London/dispersion) forces [1]. These arise from temporary fluctuations in electron distribution around I₂ molecules, creating temporary dipoles that induce dipoles in neighbouring molecules [1].
Award 2 marks. Accept "van der Waals' forces" if qualified as London forces.
9. Nitrate ion resonance.
(a) Two resonance structures: N as central atom with one N=O double bond and two N–O single bonds. Formal charges: N carries +1, singly-bonded O atoms carry –1 each, doubly-bonded O is neutral. The two structures differ in which O atom is doubly bonded.
Award 3 marks: 1 for each correct structure with formal charges, 1 for showing two distinct structures.
(b) Resonance occurs when two or more valid Lewis structures can be drawn for a molecule/ion that differ only in the distribution of electrons [1]. The actual structure is a resonance hybrid — an average of the resonance forms. In NO₃⁻, all three N–O bonds are equivalent, with bond order 1⅓ and equal bond lengths [1].
Award 2 marks.
(c) Trigonal planar [1]. There are 3 bonding regions (electron domains) around the central N atom and no lone pairs. Electron pairs repel to positions of minimum repulsion, giving bond angles of 120° [1].
Award 2 marks.
10. Ammonia–boron trifluoride adduct.
(a) Dot-and-cross diagram: N with 5 outer electrons, 3 H atoms each with 1 electron. Three N–H bonding pairs shown, one lone pair on N.
Award 1 mark for correct diagram.
(b) Ammonia has 3 bonding pairs and 1 lone pair around N [1]. Lone pair–bonding pair repulsion > bonding pair–bonding pair repulsion, so the H–N–H bond angle is compressed from the ideal tetrahedral angle of 109.5° to 107° [1].
Award 2 marks.
(c) Dative covalent (coordinate) bond [1]. The lone pair on the N atom of NH₃ is donated to the electron-deficient B atom of BF₃ (which has an empty p orbital), forming a shared pair of electrons [1].
Award 2 marks.
(d) Around N: Tetrahedral [1]. Around B: Tetrahedral [1].
Award 2 marks.
Section C: Structured Questions (15 marks)
11. Boiling points of hydrogen halides.
(a) HF molecules form hydrogen bonds with each other (H–F···H–F) [1]. Hydrogen bonds are the strongest type of intermolecular force, requiring significantly more energy to overcome than the London forces present in the other hydrogen halides, resulting in a much higher boiling point [1].
Award 2 marks.
(b) From HCl to HI, the number of electrons in the molecule increases (HCl: 18 e⁻, HBr: 36 e⁻, HI: 54 e⁻) [1]. This leads to stronger instantaneous dipole–induced dipole (London) forces, as larger electron clouds are more easily polarised. More energy is required to overcome these forces, so boiling point increases [1].
Award 2 marks.
12. Amphoteric nature of aluminium oxide.
(a) An amphoteric substance is one that can react with both acids and bases (to form a salt and water).
Award 1 mark.
(b) Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2[Al(OH)₄]⁻(aq)
Award 1 mark. Accept Al₂O₃(s) + 2OH⁻(aq) → 2AlO₂⁻(aq) + H₂O(l).
(c) Al₂O₃(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂O(l)
Award 1 mark.
13. Isotopes of magnesium.
(a) Isotopes are atoms of the same element with the same number of protons (atomic number) but different numbers of neutrons (different mass numbers).
Award 1 mark.
(b) The relative atomic mass is the weighted average mass of the naturally occurring isotopes of an element, taking into account their relative abundances [1]. Since it is an average of the masses of the isotopes (which are whole numbers), the result is generally not a whole number [1].
Award 2 marks.
(c) Protons: 12 Neutrons: 14 Electrons: 10
Award 2 marks: 1 for protons and neutrons, 1 for electrons.
14. VSEPR theory.
(a) Valence Shell Electron Pair Repulsion
Award 1 mark.
(b) Shape: Octahedral [1]. Bond angle: 90° [1].
Award 2 marks.
(c) The central I atom has 2 bonding pairs and 3 lone pairs [1]. The electron pairs arrange themselves in a trigonal bipyramidal arrangement to minimise repulsion. The three lone pairs occupy the equatorial positions, leaving the two bonding pairs in the axial positions, resulting in a linear shape [1].
Award 2 marks.
15. Lattice energy.
(a) Lattice energy is the enthalpy change when one mole of an ionic solid is formed from its gaseous ions under standard conditions.
Award 1 mark.
(b) Exothermic [1]. Energy is released when gaseous ions come together to form the solid ionic lattice due to strong electrostatic attractions between oppositely charged ions [1].
Award 2 marks.
(c) MgO contains Mg²⁺ and O²⁻ ions, while NaCl contains Na⁺ and Cl⁻ ions [1]. The charges on the ions in MgO are greater (2+ and 2− vs 1+ and 1−), and the ionic radii of Mg²⁺ and O²⁻ are smaller than those of Na⁺ and Cl⁻. The greater charge and smaller size lead to stronger electrostatic attractions and a more exothermic lattice energy [1].
Award 2 marks.
Section D: Data-Based and Integrated Questions (10 marks)
16. First ionisation energies Li to Ne.
(a) Beryllium has electron configuration 1s² 2s²; boron has 1s² 2s² 2p¹ [1]. The electron removed from boron is in a 2p orbital, which is higher in energy (further from nucleus) than the 2s orbital in beryllium. The 2s electron in Be experiences greater effective nuclear charge, so more energy is required to remove it [1].
Award 2 marks.
(b) Nitrogen has electron configuration 1s² 2s² 2p³ (half-filled p subshell); oxygen has 1s² 2s² 2p⁴ [1]. In oxygen, one of the 2p orbitals contains a pair of electrons. The repulsion between these paired electrons makes it easier to remove one of them, so the first IE of oxygen is lower than that of nitrogen [1].
Award 2 marks.
(c) Higher [1]. Neon has a greater nuclear charge than helium, and the electrons are in the same principal quantum shell (n=2 vs n=1 for He), but the increased nuclear charge outweighs the increased distance, resulting in a higher IE. (Or: Ne has a smaller atomic radius and greater effective nuclear charge than He.)
Award 1 mark for correct prediction with valid reason.
17. Group IV oxides.
(a) CO₂: Simple molecular (covalent) structure with weak intermolecular forces [1].
SiO₂: Giant covalent (macromolecular) structure with strong covalent bonds throughout [1].
Award 2 marks.
(b) CO₂ consists of discrete CO₂ molecules held together by weak London forces, which require little energy to overcome, so it sublimes at a low temperature [1]. SiO₂ has a giant covalent structure where all Si and O atoms are held together by strong covalent bonds. A large amount of energy is required to break these bonds, resulting in a very high melting point [1].
Award 2 marks.
(c) SiO₂(s) + 2NaOH(aq) → Na₂SiO₃(aq) + H₂O(l)
Award 1 mark. Accept SiO₂(s) + 2OH⁻(aq) → SiO₃²⁻(aq) + H₂O(l).
18. Electronegativity and polarity.
(a) Electronegativity is the ability of an atom in a covalent bond to attract the shared pair of electrons towards itself.
Award 1 mark.
(b) F–H is the most polar bond [1]. It has the largest difference in electronegativity (4.0 − 2.1 = 1.9) compared to the other bonds (C–H: 0.4, N–H: 0.9, O–H: 1.4) [1].
Award 2 marks.
(c) In CCl₄, the molecule is tetrahedral and symmetrical; the individual C–Cl bond dipoles cancel out, resulting in no net dipole moment [1]. In CH₃Cl, the molecule is not symmetrical (tetrahedral with different atoms attached); the bond dipoles do not cancel, resulting in a net dipole moment [1].
Award 2 marks.
19. Melting points of Period 3 elements.
(a) From Na to Al, melting point increases [1]. This is due to the increasing strength of metallic bonding. The number of delocalised valence electrons per atom increases (Na: 1, Mg: 2, Al: 3), and the ionic radius decreases, leading to stronger electrostatic attractions between metal cations and the sea of delocalised electrons [1].
Award 2 marks.
(b) Silicon has a giant covalent (macromolecular) structure, similar to diamond [1]. Each Si atom is covalently bonded to four other Si atoms in a tetrahedral arrangement. A large amount of energy is required to break these strong covalent bonds throughout the lattice, resulting in a very high melting point [1].
Award 2 marks.
(c) Phosphorus exists as discrete P₄ molecules, while sulfur exists as S₈ molecules [1]. S₈ molecules are larger and have more electrons, resulting in stronger London forces between molecules compared to P₄. More energy is required to overcome these forces, so sulfur has a higher melting point.
Award 1 mark.
20. Isoelectronic species.
(a) Isoelectronic species are atoms, ions, or molecules that have the same number of electrons (and the same electronic configuration).
Award 1 mark.
(b) Lewis structure: C≡N with a lone pair on C and a lone pair on N. Formal charges: C carries −1, N carries 0. (Triple bond, one lone pair on each atom; total 10 valence electrons.)
Award 2 marks: 1 for correct connectivity and electrons, 1 for correct formal charges.
(c) CO is polar because the electronegativity difference between C and O creates a bond dipole [1]. Although CO is isoelectronic with N₂ (both have 10 electrons), N₂ consists of two identical atoms with no electronegativity difference, so the bond is non-polar. In CO, the unequal sharing of electrons results in a net dipole moment [1].
Award 2 marks.
END OF ANSWER KEY