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A Level H2 Chemistry Atomic Structure Bonding Quiz

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A-Level Chemistry H2 Quiz - Atomic Structure Bonding

Name: __________________________
Class: __________________________
Date: __________________________
Score: _______ / 45

Duration: 45 minutes
Total Marks: 45

Instructions:

Answer all questions.
Write your answers in the spaces provided.
The use of a Data Booklet is relevant to this quiz.
Marks are indicated in brackets [ ] at the end of each question or part question.

Section A: Atomic Structure and Periodicity (Questions 1–5)

1. The first seven ionisation energies of an element X are shown below.

Ionisation Number	1st	2nd	3rd	4th	5th	6th	7th
Energy / kJ mol⁻¹	590	1145	4912	6474	8144	10496	12320

(a) To which group of the Periodic Table does element X belong? Explain your answer.
[2]

(b) Write the equation, including state symbols, for the third ionisation energy of element X.
[1]

2. Explain why the first ionisation energy of aluminium (Al) is lower than that of magnesium (Mg), despite aluminium having a higher nuclear charge.
[2]

3. The diagram below shows the variation in atomic radius across Period 3 elements (Na to Cl).

(a) State and explain the general trend in atomic radius across Period 3.
[2]

(b) Explain why the ionic radius of the sulfide ion (S²⁻) is larger than the ionic radius of the chloride ion (Cl⁻), even though they are isoelectronic.
[2]

4. An ion $Y^{3+}$ has the electron configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5$ .

(a) Deduce the atomic number of element Y.
[1]

(b) Write the electron configuration of the neutral atom Y.
[1]

5. Successive ionisation energies provide evidence for the existence of electron shells and sub-shells.

(a) Explain how the large jump between the second and third ionisation energies of magnesium supports the existence of principal quantum shells.
[2]

(b) Explain why there is a smaller increase in ionisation energy between the 2s and 2p electrons in carbon compared to the jump between 1s and 2s electrons.
[1]

Section B: Chemical Bonding and Structure (Questions 6–12)

6. Draw the 'dot-and-cross' diagram for the bonding in magnesium chloride ( $MgCl_2$ ). Show outer electrons only.
[2]

7. Aluminium chloride ( $AlCl_3$ ) exhibits different bonding characteristics depending on its state.

(a) In the solid state at low temperatures, $AlCl_3$ exists as a lattice. Describe the type of bonding and structure present.
[2]

(b) In the gas phase at high temperatures, $AlCl_3$ exists as discrete molecules. At intermediate temperatures, it dimerises to form $Al_2Cl_6$ . Draw the structure of $Al_2Cl_6$ , clearly showing any coordinate (dative covalent) bonds.
[2]

8. Consider the molecules $BF_3$ and $NF_3$ .

(a) Draw the shape of the $BF_3$ molecule. State the bond angle.
[2]

(b) Draw the shape of the $NF_3$ molecule. State the bond angle and explain why it differs from the ideal tetrahedral angle of 109.5°.
[3]

9. Silicon(IV) oxide ( $SiO_2$ ) and carbon dioxide ( $CO_2$ ) are both oxides of Group 14 elements.

(a) Explain why $SiO_2$ has a very high melting point while $CO_2$ is a gas at room temperature. Refer to the structure and bonding in your answer.
[3]

(b) Explain why $CO_2$ is a non-polar molecule despite containing polar C=O bonds.
[2]

10. The table below shows the boiling points of the hydrogen halides.

Compound	HF	HCl	HBr	HI
Boiling Point / K	293	188	206	238

(a) Explain the trend in boiling points from HCl to HI.
[2]

(b) Explain why HF has a significantly higher boiling point than HCl, despite having a lower molecular mass.
[2]

11. Benzene ( $C_6H_6$ ) is a planar molecule.

(a) Describe the bonding in benzene in terms of sigma ( $\sigma$ ) and pi ( $\pi$ ) bonds.
[2]

(b) Explain why all carbon-carbon bond lengths in benzene are identical (0.139 nm), whereas typical C-C single bonds are 0.154 nm and C=C double bonds are 0.134 nm.
[2]

12. Metallic bonding explains the physical properties of metals.

(a) Describe the structure and bonding in a typical metal such as copper.
[2]

(b) Explain why magnesium has a higher melting point than sodium.
[2]

Section C: Intermolecular Forces and Properties (Questions 13–20)

13. Which of the following species exhibits hydrogen bonding? A. $CH_4$ B. $CH_3OH$ C. $CH_3OCH_3$ D. $CH_3Cl$

Answer: ______ [1]

14. Graphite and diamond are allotropes of carbon.

(a) Explain why graphite conducts electricity whereas diamond does not.
[2]

(b) Explain why graphite is soft and slippery, making it suitable as a lubricant.
[2]

15. Water ( $H_2O$ ) expands upon freezing.

(a) Explain this phenomenon in terms of hydrogen bonding and structure.
[2]

(b) Why is this property important for aquatic life?
[1]

16. Consider the isomers propanone ( $CH_3COCH_3$ ) and propanal ( $CH_3CH_2CHO$ ).

(a) Identify the dominant intermolecular force present in pure propanone.
[1]

(b) Both compounds have the same molecular mass ( $M_r = 58$ ). Propanal has a boiling point of 49°C, while propanone has a boiling point of 56°C. Suggest a reason for this difference based on molecular shape and surface area contact.
[2]

17. The ammonium ion ( $NH_4^+$ ) is formed when ammonia reacts with a proton ( $H^+$ ).

(a) What type of bond is formed between the nitrogen atom and the fourth hydrogen atom?
[1]

(b) Draw the shape of the ammonium ion and state the bond angle.
[2]

18. Sulfur hexafluoride ( $SF_6$ ) is an octahedral molecule.

(a) Draw the shape of $SF_6$ .
[1]

(b) Explain why $SF_6$ is chemically inert and non-polar.
[2]

19. Iron(III) chloride ( $FeCl_3$ ) has a lower melting point (306°C) than sodium chloride ( $NaCl$ , 801°C).

(a) Explain this difference in terms of bonding character.
[2]

(b) $FeCl_3$ dissolves in water to form an acidic solution. Write an equation to show the reaction of the hexaaquairon(III) ion with water acting as a base.
[2]

20. Which of the following statements about the bonding in ethene ( $C_2H_4$ ) is correct? A. It contains one $\sigma$ bond and five $\pi$ bonds. B. It contains five $\sigma$ bonds and one $\pi$ bond. C. It contains four $\sigma$ bonds and two $\pi$ bonds. D. It contains six $\sigma$ bonds.

Answer: ______ [1]

Answers

A-Level Chemistry H2 Quiz - Atomic Structure Bonding: Answer Key

1. (a) Group 2. [1] There is a large jump in ionisation energy between the 2nd and 3rd electrons, indicating the 3rd electron is removed from an inner shell closer to the nucleus. [1] (b) $Mg^{2+}(g) \rightarrow Mg^{3+}(g) + e^-$ [1] (Must include state symbols and correct charges).

2. The outer electron in Al is in a 3p orbital, while in Mg it is in a 3s orbital. [1] The 3p orbital is higher in energy and further from the nucleus (and experiences more shielding) than the 3s orbital, making it easier to remove. [1]

3. (a) Atomic radius decreases across the period. [1] Nuclear charge increases while shielding remains constant (electrons added to same shell), pulling the outer electrons closer. [1] (b) S²⁻ and Cl⁻ have the same number of electrons (18). [1] Sulfur has fewer protons (16) than chlorine (17), so the nuclear attraction for the outer electrons is weaker in S²⁻, resulting in a larger radius. [1]

4. (a) Atomic number = 26 (Iron). [1] (Configuration ends in $3d^5$ , implying neutral was $3d^6 4s^2$ , total 26 electrons). (b) $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2$ [1]

5. (a) The large jump indicates the 3rd electron is removed from a principal quantum shell (n=2) closer to the nucleus than the first two (n=3). [1] This requires significantly more energy due to less shielding and stronger nuclear attraction. [1] (b) 2s and 2p electrons are in the same principal quantum shell (n=2), so the energy difference is small compared to changing shells (n=1 to n=2). [1]

6. Diagram showing $[Mg]^{2+}$ and two $[Cl]^-$ ions. [1] Correct transfer of electrons: Mg loses 2, each Cl gains 1. Outer shells of Cl should show 8 electrons (crosses/dots). [1]

7. (a) Giant ionic lattice. [1] Strong electrostatic forces of attraction between oppositely charged ions ( $Al^{3+}$ and $Cl^-$ ). [1] (b) Structure showing two Al atoms bridged by two Cl atoms. [1] Arrows or clear indication of dative bonds from Cl lone pairs to empty orbitals of Al. [1]

8. (a) Trigonal planar. [1] Bond angle: 120°. [1] (b) Trigonal pyramidal. [1] Bond angle: ~107°. [1] Lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, compressing the angle. [1]

9. (a) $SiO_2$ has a giant covalent (macromolecular) structure with strong covalent bonds throughout the lattice requiring much energy to break. [1] $CO_2$ consists of simple molecules held by weak van der Waals forces. [1] Little energy is needed to overcome these intermolecular forces. [1] (b) The molecule is linear (O=C=O). [1] The bond dipoles are equal and opposite, so they cancel out, resulting in no net dipole moment. [1]

10. (a) Boiling point increases from HCl to HI. [1] Molecular size/mass increases, leading to stronger van der Waals (London dispersion) forces. [1] (b) HF molecules form hydrogen bonds due to the high electronegativity of F and the H-F bond polarity. [1] Hydrogen bonds are stronger than the van der Waals forces in HCl. [1]

11. (a) Each carbon forms 3 $\sigma$ bonds (2 to C, 1 to H) using $sp^2$ hybrid orbitals. [1] Unhybridized p-orbitals overlap sideways to form a delocalized $\pi$ system above and below the ring. [1] (b) The $\pi$ electrons are delocalized over all 6 carbon atoms. [1] This results in bond orders of 1.5, making bond lengths intermediate between single and double bonds. [1]

12. (a) Lattice of positive metal ions/cations in a 'sea' of delocalized electrons. [1] Strong electrostatic attraction between cations and delocalized electrons. [1] (b) Mg²⁺ has a higher charge density than Na⁺. [1] Mg contributes 2 electrons to the sea of electrons vs 1 for Na, leading to stronger metallic bonding. [1]

13. B [1]

14. (a) Graphite has delocalized electrons between layers that can move and carry charge. [1] Diamond has all electrons localized in covalent bonds. [1] (b) Graphite has weak van der Waals forces between layers. [1] These layers can slide over each other easily. [1]

15. (a) In ice, hydrogen bonds hold water molecules in an open, tetrahedral lattice structure. [1] This structure has more empty space than liquid water, making ice less dense. [1] (b) Ice floats, insulating the water below and allowing aquatic life to survive in winter. [1]

16. (a) Permanent dipole-permanent dipole interactions. [1] (b) Propanone is more compact/spherical, while propanal is more linear/elongated. [1] Propanal has a larger surface area for contact, leading to stronger van der Waals forces? Correction: Actually, propanone usually has a slightly higher BP due to the carbonyl group being more exposed/polarizable or specific packing. However, standard A-Level logic often attributes BP differences in isomers to surface area. Alternative Acceptable Answer: Propanone has a more polar C=O bond exposed, leading to stronger dipole-dipole interactions than propanal where the dipole is partially shielded by the ethyl group rotation. [2] (Accept valid reasoning regarding surface area or dipole exposure).

17. (a) Dative covalent (coordinate) bond. [1] (b) Tetrahedral. [1] 109.5°. [1]

18. (a) Octahedral shape drawn correctly. [1] (b) Symmetrical shape causes bond dipoles to cancel. [1] S-F bonds are strong and F is small, protecting the S atom from attack (kinetic inertness). [1]

19. (a) $Fe^{3+}$ has a high charge density, causing significant polarization of the chloride electron cloud. [1] This introduces covalent character to the bonding, weakening the lattice energy compared to pure ionic NaCl. [1] (b) $[Fe(H_2O)_6]^{3+} + H_2O \rightleftharpoons [Fe(H_2O)_5(OH)]^{2+} + H_3O^+$ [2] (Correct species and equilibrium arrow).

20. B [1] (4 C-H sigma, 1 C-C sigma, 1 C-C pi).