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A Level H1 Chemistry Atomic Structure Bonding Quiz

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Questions

A-Level Chemistry H1 Quiz - Atomic Structure Bonding

Name: ________________________
Class: ________________________
Date: ________________________
Score: ______ / 50

Duration: 45 minutes
Total Marks: 50

Instructions:

Answer ALL questions in the spaces provided.
Show all working for calculation questions.
Marks are indicated in square brackets [ ].
You may use a scientific calculator and the Data Booklet.

Section A: Atomic Structure (Questions 1–5)

Total: 12 marks

1. Define the term isotope.
[2]

2. Complete the table below for the species shown.

Species	Number of protons	Number of neutrons	Number of electrons
²³Na⁺
¹⁸O²⁻

[4]

3. A sample of chlorine contains two isotopes: ³⁵Cl (75.0% abundance) and ³⁷Cl (25.0% abundance). Calculate the relative atomic mass of chlorine in this sample.
[2]

4. Write the electronic configuration of a phosphorus atom (atomic number 15) using: (a) the 1s² notation
[1]

(b) orbital box notation (electrons-in-boxes)
[2]

5. State the number of orbitals in the n = 3 principal quantum shell and name the types of orbitals present.
[1]

Section B: Chemical Bonding (Questions 6–15)

Total: 26 marks

6. Draw a dot-and-cross diagram for magnesium oxide, MgO. Show outer electrons only.
[3]

7. Explain why solid sodium chloride does not conduct electricity, but molten sodium chloride does.
[2]

8. Draw the dot-and-cross diagram for the ammonia molecule, NH₃. Hence, predict and explain the shape of the ammonia molecule using VSEPR theory.
[4]

9. Explain why the H–N–H bond angle in ammonia (107°) is smaller than the H–C–H bond angle in methane (109.5°).
[2]

10. State the type of bonding present in solid aluminium and explain why aluminium is malleable.
[2]

11. The electronegativity values of some elements are:

Hydrogen: 2.1
Carbon: 2.5
Nitrogen: 3.0
Oxygen: 3.5
Fluorine: 4.0

(a) Explain what is meant by the term electronegativity.
[1]

(b) Using the data, identify the most polar bond from the following pairs and explain your choice:

C–H
N–H
O–H
F–H
[2]

12. Boron trifluoride, BF₃, reacts with ammonia, NH₃, to form the compound F₃BNH₃.

(a) Draw a dot-and-cross diagram of BF₃, showing outer electrons only.
[2]

(b) Name the type of bond formed between BF₃ and NH₃ in F₃BNH₃. Explain how this bond is formed.
[2]

13. Explain why the boiling point of water (H₂O, 100 °C) is much higher than that of hydrogen sulfide (H₂S, –60 °C), despite both molecules having a bent shape.
[3]

14. Carbon dioxide, CO₂, and silicon dioxide, SiO₂, are both oxides of Group 14 elements.

(a) Draw a dot-and-cross diagram for CO₂, showing outer electrons only.
[2]

(b) Explain why CO₂ is a gas at room temperature while SiO₂ is a solid with a very high melting point.
[3]

15. State the type of intermolecular force that exists between molecules of each of the following substances: (a) CH₄
[1]

(b) HCl
[1]

Section C: Integrated Questions (Questions 16–20)

Total: 12 marks

16. The nitrate ion, NO₃⁻, contains one coordinate covalent (dative) bond.

(a) Draw a dot-and-cross diagram for the nitrate ion, showing outer electrons only.
[2]

(b) State the shape of the nitrate ion and the O–N–O bond angle.
[1]

17. The table below shows the boiling points of the noble gases.

Noble gas	Boiling point / K
He	4.2
Ne	27.1
Ar	87.3
Kr	119.9
Xe	165.0

Explain the trend in boiling points down the group.
[2]

18. Magnesium chloride, MgCl₂, and silicon tetrachloride, SiCl₄, are both chlorides of Period 3 elements.

(a) State the type of bonding present in each compound.
[2]

(b) Predict which compound has the higher melting point. Explain your answer.
[2]

19. The hydrazine molecule, N₂H₄, has the following structure:

    H   H
     \ /
      N–N
     / \
    H   H

(a) State the total number of bonding pairs and lone pairs of electrons in one molecule of N₂H₄.
[1]

(b) Predict the H–N–H bond angle in hydrazine. Explain your answer.
[2]

20. Explain why graphite conducts electricity but diamond does not, even though both are allotropes of carbon.
[2]

END OF QUIZ

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Answers

A-Level Chemistry H1 Quiz - Atomic Structure Bonding — Answer Key

Section A: Atomic Structure

1. Define the term isotope.

Isotopes are atoms of the same element [1] that have the same number of protons (same atomic number) but different numbers of neutrons (different mass numbers). [1]

2. Complete the table:

Species	Number of protons	Number of neutrons	Number of electrons
²³Na⁺	11 [1]	12 (23–11) [1]	10 (11–1) [1]
¹⁸O²⁻	8 [1]	10 (18–8)	10 (8+2) [1]

Award 1 mark for each correct row (protons, neutrons, electrons must all be correct). Total: 4 marks.

3. Calculate relative atomic mass of chlorine:

Ar = (35 × 75.0/100) + (37 × 25.0/100) [1]
Ar = 26.25 + 9.25 = 35.5 [1]
Accept 35.5 with correct working. Total: 2 marks.

4. Electronic configuration of phosphorus (Z = 15): (a) 1s² 2s² 2p⁶ 3s² 3p³ [1] (b) Orbital box notation:

1s: ↑↓; 2s: ↑↓; 2p: ↑↓ ↑↓ ↑↓; 3s: ↑↓; 3p: ↑ ↑ ↑ (each in separate box) [2]
Award 1 mark for correct number of electrons, 1 mark for correct orbital filling (Hund's rule applied to 3p). Total: 3 marks.

5. Number of orbitals in n = 3:

9 orbitals [1]: one 3s, three 3p, and five 3d orbitals.
Accept "9 orbitals (1 × s + 3 × p + 5 × d)". Total: 1 mark.

Section B: Chemical Bonding

6. Dot-and-cross diagram for MgO:

Mg loses 2 electrons → Mg²⁺ (no outer electrons shown, or [2,8]²⁺) [1]
O gains 2 electrons → O²⁻ with 8 outer electrons (dots and crosses distinct) [1]
Correct charges and brackets: [Mg]²⁺ [O]²⁻ [1]
Total: 3 marks.

7. Electrical conductivity of NaCl:

In solid NaCl, ions are held in fixed positions in the giant ionic lattice and cannot move. [1]
In molten NaCl, the ions are free to move and carry charge. [1]
Total: 2 marks.

8. NH₃ dot-and-cross diagram and shape:

Dot-and-cross diagram: N with 5 outer electrons; 3 bonding pairs shared with 3 H atoms; 1 lone pair on N. [1]
Shape: Trigonal pyramidal. [1]
Explanation: 4 electron pairs around N (3 bond pairs + 1 lone pair) arrange tetrahedrally for maximum separation. [1]
Lone pair repels more strongly than bond pairs, compressing H–N–H angle to ~107°. Molecular shape (atoms only) is trigonal pyramidal. [1]
Total: 4 marks.

9. Bond angle comparison:

NH₃ has 1 lone pair + 3 bond pairs; CH₄ has 4 bond pairs and no lone pair. [1]
Lone pair–bond pair repulsion > bond pair–bond pair repulsion, so the H–N–H angle is compressed from the tetrahedral 109.5° to ~107°. [1]
Total: 2 marks.

10. Bonding and malleability of aluminium:

Type of bonding: Metallic bonding — lattice of Al³⁺ cations surrounded by a sea of delocalised electrons. [1]
Malleability: When a force is applied, layers of cations can slide over each other without breaking the metallic bonding because the delocalised electrons can move and maintain the electrostatic attraction. [1]
Total: 2 marks.

11. Electronegativity: (a) Electronegativity is the ability of an atom in a covalent bond to attract the bonding pair of electrons towards itself. [1] (b) Most polar bond: F–H. [1]

Explanation: Fluorine has the highest electronegativity (4.0) and hydrogen the lowest (2.1) among the pairs, giving the greatest electronegativity difference (1.9). [1]
Total: 3 marks.

12. BF₃ and NH₃ reaction: (a) Dot-and-cross diagram of BF₃:

B has 3 outer electrons, each shared with one F atom (F has 7 outer electrons, 6 as lone pairs + 1 bonding). [1]
B has only 6 electrons in its outer shell (electron-deficient). [1] (b) Bond type: Coordinate covalent (dative) bond. [1]
Formation: The lone pair on the N atom of NH₃ is donated to the empty orbital on the B atom of BF₃, forming a bond where both electrons come from nitrogen. [1]
Total: 4 marks.

13. Boiling point comparison (H₂O vs H₂S):

H₂O molecules form hydrogen bonds with each other (due to O–H bond and high electronegativity of O). [1]
H₂S molecules have only weak instantaneous dipole–induced dipole (id–id) interactions. [1]
Hydrogen bonds are much stronger than id–id interactions, so more energy is required to separate H₂O molecules, resulting in a much higher boiling point. [1]
Total: 3 marks.

14. CO₂ and SiO₂: (a) Dot-and-cross diagram of CO₂:

C shares 2 pairs of electrons with each O atom (double bonds). [1]
Each O has 2 lone pairs; C has no lone pairs. Linear molecule. [1] (b) Explanation:
CO₂ has a simple molecular structure with weak intermolecular forces (id–id) between molecules, so little energy is needed to separate them. [1]
SiO₂ has a giant covalent (network) structure with strong covalent bonds between Si and O atoms throughout the lattice. [1]
A large amount of energy is required to break these strong covalent bonds, giving SiO₂ a very high melting point. [1]
Total: 5 marks.

15. Intermolecular forces: (a) CH₄: Instantaneous dipole–induced dipole (id–id) interactions / van der Waals forces. [1] (b) HCl: Permanent dipole–dipole interactions (and id–id). [1] (c) CH₃OH: Hydrogen bonding (due to O–H group). [1]

Total: 3 marks.

Section C: Integrated Questions

16. Nitrate ion, NO₃⁻: (a) Dot-and-cross diagram:

N is central, bonded to three O atoms. [1]
One N=O double bond (4 shared electrons), two N→O dative bonds (one shown with arrow or both electrons from N). Overall charge –1 shown. Resonance structures acceptable. [1] (b) Shape: Trigonal planar. Bond angle: 120°. [1]
Total: 3 marks.

17. Noble gas boiling point trend:

Down the group, the number of electrons and size of atoms increase. [1]
This increases the strength of instantaneous dipole–induced dipole (id–id) interactions between atoms, requiring more energy to overcome, so boiling points increase. [1]
Total: 2 marks.

18. MgCl₂ and SiCl₄: (a) Bonding:

MgCl₂: Ionic bonding. [1]
SiCl₄: Covalent bonding (simple molecular). [1] (b) Higher melting point: MgCl₂. [1]
Explanation: MgCl₂ has a giant ionic lattice with strong electrostatic forces between Mg²⁺ and Cl⁻ ions, requiring a lot of energy to break. SiCl₄ has a simple molecular structure with weak intermolecular forces between molecules. [1]
Total: 4 marks.

19. Hydrazine, N₂H₄: (a) Bonding pairs: 5 (one N–N bond + four N–H bonds). Lone pairs: 2 (one on each N atom). [1] (b) H–N–H bond angle: Approximately 107°. [1]

Each N atom has 3 bond pairs and 1 lone pair (4 electron pairs total). Electron pairs arrange tetrahedrally; lone pair repulsion compresses the bond angle from 109.5° to ~107°, similar to NH₃. [1]
Total: 3 marks.

20. Graphite vs diamond conductivity:

In graphite, each C atom is bonded to 3 other C atoms, leaving one delocalised electron per C atom. These delocalised electrons can move between the layers and carry charge. [1]
In diamond, each C atom is bonded to 4 other C atoms; all outer electrons are used in covalent bonds. There are no free/mobile charge carriers. [1]
Total: 2 marks.

END OF ANSWER KEY