A Level H2 Chemistry Atomic Structure Bonding Quiz

A-Level Chemistry H2 Quiz - Atomic Structure Bonding - ANSWERS

Total Marks: 50

Section A: Multiple Choice (10 marks)

1. Which of the following species has the same number of electrons as a neon atom? Answer: B. F⁻ Neon has 10 electrons. F⁻ has 9+1=10 electrons. Na⁺ has 10, Mg²⁺ has 10, O²⁻ has 10. However, only one answer is typically correct in this format; F⁻ is a common choice, but note that Na⁺, Mg²⁺, and O²⁻ also have 10 electrons. The question likely expects F⁻ as the most direct isoelectronic species, or there is an error in the question. Given standard A-level questions, F⁻ is the intended answer.

2. The successive ionisation energies (in kJ mol⁻¹) of an element are: 590, 1145, 4912, 6491, 8153. The element is most likely to be in: Answer: B. Group II Large jump between 2nd and 3rd IE indicates removal of electron from inner shell after 2 valence electrons are removed.

3. Which molecule has a permanent dipole moment? Answer: C. SO₂ SO₂ is bent (V-shaped) and polar. CO₂, CCl₄, BeCl₂ are linear/tetrahedral and non-polar.

4. The shape of the IF₄⁺ ion is best described as: Answer: C. see-saw I has 7 valence electrons, plus 4 from F, minus 1 for positive charge = 10 electrons, 5 pairs. 4 bonding pairs, 1 lone pair -> see-saw.

5. Which statement about metallic bonding is correct? Answer: A. Metals are malleable because the positive ions can slide over each other.

6. Which pair of species has the same bond order? Answer: D. N₂ and CO Both have bond order 3 (isoelectronic).

7. The boiling point of HF (19.5 °C) is anomalously high compared to HCl (−85 °C). This is best explained by: Answer: C. hydrogen bonding between HF molecules

8. Which of the following does NOT have a linear shape? Answer: D. NO₂⁻ NO₂⁻ is bent (V-shaped) due to lone pair on N. I₃⁻, XeF₂, NO₂⁺ are linear.

9. The electron configuration of a transition metal ion is [Ar] 3d⁵. The ion could be: Answer: D. Fe³⁺ Fe is [Ar] 3d⁶4s², Fe³⁺ is [Ar] 3d⁵. Mn²⁺ is also [Ar] 3d⁵, but Fe³⁺ is a common answer.

10. Which substance has the highest melting point? Answer: A. SiO₂ Giant covalent structure. CO₂, SO₂, P₄O₁₀ are simple molecular.

Section B: Structured Questions (25 marks)

11. (a) The large increase between the 2nd and 3rd ionisation energies indicates that the 3rd electron is being removed from an inner principal quantum shell (closer to the nucleus) where it experiences a much greater effective nuclear charge and is not shielded by inner electrons. [2 marks]

(b) Element X is magnesium (Mg). The first two ionisation energies are relatively low and close together, indicating removal of two valence electrons. The large jump to the 3rd IE shows that the 3rd electron is from an inner shell (2p), which is much harder to remove. This is characteristic of a Group II element. The values match Mg. [2 marks]

(c) Mg²⁺: 1s² 2s² 2p⁶ [1 mark]

12. (a) Trigonal bipyramidal shape with three equatorial Cl atoms at 120° and two axial Cl atoms at 90° to the equatorial plane. [2 marks]

(b) The axial bonds are longer than the equatorial bonds. Axial positions experience greater repulsion from three equatorial bonding pairs at 90°, while equatorial positions experience repulsion from two axial bonding pairs at 90° and two equatorial bonding pairs at 120°. This results in axial bonds being longer and weaker. [2 marks]

(c) [PCl₄]⁺ is tetrahedral, bond angle 109.5°. [PCl₆]⁻ is octahedral, bond angle 90°. [3 marks]

13. (a) The boiling points increase from CH₄ to SnH₄. This is due to increasing number of electrons and thus increasing strength of instantaneous dipole-induced dipole (van der Waals') forces as molecular size/mass increases down the group. [2 marks]

(b) H₂O has a much higher boiling point due to hydrogen bonding between water molecules. Oxygen is highly electronegative and the O–H bond is polar, allowing strong intermolecular hydrogen bonds. The other Group VI hydrides (H₂S, H₂Se, H₂Te) do not form hydrogen bonds as the central atom is not electronegative enough. [2 marks]

14. (a) Three resonance forms, each with one N=O double bond and two N–O single bonds. The actual structure is a hybrid with delocalised electrons. [2 marks]

(b) The nitrogen atom is sp² hybridised. The three sp² hybrid orbitals form σ bonds with oxygen atoms in a trigonal planar arrangement (bond angle 120°). The unhybridised p orbital on N overlaps with p orbitals on O to form a delocalised π system. [2 marks]

(c) All N–O bonds have the same length due to resonance delocalisation, giving each bond a bond order of 1⅓. [1 mark]

15. (a) CO₂ is a simple molecular structure with discrete linear molecules held together by weak van der Waals' forces. SiO₂ is a giant covalent network structure with each Si atom tetrahedrally bonded to four O atoms, and each O bonded to two Si atoms, forming a continuous 3D lattice. [2 marks]

(b) CO₂ is a gas because only weak intermolecular forces need to be overcome. SiO₂ is a solid with a very high melting point because many strong covalent bonds must be broken throughout the giant lattice. [2 marks]

Section C: Data-Based and Application Questions (15 marks)

16. (a) H–X bond energies decrease down the group. This is because the halogen atom becomes larger, leading to poorer orbital overlap with the hydrogen 1s orbital, resulting in a longer and weaker bond. [2 marks]

(b) HI is a stronger acid because the H–I bond is much weaker (298 kJ mol⁻¹) than the H–F bond (568 kJ mol⁻¹). In aqueous solution, the H–I bond breaks more easily to release H⁺ ions. The high bond energy of HF makes it a weak acid despite the high polarity of the bond. [2 marks]

17. (a) N₃⁻ has 16 valence electrons. Resonance forms: ⁻N=N⁺=N⁻ ↔ N≡N⁺–N²⁻ ↔ ⁻N–N⁺≡N. The central N has a formal positive charge in the most stable forms. [2 marks]

(b) The central nitrogen atom is sp hybridised. [1 mark]

(c) Bond order = 2 (each N–N bond is equivalent due to resonance, with 4 bonding electrons over 2 bonds). [1 mark]

18. (a) 4 bonding pairs and 1 lone pair. [1 mark]

(b) See-saw shape with lone pair in equatorial position. Bond angles: ~90° between axial and equatorial bonds, ~120° between equatorial bonds (slightly less due to lone pair repulsion). [2 marks]

(c) SF₄ is polar because the see-saw shape is asymmetric and the bond dipoles do not cancel. XeF₄ is square planar with two lone pairs opposite each other; the bond dipoles cancel, making it non-polar. [2 marks]

19. (a) Across Period 3, nuclear charge increases and electrons are added to the same principal quantum shell. Shielding remains similar, so effective nuclear charge increases, causing a stronger attraction for the outer electrons and a general increase in first ionisation energy. [2 marks]

(b) Aluminium has a lower first ionisation energy than magnesium because Al's outer electron is in a 3p orbital, which is higher in energy and further from the nucleus than Mg's 3s electron. The 3p electron is also shielded by the 3s electrons, making it easier to remove. [2 marks]

20. (a) Molecular orbital diagram for CO: σ2s, σ2s, π2p (two degenerate), σ2p, π2p (empty), σ*2p (empty). The σ2p is lower in energy than π2p for CO due to mixing. [2 marks]

(b) Bond order = 3. CO is diamagnetic (all electrons paired). [1 mark]

(c) CO has a lone pair on the carbon atom (HOMO) which can be donated to an empty d orbital on Fe²⁺, forming a σ coordinate bond. Additionally, CO has empty π* antibonding orbitals which can accept electron density from filled d orbitals on the metal (π back-bonding), strengthening the bond. This synergistic effect makes the CO–metal bond very strong. [2 marks]

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