A Level Chemistry H3 Atomic Structure Bonding Quiz

A-Level Chemistry H3 Quiz - Atomic Structure Bonding: Answer Key

Total Marks: 50

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Section A: Multiple Choice Questions (Questions 1–5, 10 marks)

1. B) The HOMO is a π orbital and the LUMO is a σ orbital.** [2 marks]

Explanation: The molecular orbital diagram for O₂ has the order: σ(2s), σ*(2s), σ(2pz), π(2px)=π(2py), π*(2px)=π*(2py), σ*(2pz). O₂ has 12 valence electrons. The first 10 electrons fill up to π(2px)=π(2py). The remaining 2 electrons go into the π*(2px)=π*(2py) orbitals (Hund's rule), making the HOMO a π* orbital. The LUMO is the next empty orbital, which is σ*(2pz).

Common mistake: Students often confuse the MO order for O₂ with that of N₂. For O₂ and F₂, the π orbitals are lower in energy than the σ(2pz) orbital.

2. A) Halved [2 marks]

Explanation: Energy of a photon is given by E = hf = hc/λ. If λ is doubled, E is halved because E ∝ 1/λ.

3. D) CH₄ [2 marks]

Explanation: CO₂ is linear and has 4 vibrational modes (3N-5 = 4), but only 2 are IR-active (the asymmetric stretch and the bending modes). SO₂ is bent and has 3 vibrational modes (3N-6 = 3), all IR-active. H₂O is bent and has 3 vibrational modes, all IR-active. CH₄ is tetrahedral and has 9 vibrational modes (3N-6 = 9), of which 4 are IR-active. The question asks for the highest number of IR-active stretching vibrations. CH₄ has 4 IR-active modes, which is the highest among the options.

4. B) The electronegativity of nearby atoms [2 marks]

Explanation: Chemical shift is primarily affected by the electron density around the proton. Electronegative atoms withdraw electron density, deshielding the proton and shifting it downfield (higher δ). While the number of neighbouring protons affects splitting, it does not directly affect chemical shift.

5. C) The ion formed by loss of one electron from the molecule [2 marks]

Explanation: The molecular ion (M⁺•) is formed when the molecule loses one electron during ionisation. It has the same mass as the molecule and appears at the highest m/z value (excluding isotopic peaks).

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Section B: Short Answer Questions (Questions 6–10, 15 marks)

6. (a) MO diagram for H₂: [3 marks]

The diagram should show:

• Two 1s atomic orbitals (left and right) at the same energy level
• A σ bonding MO (lower energy) and a σ* antibonding MO (higher energy) in the centre
• Two electrons (arrows) in the σ bonding MO
• HOMO = σ bonding MO (filled)
• LUMO = σ* antibonding MO (empty)

Marking scheme:

• Correct atomic orbitals and MOs: 1 mark
• Correct electron filling: 1 mark
• Correct HOMO/LUMO labels: 1 mark

(b) Bond order of H₂ = (2 - 0)/2 = 1. [2 marks]

Explanation: Bond order = (number of bonding electrons - number of antibonding electrons)/2. For H₂, there are 2 bonding electrons and 0 antibonding electrons. A bond order of 1 indicates a stable single bond.

7. (a) A chromophore is a part of a molecule responsible for its colour, typically a functional group containing double or triple bonds, delocalised systems, or lone pairs that can absorb UV/visible light. [1 mark]

(b) Increasing conjugation extends the delocalised π-electron system, which raises the energy of the HOMO and lowers the energy of the LUMO, decreasing the HOMO-LUMO energy gap. A smaller energy gap means lower energy photons are absorbed, corresponding to longer wavelengths (λmax shifts to longer wavelengths). [2 marks]

Explanation: Conjugation creates a series of molecular orbitals with smaller energy gaps between them. The lowest energy transition (HOMO→LUMO) requires less energy, so it occurs at longer wavelengths (bathochromic shift).

8. Using A = εcl: [3 marks]

c = A/(εl) = 0.450 / (1.50 × 10⁴ dm³ mol⁻¹ cm⁻¹ × 1.00 cm)

c = 0.450 / 1.50 × 10⁴

c = 3.00 × 10⁻⁵ mol dm⁻³

Marking scheme:

• Correct rearrangement of Beer-Lambert Law: 1 mark
• Correct substitution: 1 mark
• Correct final answer with units: 1 mark

Common mistake: Forgetting to include the path length (l) in the calculation.

9. (a) CO₂ is a linear molecule (O=C=O). It has 4 vibrational modes (3N-5 = 4): symmetric stretch, asymmetric stretch, and two bending modes (degenerate). The symmetric stretch does not change the dipole moment, so it is IR-inactive. Therefore, CO₂ has 2 IR-active stretching vibrations (asymmetric stretch and bending). [2 marks]

Marking scheme:

• Correct number of IR-active vibrations: 1 mark
• Correct reasoning: 1 mark

(b) Polyatomic gases like CO₂ and H₂O absorb IR radiation because their vibrational modes involve changes in dipole moment. The absorbed IR radiation is re-emitted in all directions, including back towards the Earth's surface. This traps heat in the atmosphere, contributing to the greenhouse effect. [2 marks]

Explanation: The key concept is that IR-active vibrations absorb specific wavelengths of IR radiation emitted by the Earth. The re-emission of this energy warms the lower atmosphere.

10. (a) The compound is ethyl acetate (CH₃COOCH₂CH₃). [2 marks]

Explanation:

• δ 1.2 (singlet, 3H): CH₃ group adjacent to carbonyl (CH₃CO-)
• δ 2.3 (quartet, 2H): CH₂ group adjacent to CH₃ (CH₃CH₂-)
• δ 4.1 (triplet, 3H): This is inconsistent with the formula. The triplet at δ 4.1 should be 2H (CH₂ adjacent to CH₃ and O). The integration suggests 3H, but this is likely a typo in the question. The correct structure is CH₃COOCH₂CH₃.

Marking scheme:

• Correct identification of ethyl group: 1 mark
• Correct identification of acetate group: 1 mark

(b) The signal at δ 2.3 is a quartet because the CH₂ group is adjacent to a CH₃ group (3 equivalent protons). According to the n+1 rule, n = 3, so the signal is split into 4 peaks (quartet). [1 mark]

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Section C: Structured Questions (Questions 11–15, 15 marks)

11. (a) MO diagram for F₂: [3 marks]

The diagram should show:

• Two 2s and two 2p atomic orbitals (left and right)
• MOs in order: σ(2s), σ*(2s), σ(2pz), π(2px)=π(2py), π*(2px)=π*(2py), σ*(2pz)
• 14 valence electrons filled in order: σ(2s)², σ*(2s)², σ(2pz)², π(2px)²=π(2py)², π*(2px)²=π*(2py)²
• HOMO = π*(2px)=π*(2py) (filled)
• LUMO = σ*(2pz) (empty)

Marking scheme:

• Correct atomic orbitals and MOs: 1 mark
• Correct energy ordering: 1 mark
• Correct electron filling and HOMO/LUMO: 1 mark

(b) HOMO of F₂: π*(2px)=π*(2py) (degenerate, filled). LUMO of F₂: σ*(2pz) (empty). [1 mark]

Bond order of F₂ = (8 - 6)/2 = 1. [1 mark]

Explanation: Bonding electrons: σ(2s)², σ(2pz)², π(2px)²=π(2py)² = 8 electrons. Antibonding electrons: σ*(2s)², π*(2px)²=π*(2py)² = 6 electrons. Bond order = (8-6)/2 = 1.

12. (a) Allowed transitions are those that obey the selection rules (e.g., Δl = ±1, no change in spin multiplicity). Forbidden transitions violate these rules and have much lower probability (lower intensity). [2 marks]

(b) An example of a forbidden transition is n→π* in a carbonyl group. This transition is forbidden because it involves a change in orbital symmetry (n is non-bonding, π* is antibonding) and does not involve a change in angular momentum quantum number. [1 mark]

Explanation: The n→π* transition in carbonyls is actually observed but with low intensity because it is "forbidden" by symmetry but becomes weakly allowed through vibronic coupling.

13. (a) Chlorine has two naturally occurring isotopes: ³⁵Cl (75.8%) and ³⁷Cl (24.2%). The peak at m/z 80 corresponds to the molecular ion containing ³⁷Cl instead of ³⁵Cl. The intensity ratio of approximately 3:1 (m/z 78 : m/z 80) reflects the natural abundance ratio of ³⁵Cl:³⁷Cl. [2 marks]

(b) A possible molecular formula is CH₃Cl (methyl chloride). [1 mark]

Explanation: M(CH₃³⁵Cl) = 12 + 3 + 35 = 50. M(CH₃³⁷Cl) = 12 + 3 + 37 = 52. This does not match m/z 78 and 80. A better suggestion is C₂H₅Cl (ethyl chloride): M(C₂H₅³⁵Cl) = 24 + 5 + 35 = 64. M(C₂H₅³⁷Cl) = 24 + 5 + 37 = 66. This also does not match. The correct formula should be C₂H₃Cl₃ (trichloroethene): M(C₂H₃³⁵Cl₃) = 24 + 3 + 105 = 132. This is too high. The question has a typo. A correct example would be CH₂Cl₂ (dichloromethane): M(CH₂³⁵Cl₂) = 12 + 2 + 70 = 84. M(CH₂³⁵Cl³⁷Cl) = 12 + 2 + 72 = 86. This is close but not exact. The intended answer is likely CH₃Cl, but the m/z values are incorrect. For the purpose of this answer key, accept any reasonable formula that shows the chlorine isotope pattern.

14. (a) The most stable conformation of butane is the anti conformation. In a Newman projection, this is shown with the two methyl groups 180° apart. [2 marks]

Image pending generation: diagram for Q14.

(b) The anti conformation is most stable because the two bulky methyl groups are as far apart as possible (180° dihedral angle), minimising steric strain (torsional strain). [1 mark]

Explanation: The anti conformation has the lowest energy because the methyl groups are staggered and far apart. The gauche conformation (60° dihedral angle) has higher energy due to steric repulsion between the methyl groups.

15. (a) The Hammond Postulate states that the structure of a transition state resembles the structure of the nearest stable species (reactant or product) in terms of energy. [1 mark]

(b) In the SN1 reaction of tert-butyl bromide with water, the rate-determining step is the formation of the tert-butyl carbocation. This step is highly endothermic (the carbocation is much higher in energy than the reactant). According to the Hammond Postulate, the transition state for this step will resemble the carbocation (the high-energy intermediate), meaning it is a late transition state. [2 marks]

Explanation: The Hammond Postulate is used to predict transition state structure. For endothermic steps, the transition state is product-like (late). For exothermic steps, the transition state is reactant-like (early).

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Section D: Extended Response Questions (Questions 16–20, 10 marks)

16. (a) NMR spectroscopy is based on the principle that certain atomic nuclei (e.g., ¹H, ¹³C) have a property called nuclear spin. When placed in an external magnetic field, these nuclei can align either with (lower energy) or against (higher energy) the field. Absorption of radiofrequency radiation of the correct energy causes the nucleus to "flip" from the lower energy state to the higher energy state (resonance). [2 marks]

(b) TMS is used as a reference because: [2 marks]

• It has 12 equivalent protons, giving a single, strong signal.
• The protons are highly shielded (silicon is less electronegative than carbon), so they resonate at a very low chemical shift (δ = 0).
• It is chemically inert and non-toxic.
• It is volatile, making it easy to remove from the sample.

Marking scheme:

• Any two valid reasons: 2 marks (1 mark each)

17. (a) Optical purity = ([α]obs / [α]pure) × 100% = (+10.5° / +13.8°) × 100% = 76.1% [2 marks]

(b) The sample contains 76.1% (R)-2-butanol and 23.9% (S)-2-butanol. [1 mark]

Explanation: Optical purity (also called enantiomeric excess) is the excess of one enantiomer over the racemic mixture. If the optical purity is 76.1%, then 76.1% of the sample is the pure (R)-enantiomer and the remaining 23.9% is a racemic mixture (equal parts R and S). So the percentage of (R)-enantiomer = 76.1% + (23.9%/2) = 88.05%. The percentage of (S)-enantiomer = 23.9%/2 = 11.95%. However, the simpler interpretation is that the optical purity directly gives the enantiomeric excess: 76.1% ee, meaning 76.1% excess of (R)-enantiomer over the racemic mixture. This means the sample is 88.05% (R) and 11.95% (S).

Common mistake: Confusing optical purity with percentage composition. Optical purity is the enantiomeric excess, not the percentage of one enantiomer.

18. (a) Enantiomers are non-superimposable mirror images of each other. They have identical physical properties (except for optical rotation) and identical chemical properties in achiral environments. Diastereoisomers are stereoisomers that are not mirror images of each other. They have different physical and chemical properties. [2 marks]

(b) An example of a pair of diastereoisomers is (R,R)-tartaric acid and (R,S)-tartaric acid (meso-tartaric acid). [1 mark]

Explanation: The key difference is that enantiomers are mirror images, while diastereoisomers are not. Diastereoisomers arise when a molecule has two or more chiral centres.

19. (a) SN1 reactions follow first-order kinetics: rate = k[alkyl halide]. The rate depends only on the concentration of the alkyl halide. SN2 reactions follow second-order kinetics: rate = k[alkyl halide][nucleophile]. The rate depends on the concentrations of both the alkyl halide and the nucleophile. [2 marks]

(b) The rate of SN2 reactions is strongly affected by steric hindrance around the electrophilic carbon. Primary alkyl halides react fastest because the carbon is least hindered. Tertiary alkyl halides react very slowly or not at all via SN2 because the bulky alkyl groups block the nucleophile's approach. [1 mark]

Explanation: SN2 reactions involve a backside attack, so the nucleophile needs access to the carbon atom. Steric hindrance from alkyl groups slows the reaction.

20. (a) Zaitsev's rule states that in elimination reactions, the more substituted alkene (the alkene with more alkyl groups attached to the double bond) is the major product. [1 mark]

(b) In the E2 reaction of 2-bromobutane with a strong base, the reaction proceeds through a concerted anti-periplanar elimination. The transition state leading to the more substituted alkene (2-butene) is more stable because the developing double bond is stabilised by hyperconjugation from the alkyl groups. According to the Hammond Postulate, the transition state resembles the alkene product, so the more stable alkene (more substituted) is formed faster. [2 marks]

Explanation: Zaitsev's rule is a consequence of the greater stability of more substituted alkenes. The transition state leading to the more substituted alkene is lower in energy, so the reaction is faster.

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End of Answer Key