AQA A-Level Chemistry Paper 1, 2019: Question 8

11 marks · Medium difficulty · State/Explain/Describe

Draw hydrogen bonding between ethanol molecules, explain the difference in boiling points between ethanol and methoxymethane, and determine the shapes and bond angles of POCl3 and ClF4-.

Practise this question

Question

Question 08 contains three parts about structure and bonding. Part 08.1 asks to draw a diagram showing the strongest intermolecular interaction between two molecules of ethanol in the liquid phase, including all lone pairs and partial charges, worth 3 marks. Part 08.2 presents Table 5 with boiling points: ethanol at 78 °C and methoxymethane at -24 °C, asking to explain the difference in boiling points in terms of intermolecular forces, worth 3 marks. Part 08.3 asks to draw the shapes of the POCl3 molecule and the ClF4- ion including lone pairs that influence shape, name each shape, and suggest the bond angle in ClF4-, worth 5 marks.
Question text

08 This question is about structure and bonding.

08.1 Draw a diagram to show the strongest type of interaction between two molecules of

ethanol (C2H5OH) in the liquid phase.

Include all lone pairs and partial charges in your diagram.

[3 marks]

08.2 Methoxymethane (CH3OCH3) is an isomer of ethanol.

Table 5 shows the boiling points of ethanol and methoxymethane.

Table 5

Compound Boiling point / °C

ethanol 78

methoxymethane –24

In terms of the intermolecular forces involved, explain the difference in boiling points.

[3 marks]

Extra space

08.3 Draw the shape of the POCl molecule and the shape of the ClF – ion.

Include any lone pairs of electrons that influence the shapes.

In a POCl3 molecule the oxygen atom is attached to the phosphorus atom by a

double bond that uses two electrons from phosphorus.

Name each shape.

Suggest a value for the bond angle in ClF4 ̄

Shape of POCl3 Shape of ClF4 ̄

[5 marks]

Name of shape of POCl3

Name of shape of ClF4 ̄

Bond angle in ClF4 ̄

Mark scheme

Show the mark scheme Mark scheme for question 08. For 08.1 (3 marks): M1 for two lone pairs on each O atom and delta-plus on H, delta-minus on O of each O-H bond; M2 for a dotted/broken line between a lone pair on one O and the H of another; M3 for O···H-O shown in a straight line. For 08.2 (3 marks): 1 mark for hydrogen bonds between ethanol molecules, 1 mark for permanent dipole-dipole or van der Waals forces between methoxymethane molecules, and 1 mark for stating hydrogen bonds are stronger. For 08.3 (5 marks): 1 mark for tetrahedral-like diagram of POCl3 with one P=O and three P-Cl bonds; 1 mark for square planar diagram of ClF4- with 4 Cl-F bonds and 2 lone pairs; 1 mark for name (distorted) tetrahedral; 1 mark for name square planar; 1 mark for bond angle of 90 degrees.

Question Answers Additional Comments/Guidelines Mark

M1 two lone pairs on each O atom 1

and

δ+ and δ− on each H-O bond

O

ᵟ –

C H M2 dotted/broken line shown between lone pair on 1

25 Hᵟ+

one molecule and the correct H on another

……. 1

M3 O H–O in straight line, dependent on M2

ᵟ –

08.1

O Ignore any partial charges on C–H or C–O bonds

ᵟ+

C2H5

H For straight line in M3, allow a deviation of up to

15°

If a different molecule containing hydrogen bonding

due to O–H bond drawn (e.g. methanol, water) or

an incorrect attempt at the structure of ethanol,

then maximum of 2 marks (i.e. only penalise if

would score all three marks otherwise)

Hydrogen bonds (between ethanol molecules) 1

(permanent) dipole-dipole OR van der Waals force (between Allow vdW 1

methoxymethane molecules)

08.2

Hydrogen bonds are stronger/est intermolecular force Allow more energy to break/overcome hydrogen 1

bonding

Allow converse arguments

O 23

POCl3: allow any shape showing 1 double bond

F between P and O and 3 P-Cl bonds 1

F

P Cl ClF -: allow any shape showing 4 Cl-F bonds and 2

Cl Cl F lone pairs 1

F

Cl

08.3

(distorted) Tetrahedral 1

Square planar

o 1

How to answer it

Intermolecular Bonding, VSEPR & Molecular Geometry

What this question tests

Fundamental principles of chemical structure and bonding:

  • Hydrogen bonding diagrams: Precise representation of dipoles (δ+ / δ-), lone pairs, dotted interaction lines, and linear geometry around the bridge.
  • Comparing intermolecular forces: Linking differences in boiling point between functional group isomers to the relative strengths of hydrogen bonding vs permanent dipole-dipole / van der Waals forces.
  • VSEPR theory & complex shapes: Deducing electron pair geometry, molecular shapes, lone pair positioning, and bond angles for species with expanded octets (e.g., POCl₃ and ClF₄⁻).

Question 08.1

Diagram of Hydrogen Bonding in Ethanol [3 Marks]

✅ What the Diagram Must Show

A diagram showing two separate ethanol molecules (C₂H₅OH) linked together:

1. Partial charges & lone pairs:
• Show two distinct lone pairs on each oxygen atom (drawn as lobes or pairs of dots/crosses).
• Show δ- on both oxygen atoms and δ+ on both hydroxyl hydrogen atoms (H attached directly to O).

2. Hydrogen bond:
• A dotted or dashed line going cleanly from a lone pair on one oxygen directly to the Hδ+ of the other ethanol molecule.

3. Alignment (Linearity):
• The O—H····O sequence must be completely in a straight line (examiner allows up to a 15° deviation).

🧠 Exam Technique & Mark Breakdown

  • M1 (1 mark): Both lone pairs shown on each O atom AND both δ+ / δ- dipoles labelled correctly on both O—H bonds.
  • M2 (1 mark): A clearly dotted or broken line between a lone pair on one O and the correct H of the second molecule. (A solid line scores 0).
  • M3 (1 mark): The three atoms involved in the bridge ( O—H····O ) form a straight 180° line. Dependent on M2.

💡 Key Knowledge

  • Hydrogen bonding occurs when hydrogen is bonded to a highly electronegative atom (N, O, or F) with at least one available lone pair.
  • The bond is linear (180°) around the hydrogen because the repulsion between the bonding pair and the lone pair of the accepting atom is minimised in this alignment.

❌ Common Errors to Avoid

  • Forgetting one molecule's lone pairs: Both oxygen atoms must show their two lone pairs (4 lone pairs in total across the diagram).
  • Bending the H-bond: Drawing the O—H····O bond bent at an angle loses mark 3 immediately.
  • Solid interaction line: Drawing a continuous line instead of dashed/dotted makes it look like a covalent bond.
  • Wrong charges: Labelling the ethyl group ( C₂H₅ ) with partial charges instead of the O—H group.
Examiner Note: If an incorrect molecule containing an O—H bond was drawn (e.g. water or methanol), a maximum of 2 marks could be awarded (M1 and M3 only).

Question 08.2

Explaining Differences in Boiling Point [3 Marks]

✅ Model Answer

Ethanol has a much higher boiling point (78 °C) than methoxymethane (-24 °C) because:

  • Ethanol has hydrogen bonds between molecules.
  • Methoxymethane has permanent dipole-dipole forces (or van der Waals forces) between molecules.
  • Hydrogen bonds are stronger than dipole-dipole / van der Waals forces, requiring more energy to overcome.

🧠 Exam Technique: Name, Name, Compare

  • Step 1: Identify and name the primary intermolecular force in Compound A.
  • Step 2: Identify and name the primary intermolecular force in Compound B.
  • Step 3: Explicitly state which force is stronger and link to the energy needed to separate molecules.

💡 Key Knowledge

  • Both ethanol and methoxymethane have the same molecular formula ( C₂H₆O ) and identical Mᵣ (46.0), meaning their London dispersion (van der Waals) forces are very similar in scale.
  • Methoxymethane is an ether ( CH₃–O–CH₃ ). Although it has an oxygen atom with lone pairs, it lacks an H bonded to O, so it cannot form hydrogen bonds with itself.

❌ Common Errors

  • "Breaking covalent bonds": Never state that covalent bonds are broken upon boiling. Only intermolecular forces are overcome!
  • Vague comparisons: Stating simply "ethanol has stronger bonds" without specifying that it is hydrogen bonding that is stronger than dipole-dipole/van der Waals forces.

Question 08.3

Shapes of POCl₃ and ClF₄⁻ [5 Marks]

✅ Model Answers & Diagram Details

1. Shape of POCl₃ (Tetrahedral):
• Central atom: Phosphorus (P).
• Outer atoms: Double bond to oxygen ( P=O ) pointing upwards, and 3 single bonds to chlorine ( P—Cl ) spreading down in a tripod arrangement.
• Name: (Distorted) Tetrahedral [1 mark]
2. Shape of ClF₄⁻ (Square Planar):
• Central atom: Chlorine (Cl).
• 4 fluorine atoms ( Cl—F ) arranged symmetrically in a square horizontal plane.
• 2 lone pairs: One vertically above and one vertically below the central Cl atom (180° apart to minimise repulsion).
• Name: Square planar [1 mark]
• Bond angle: 90° [1 mark]

📐 Step-by-Step VSEPR Deduction

For POCl₃:
  1. P is in Group 5 (5 outer electrons).
  2. O forms a double bond using 2 electrons from P (stated in question).
  3. 3 Cl atoms share 1 electron each (3 electrons used).
  4. Total electrons used by P = 2 + 3 = 5. Lone pairs = 0.
  5. 4 bonding regions = Tetrahedral.
For ClF₄⁻:
  1. Central Cl has 7 electrons + 1 (from negative charge) = 8 electrons.
  2. 4 F atoms form 4 single bonds = 4 bonding pairs.
  3. Remaining electrons: 8 - 4 = 4 electrons = 2 lone pairs.
  4. Total regions = 6 (Octahedral arrangement).
  5. To minimise lp-lp repulsion, lone pairs sit opposite each other (axial positions), giving a Square Planar molecular geometry with 90° bond angles.

❌ Common Errors

  • Missing the lone pairs on Cl: Drawing ClF₄⁻ as a simple cross without showing both lone pairs loses the diagram mark!
  • Incorrect bond angle: Suggesting angles like 109.5° or 120° for ClF₄⁻. Because the four F atoms lie in a flat plane, all adjacent F—Cl—F angles are exactly 90°.
  • Confusing electron pair geometry with shape name: Writing "octahedral" for ClF₄⁻ instead of the shape defined solely by the atoms: square planar.

🧠 Summary of Awarded Marks

  • Mark 1: Correct 3D representation/diagram of POCl₃ showing 1 P=O and 3 P—Cl bonds.
  • Mark 2: Correct diagram of ClF₄⁻ showing 4 Cl—F bonds and 2 opposing lone pairs.
  • Mark 3: POCl₃ named as tetrahedral (allow 'distorted tetrahedral').
  • Mark 4: ClF₄⁻ named as square planar.
  • Mark 5: ClF₄⁻ bond angle given as 90°.

Topics

Physical Chemistry · 3.1.3 Bonding

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.