AQA GCSE Combined Science: Trilogy Chemistry Paper 1 (Higher), 2023: Question 4

electron → proton → neutron

Marks: 1 (tick the correct option)

• Electron discovered first (earliest evidence from cathode rays).
• Proton identified next.
• Neutron discovered last (harder to detect because it has no charge).

• Putting neutron before proton (students often assume “neutral” was found early).
• Mixing up “discovery order” with “position in the atom”.

Method (ratio): times heavier = (mass of proton) ÷ (mass of electron)

1. Write the division: (1.673 × 10⁻²⁷) ÷ (9.109 × 10⁻³¹)
2. Calculate with a calculator: ≈ 1836.645…
3. Round appropriately (≥ 3 s.f. is acceptable here): ≈ 1837

Final: proton is 1837 times heavier than an electron.

1.673 × 10⁻²⁷ ÷ 9.109 × 10⁻³¹ = 1837

Marks (2):
• 1 mark for setting up the correct division
• 1 mark for a correct final value (accept 1836.645… rounded to at least 3 s.f.)

• “How many times heavier” means divide, not subtract.
• Don’t worry about the kg unit here: the answer is a ratio (unitless).
• Check your number is sensible: proton should be around 2000× heavier, not smaller.

• Subtracting masses instead of dividing.
• Dividing the wrong way round (would give a tiny decimal).
• Rounding too early or giving an over-rounded answer (e.g. “1800” with no working).
• Calculator entry mistakes with powers of ten (10⁻²⁷ and 10⁻³¹).

Number of neutrons = 46

Mark: 1

• Mass number = protons + neutrons
• Atomic number = number of protons
• So neutrons = mass number − atomic number

Here: 81 − 35 = 46

• Using 35 as neutrons (it is protons).
• Doing 35 − 81 (wrong order gives a negative number).

Number of electrons in Br⁻ = 36

Mark: 1

• A neutral atom has electrons = protons.
• Br has 35 protons → neutral Br has 35 electrons.
• Br⁻ has gained 1 extra electron → 35 + 1 = 36.

• Answering 35 (forgetting the − charge means an extra electron).
• Subtracting 1 because the ion is negative (it’s the opposite: negative means more electrons).

Weighted mean: Aᵣ = (mass × abundance + mass × abundance) ÷ 100

1. Multiply each mass number by its % abundance:
   • 35 × 75.77 = 2651.95
   • 37 × 24.23 = 896.51
2. Add: 2651.95 + 896.51 = 3548.46
3. Divide by 100: 3548.46 ÷ 100 = 35.4846
4. Round to 2 d.p.: 35.48

Aᵣ = ((35 × 75.77) + (37 × 24.23)) ÷ 100

= 35.4846 → 35.48 (2 d.p.)

Marks (3):
• 1 mark: correct set-up of weighted mean
• 1 mark: correct unrounded calculation (e.g. 35.4846)
• 1 mark: correctly rounded to 2 d.p. (35.48)

• Always divide by 100 when abundances are percentages.
• Round at the end to avoid losing accuracy.
• Write the final answer clearly as a single number (Aᵣ has no units).

Examiner insight (from the mark scheme): if you use the values given and show a sensible method, you can still gain marks even if you make a small slip, as long as your final is correctly rounded from your working.

• Doing (35 + 37) ÷ 2 (this ignores abundance).
• Forgetting to divide by 100.
• Rounding each product too early, then getting the final rounding wrong.
• Rounding to 2 s.f. instead of 2 d.p. (they are different).

• Chlorine is in Group 7 → it has 7 outer (valence) electrons.
• In Cl₂, each chlorine shares 1 electron to make a single covalent bond.
• After bonding, each Cl has a full outer shell (8 electrons):
  1 shared pair + 3 lone pairs (6 non-bonded electrons).

• In the overlap: draw one shared pair of electrons (one dot from the left Cl and one cross from the right Cl).
• Outside the overlap: on each chlorine, draw 6 non-bonded electrons (three lone pairs) on its outer shell.

Marks (2):
• 1 mark: one shared pair in the overlap
• 1 mark: each chlorine shown with 6 non-bonded outer electrons

• Use dots for one atom and crosses for the other so the examiner can see one from each in the bond.
• Only show outer shell electrons (the mark scheme says inner shells can be ignored).
• Count carefully: each Cl should “own” 7 electrons (6 as lone electrons + 1 used in the shared pair), and “see” 8 total in its shell after sharing.

Allowed: the mark scheme accepts any sensible electron symbols (dots/crosses/circles/e⁻), as long as the bonding is correct.

• Drawing two shared pairs (that would be a double bond; Cl₂ has a single bond).
• Forgetting lone pairs (leaving each Cl with too few electrons).
• Putting both shared electrons as dots (or both crosses) so it doesn’t show one from each atom.
• Trying to add inner shells—wastes time and can lead to miscounting (and isn’t needed for marks).