Edexcel A-Level Chemistry AS Paper 1, June 2023: Question 1

6 marks · Medium difficulty · Calculations

Identify fundamental particle properties, orbital shapes, calculate relative isotopic mass from isotopic abundances, and determine nuclide symbols for silicon isotopes and related atoms.

Practise this question

Question

Exam question 1 with parts (a) to (d) about silicon atoms, subatomic particles, atomic structure, and isotopic mass calculations. Part (a) is a multiple-choice question with a table giving relative charges and masses for protons and neutrons. Part (b) asks about orbital shapes and subatomic particles in the 2s orbital. Part (c) gives a table of abundances for two silicon isotopes and asks to determine the relative isotopic mass of the third isotope. Part (d) asks for the nuclide symbol of an element with specific numbers of protons and neutrons relative to a silicon isotope.
Question text

1 Atoms of silicon contain protons, neutrons and electrons.

(a) Which are the correct data for a proton and a neutron?

(1)

Proton Neutron

Relative charge Relative mass Relative charge Relative mass

A +1 1 –1 0.0005

B +1 1 0 1

C –1 0.0005 0 1

D 0 1 +1 1

(b) (i) State the shape of the 2s orbital of a silicon atom.

(1)

(ii) State the difference between the two subatomic particles in the 2s orbital.

(1)

(c) Silicon has three stable isotopes.

A sample of silicon was found to have a relative atomic mass, Ar , of 28.11 due to

the presence of only the three stable isotopes.

The abundance for two of the isotopes is given in the table.

Isotope Abundance / %

28Si 92.2

30Si 3.1

Determine the relative isotopic mass of the third stable isotope.

You must show your working.

(2)

*P71926A0228*

(d) Give the symbol, including the mass number and atomic number, for an atom of

element X that has four fewer protons and three fewer neutrons than an atom

of 28Si.

(1)

(Total for Question 1 = 6 marks)

Mark scheme

Show the mark scheme Mark scheme for question 1, showing correct answers and guidance for each part (a) through (d), allocating a total of 6 marks.

Question

Answer Mark

Number

1(a) (1)

The only correct answer is B ( +1 1 0 1 )

A is not correct because the data for the neutron are the data for an electron

C is not correct because the data for the proton are the data for an electron

D is not correct because the data for the neutron and proton have been swapped

Question Answer Additional Guidance Mark

Number

1(b)(i) An answer that makes reference to the following (1)

point

• spherical Allow a diagram of a sphere

Ignore circular / description of a circle

Number

1(b)(ii) An answer that makes reference to the following (1)

point

• (a pair of electrons) with opposite spin Allow different direction of spin

Number

1(c) Example of calculation: (2)

• calculation of the relative abundance of the

third isotope (1) 100 ‒ 92.2 ‒ 3.1 = 4.7(%) (ans 1)

• calculation of relative isotopic mass of the

third isotope (1) (28.11 x 100) ‒ (28 x 92.2) ‒ (30 x 3.1)

(ans 1)

= 29.021 = 29

Allow for M2 the assumption that the relative isotopic mass is 29

with correct calculation showing that 29 gives the correct relative

atomic mass.

Allow 3, 4, 5 SF

Allow TE on incorrect (ans 1)

Correct answer with no working scores 0

Number

1(d) An answer that makes reference to the following (1)

point:

• 21Ne Allow Ne

(Total for Question 1 = 6 marks)

How to answer it

Study Guide: Atomic Structure and Isotopes

What this question tests

This question assesses foundational AS Level knowledge of atomic structure, subatomic particle properties (relative charge and mass), electron orbital shapes, electron pairing within subshells, relative atomic mass (Ar) calculations involving isotopic abundances, and the interpretation of nuclide notation (atomic and mass numbers).

Part (a): Subatomic Particle Properties

Identifying standard relative charges and masses

✅ Correct Answer

Option B is correct.

Proton: Relative charge = +1, Relative mass = 1

Neutron: Relative charge = 0, Relative mass = 1

💡 Key Knowledge

  • Protons have a positive charge (+1) and a mass of roughly 1 atomic mass unit.
  • Neutrons are neutral, having a charge of 0, but share approximately the same mass (1) as a proton.
  • Electrons have a negligible relative mass of 0.0005 (or 1/1836).

❌ Common Errors

Students often confuse the properties of protons, neutrons, and electrons. For instance, choosing option A assigns electron properties to the neutron, while option D swaps the charges of the proton and neutron.

Mark: 1 mark

Part (b): Orbitals and Subatomic Particles

Understanding shapes and orbital occupancy

✅ Correct Answers

(i) spherical (A diagram of a sphere is also fully accepted; ignore references to "circular").

(ii) (a pair of electrons) with opposite spin .

🧠 Exam Technique

Be precise with chemical terminology. For part (i), always use spherical rather than "circular" (orbitals are 3D regions of space, not 2D shapes). For part (ii), remember that an orbital holds a maximum of two electrons, and Pauli's exclusion principle dictates they must have opposite spins.

Marks: 1 mark for (b)(i), 1 mark for (b)(ii)

Part (c): Relative Isotopic Mass Calculation

Determining missing isotopic abundance and mass

📐 Step-by-Step Calculation

  1. Find the abundance of the third isotope:
    Total percentage abundance must equal 100%.
    100 - 92.2 - 3.1 = 4.7%
  2. Set up the Ar equation:
    Ar = [(Mass₁ × Abund₁) + (Mass₂ × Abund₂) + (Mass₃ × Abund₃)] / 100
    28.11 = [(28 × 92.2) + (30 × 3.1) + (x × 4.7)] / 100
  3. Rearrange and solve for x:
    2811 = 2581.6 + 93 + 4.7x
    2811 = 2674.6 + 4.7x
    4.7x = 136.4
    x = 29.021...
  4. State final answer:
    Relative isotopic mass = 29 (as relative isotopic masses are integers/whole numbers in this context).

❌ Common Errors & Examiner Warning

Crucial rule from mark scheme: A correct final answer with no working scores 0 marks! You must show your subtraction step to find the 4.7% abundance and your formula substitution.

Marks: 2 marks total (1 for abundance calculation, 1 for isotopic mass calculation)

Part (d): Nuclide Notation

Deducing symbol, atomic number, and mass number

✅ Correct Answer

²¹₁₀Ne (or written vertically with 21 on top, 10 on bottom, next to Ne).

💡 Key Knowledge & Working

  • Starting atom: Silicon-28, written as 2814Si. This means it has 14 protons (atomic number) and 14 neutrons (28 - 14 = 14).
  • Four fewer protons: 14 - 4 = 10 protons. Element with atomic number 10 is Neon (Ne).
  • Three fewer neutrons: 14 - 3 = 11 neutrons.
  • New mass number = protons + neutrons = 10 + 11 = 21 .
Mark: 1 mark

Topics

Physical Chemistry · Topic 1: Atomic Structure and the Periodic Table

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.