AQA A-Level Chemistry Paper 3, June 2025: Question 16

Period 3 Trends: First Ionisation Energy

📌 What this question tests

Periodic Trends: The general increase in first ionisation energy across Period 3 due to increasing nuclear charge and similar shielding.
Sub-shell Dips (Exceptions): Understanding why dips occur at Group 3 (Mg to Al) and Group 6 (P to S).
Spin-Pair Repulsion: Explaining the difference between half-filled and paired p-orbital electrons.
Exam Awareness: Navigating negative-stem questions ( not correct ).

Question 16 Breakdown

Multiple Choice Analysis (1 Mark)

✅ Correct Answer: C

Statement C is NOT correct: "The first ionisation energy of P is less than that of S".

In reality, phosphorus (P) has a higher first ionisation energy than sulfur (S). Sulfur has a pair of electrons in one of its 3p orbitals; mutual repulsion between these paired electrons makes it easier to remove one, dropping sulfur's ionisation energy below that of phosphorus.

Awarded 1 mark for selecting C [AO1]

💡 Key Knowledge: Why the Dip at Sulfur Occurs

Phosphorus (P, Z = 15): 1s² 2s² 2p⁶ 3s² 3p³
Each 3p orbital contains one unpaired electron (half-filled sub-shell: stable configuration).
Sulfur (S, Z = 16): 1s² 2s² 2p⁶ 3s² 3p⁴
One 3p orbital contains a pair of electrons with opposing spins.
Repulsion: The repulsion between the two paired electrons in the same 3p orbital means less energy is required to remove one electron from S than from P.

Option-by-Option Evaluation

📐 Option A: Mg > Na (True statement)

Configuration: Na is [Ne] 3s¹, Mg is [Ne] 3s².

Mg has an extra proton (12 vs 11) with no extra shielding shell. The atomic radius is smaller and the electrostatic pull on the 3s electrons is stronger, so first IE increases. (Statement is true, so it is not the answer.)

📐 Option B: Al < Mg (True statement)

Configuration: Mg is [Ne] 3s², Al is [Ne] 3s² 3p¹.

The outer electron in Al is in a 3p orbital, which is higher in energy than the 3s orbital and shielded by the 3s² sub-shell. Therefore, Al has a lower first IE than Mg. (Statement is true, so it is not the answer.)

📐 Option D: Ar is the highest (True statement)

Configuration: Ar is [Ne] 3s² 3p⁶.

Argon has the largest nuclear charge (18 protons) in Period 3 with electrons in the same principal energy level (similar shielding) and the smallest atomic radius. Hence, it has the highest first IE in the period. (Statement is true, so it is not the answer.)

Exam Technique & Common Pitfalls

🧠 Exam Technique: Negative Stem Questions

Always circle the word "not" immediately upon reading the question.
Label each option with T (True) or F (False) down the margin.
Your target answer for this question type is the single F statement.
Sketch a quick Period 3 ionisation energy trend profile (the "up-down-up-up-down-up-up" zigzag) to visually confirm the anomalies.

❌ Common Misconceptions

Assuming a smooth monotonic trend: Thinking first IE strictly increases across the whole period without checking the group 2→3 and group 5→6 dips.
Picking Option B by mistake: Students remember that IE generally increases, so they see "Al is less than Mg" and mistakenly think it must be false. In fact, Al is lower than Mg!
Conflating shielding with sub-shell pairing: Citing shielding as the reason for the dip at sulfur. S and P have the same inner core shielding; the dip at S is strictly caused by electron-pair repulsion in a 3p orbital.