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

✅ Correct Answer

1s²2s²2p⁶3s²3p⁶

💡 Key Knowledge

• Across Period 3, first ionisation energy has a general upward trend due to increasing nuclear charge (more protons) with similar shielding.
• Argon sits at the end of the period (noble gas); it has the highest nuclear charge in Period 3 while valence electrons remain in the 3p sub-shell.

🧠 Exam Technique

• Read the prompt strictly: The question asks for the full electron configuration, not the element name or symbol. Writing just "Argon" or "Ar" scores 0.
• Do not write shorthand such as [Ne] 3s²3p⁶ when "full electron configuration" is demanded.

❌ Common Errors

• Writing [Ne]3s²3p⁶ — noble gas abbreviation is rejected when "full" configuration is requested.
• Writing Chlorine ( ...3p⁵ ) by forgetting noble gases have the maximum ionisation energy in their period.
• Capitalising orbital letters (e.g. 1S²2S²2P⁶... ).

✅ Correct Answer

Na⁺(g) → Na²⁺(g) + e⁻

💡 Key Knowledge

• Definition of n-th Ionisation Energy: The energy required to remove one mole of electrons from one mole of gaseous (n-1)⁺ ions to form one mole of gaseous n⁺ ions.
• For the 2nd IE: Start from Na⁺(g) and form Na²⁺(g) .
• State symbols on electrons (e.g. e⁻(g) ) are ignored even if incorrect, but state symbols on the ions are mandatory.

🧠 Exam Technique

Always double-check two things before moving on from an ionisation energy equation:

1. Did you start from a single positive ion for the 2nd IE? (Not neutral Na).
2. Are both sodium species marked with (g)?

❌ Common Errors

• Writing the two-step overall loss: Na(g) → Na²⁺(g) + 2e⁻ (this is the sum of 1st + 2nd IE, not the 2nd IE itself).
• Omitting state symbols or writing Na⁺(s) .
• Starting with neutral sodium: Na(g) → Na⁺(g) + e⁻ (this is 1st IE).

📐 Step-by-Step Data Analysis

• IE 1 → 6: Gradual increase from 1000 to 8490 kJ mol⁻¹ (electrons removed from same outer shell).
• IE 6 → 7: Massive jump from 8490 to 27100 kJ mol⁻¹ (an increase of +18610 kJ mol⁻¹!).
• Deduction: 6 electrons are easily removed from the valence shell. The 7th electron comes from an inner principal energy level.
• 6 valence electrons in Period 3 = Group 16 (Group 6) → Sulfur (S).

✅ Mark Scheme Breakdown

Element: Sulfur / S [M1]

Explanation:

• [M2] Large jump after the sixth electron is removed / large difference between 6th and 7th ionisation energies.
• [M3] The 7th electron is removed from the inner shell / 2nd shell / 2p orbital AND is closer to the nucleus / lower in energy / experiences less shielding.

🧠 Exam Technique: Securing M3

The mark scheme notes: "Both ideas needed for mark" for M3. To guarantee the mark, your answer must explicitly state:

1. Where the electron is removed from: specify inner shell / 2nd shell / 2p sub-shell.
2. Why it takes more energy: state that it is closer to the nucleus OR experiences less shielding.

❌ Common Misconceptions & Lost Marks

• Counting error: Misidentifying the jump as between the 7th and 8th electron, leading to Chlorine.
• Incomplete M3: Stating only that the 7th electron is "in an inner shell" without mentioning less shielding or closer to the nucleus.
• Vague shell references: Saying "it is in a new shell" rather than specifying it comes from the second shell / 2p orbital.
• Full outer shell fallacy: Claiming "it takes more energy because you are breaking a stable octet/noble gas configuration" — examiners award 0 marks for octet/stability arguments without electrostatic reasoning.