AQA A-Level Chemistry Paper 1, 2018: Question 4

✅ Correct Answer

1s² 2s² 2p⁶ 3s² 3p⁶

(Acceptable: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ )

❌ Common Errors

• Writing the neutral atom configuration: ...3p⁶ 4s² .
• Using noble gas shorthand [Ar] when the question explicitly demands the full electron configuration.

✅ Model Answer

Point 1: In Ca⁺, the electron is removed from a higher energy orbital (4s), which is further from the nucleus than the electron being removed from K⁺ (which is in a 3p orbital closer to the nucleus).

Point 2: The outer electron in Ca⁺ experiences more electron shielding than the outer electron in K⁺.

🧠 Exam Technique: Be Comparative

• Always state which species is being compared: state both Ca⁺ and K⁺ (or converse).
• Mention the specific sub-levels: Ca⁺ loses from 4s (4th shell), whereas K⁺ loses from 3p (3rd shell).
• Use comparative language: further, higher energy, more shielding. Absolute statements like "Ca has shielding" lose marks.

✅ 04.3 Highest First IE in s-Block

Beryllium (or Be )

Note: The question specifies an s-block metal. Helium and Hydrogen are non-metals. Be is at the top of Group 2, having fewer shells and the smallest atomic radius among s-block metals.

✅ 04.4 Least Soluble Group 2 Hydroxide

Mg(OH)₂ (or Magnesium hydroxide)

Trend: Solubility of Group 2 hydroxides increases down the group: Mg(OH)₂ is sparingly soluble (least soluble from Mg to Ba), whereas Ba(OH)₂ is the most soluble.

✅ Required Answers

Ionic equation: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Reagent in excess: Barium chloride ( BaCl₂ )

Total volume of other reagent: 10 cm³ (or 0.010 dm³)

📐 Step-by-Step Calculation

1. Find moles of each reactant:
   Moles of BaCl₂ = (6 / 1000) × 0.25 = 1.5 × 10⁻³ mol
   Moles of Na₂SO₄ = (8 / 1000) × 0.15 = 1.2 × 10⁻³ mol
2. Identify excess reagent:
   Reacting ratio is 1:1. Since 1.5 × 10⁻³ > 1.2 × 10⁻³, BaCl₂ is in excess (by 3 × 10⁻⁴ mol).
3. Calculate total volume of Na₂SO₄ needed:
   For filtrate to contain only one solute (NaCl), BaCl₂ must react completely with no excess Ba²⁺ remaining.
   Required moles of Na₂SO₄ = 1.5 × 10⁻³ mol.
   Volume = moles / concentration = (1.5 × 10⁻³) / 0.15 = 0.010 dm³ = 10 cm³.

💡 Why Isotopes Have Identical Chemical Properties

They have the same electronic configuration (or same number of valence electrons / all have 38 electrons).

Examiner note: Chemical reactions involve electrons, so identical electron arrangements mean identical chemical behavior.

📐 Calculating Percentage Abundance of ⁸⁸Sr

1. Define variables based on the 1:1 ratio:
   Let % abundance of ⁸⁶Sr = x
   % abundance of ⁸⁷Sr = x
   % abundance of ⁸⁸Sr = 100 − 2x
2. Set up the relative atomic mass expression:
   [86x + 87x + 88(100 − 2x)] / 100 = 87.7
3. Expand and solve for x:
   173x + 8800 − 176x = 8770
   −3x = 8770 − 8800 = −30
   x = 10%
4. Find percentage of ⁸⁸Sr:
   % abundance = 100 − 2(10) = 80.0%

✅ Correct Answer

¹³⁸Ba⁺

💡 TOF Principle: Why ¹³⁸Ba⁺ takes the longest?

All ions are accelerated to have the same kinetic energy ( KE = ½mv² ).

Because v = √(2KE / m) , heavier ions have a lower velocity and therefore take a longer time of flight to reach the detector.

❌ Critical Traps to Avoid

• Forgetting to convert grams to kilograms: Mass must be in kg to be compatible with Joules (1 J = 1 kg m² s⁻²). Forgetting the factor of 10⁻³ gives d = 0.0485 m (caps score at 4 marks).
• Forgetting to divide by Avogadro's constant: Using molar mass instead of the mass of a single particle loses marks M1 and M5.
• Rounding too early: Keep intermediate values in your calculator memory to prevent rounding errors in the final significant figure.

🧠 Alternative Single Formula Method

Combine formulas before substituting values:

d = t × √(2KE / m)

Substituting values directly:

d = (2.71 × 10⁻⁵) × √[ (2 × 3.65 × 10⁻¹⁶) / 2.275 × 10⁻²⁵ ] = 1.54 m