Edexcel A-Level Chemistry Paper 1, June 2025: Question 2

Periodic Trends, Ionisation Energies & Intermolecular Forces

What this question tests

This 16-mark Edexcel exam question tests foundational understanding of periodicity, electronic configuration, ionisation energy definitions, ionic radii of isoelectronic species, and comparative intermolecular forces.

Graphing Periodicity: Plotting first ionisation energies (Period 2 trends and the Period 3 drop).
Explaining Anomalies: Nuclear charge vs shielding and subshell energy differences (Be vs Li and Be vs B).
Ionisation Equations & Orbitals: Defining successive ionisation energies and identifying electron subshells.
Isoelectronic Trends: Comparing ionic radius based on proton-to-electron ratios.
Extended Explanation (6 Marks): Dissecting hydrogen bonding vs London dispersion vs permanent dipole-dipole forces to explain boiling temperatures.

Part (a) • 3 Marks

Completing the First Ionisation Energy Graph

Plotting points for Nitrogen (N), Oxygen (O), and Sodium (Na)

✅ Correct Coordinates / Positions

Nitrogen (N): Plotted above Carbon (~1090 kJ mol⁻¹) and below Fluorine (~1680 kJ mol⁻¹), typically around 1400 kJ mol⁻¹. Must be higher than Oxygen [1 mark].
Oxygen (O): Plotted below Nitrogen and below Fluorine, but strictly above Carbon, typically around 1310 kJ mol⁻¹ [1 mark].
Sodium (Na): Plotted strictly below Lithium (Li is ~520 kJ mol⁻¹), typically around 496 kJ mol⁻¹ [1 mark].

🧠 Exam Technique & Visual Check

Remember the distinct "up, up, down, up, down, up, down, plunge" periodic pattern across Period 2.
N to O dip: Spin-pairing repulsion in the 2p⁴ configuration of O causes a slight drop below N.
Ne to Na plunge: Na begins Period 3; its 3s¹ electron enters a new principal quantum shell with significantly increased shielding and distance from the nucleus.

❌ Common Errors

Plotting Na higher than Li: Na has greater distance and inner shielding, so its IE is lower than Li.
Plotting Oxygen lower than Carbon: Oxygen drops relative to Nitrogen, but remains higher than Carbon and Boron.
Plotting markers in the spaces between grid lines rather than precisely aligned with the element labels on the x-axis.

Part (b) • 4 Marks

Explaining Anomalies: Beryllium vs Lithium & Boron

Explaining why 1st IE of Be is higher than both Li and B

✅ Mark Scheme Breakdown

Why Be is higher than Li (2 marks):

Be has one more proton / higher nuclear charge than Li [1 mark].
Shielding is the same / similar in Be and Li (outer electron is in the same 2s orbital / same distance from nucleus) [1 mark].

Why Be is higher than B (2 marks):

The outer electron in B is in a 2p subshell/orbital, whereas in Be it is in a 2s subshell/orbital [1 mark].
The 2p electron in B is higher in energy / experience increased shielding from the 2s² inner subshell, which outweighs the increased nuclear charge [1 mark].

💡 Key Scientific Phrasing

Always specify the exact word "outer electron". The mark scheme penalises omitting "outer" or "valence".
For Be vs B: State clearly that the 2p subshell is higher in energy than the 2s subshell (or more shielded by 2s²), requiring less energy to remove.
Always state which factor outweighs the other: increased shielding/distance outweighs the extra proton in Boron.

❌ Examiner Trap Alert

Do not confuse the Be vs B drop (2s vs 2p subshell change) with the N vs O drop (electron-pair repulsion in the same p orbital). Mentioning pairing repulsion here earns 0 marks for the Be/B comparison!

Parts (c), (d) & (e) • 3 Marks Total

Equations, Orbitals & Isoelectronic Radii

Part (c): 2nd IE of Beryllium [1 Mark]

Be⁺(g) → Be²⁺(g) + e⁻

Must include: Gaseous state symbols (g) on both ions. State symbol on electron is optional. Reversible arrows ( ⇌ ) are not allowed.

Part (d): Orbital for 2nd IE of Calcium [1 Mark]

Correct Answer: C (4s)

Reason: Ground-state Ca is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² . The first electron is removed from the 4s orbital to form Ca⁺ ( 4s¹ ). The second electron is also removed from the 4s orbital to form Ca²⁺ ( 4s⁰ ).

Part (e): Largest Isoelectronic Ion [1 Mark]

Correct Answer: A (S²⁻)

All four ions (S²⁻, Cl⁻, K⁺, Ca²⁺) are isoelectronic with 18 electrons:

Ion	Protons	Electrons
S²⁻	16	18 (Weakest pull → Largest)
Cl⁻	17	18
K⁺	19	18
Ca²⁺	20	18 (Strongest pull → Smallest)

Part (f) • 6 Marks

Extended Response: Boiling Temperatures & Intermolecular Forces

Explaining boiling points of H₂O (373 K), HF (293 K), HCl (188 K), and HI (238 K)

✅ 6-Mark Comprehensive Structure

Point 1: Hydrogen bonding is the strongest intermolecular force [1 mark].
Point 2: There is hydrogen bonding between H₂O molecules and between HF molecules [1 mark].
Point 3 (H₂O vs HF): H₂O has two hydrogen bonds per molecule (because O has 2 lone pairs and 2 δ+ H atoms) whereas HF has only one hydrogen bond per molecule on average (limited by only 1 H atom). Thus, H₂O has a higher boiling temperature than HF [1 mark].
Point 4 (HCl): There are permanent dipole-dipole forces between HCl molecules (as well as London forces) [1 mark].
Point 5 (HI vs HCl): HI has stronger / more London forces than HCl because HI molecules have more electrons (larger electron cloud) [1 mark].
Point 6 (Outweighing factor): The increased London forces in HI outweigh the permanent dipole-dipole forces in HCl, making the boiling temperature of HI higher than HCl [1 mark].

🧠 Structuring Your 6-Mark Answer

Divide your response clearly into two comparisons:

Comparison 1: H₂O vs HF (Hydrogen Bonding)
Establish that both have H-bonds (hence much higher than HCl/HI). Explain the number of H-bonds per molecule: Water can form 4 bonds per molecule (2 donated, 2 accepted → network of 2 per molecule), whereas HF is limited to 1 per molecule despite F being more electronegative.
Comparison 2: HI vs HCl (London vs Dipole-Dipole)
Identify permanent dipole-dipole in HCl. Contrast with HI having vastly more electrons (54 vs 18), creating significantly stronger London dispersion forces that outweigh HCl's dipole forces.

❌ Critical Examiner Warning: Do NOT Break Covalent Bonds!

The mark scheme contains a strict penalty: "Penalise reference to covalent bonds breaking once only."

Boiling is a physical change! Always state that intermolecular forces (hydrogen bonds, dipole-dipole forces, or London forces) are overcome or broken, never the intramolecular H–O, H–F, H–Cl, or H–I covalent bonds.