Edexcel A-Level Chemistry AS Paper 2, November 2020: Question 5

12 marks · Hard difficulty · Calculations

Calculate the bond energy of I-Cl, draw bond dipoles and reaction mechanisms for addition to alkenes, calculate moles in titrations related to iodine numbers, and explain reaction rates of polar interhalogen molecules compared to halogens.

Practise this question

Question

A multi-part question about iodine monochloride (ICl). Part (a) asks to calculate the I-Cl bond energy using given bond energies and a reaction enthalpy equation. Part (b)(i) asks to add a dipole to an I-Cl bond diagram, and (b)(ii) asks to draw the mechanism for the addition of ICl to propene. Part (c)(i) and (c)(ii) involve titration calculations using sodium thiosulfate to determine moles, iodine number, and identify an oil from a table. Part (c)(iii) asks why the reaction of ICl is faster than iodine.
Question text

5 This question concerns iodine monochloride, ICl, a red‑brown solid which melts at 27°C

to form a red‑brown liquid.

Iodine monochloride is used in measuring unsaturation in organic compounds.

(a) Iodine monochloride gas can be produced by the reaction between iodine vapour

and chlorine gas. The reaction is exothermic.

I (g) + Cl (g) → 2ICl(g) Δ H = −30 kJ mol−1

22 r

The table shows bond energy values for the bonds in iodine and chlorine.

Calculate the value of the bond energy of the I Cl bond using these data and

the equation.

Bond Energy / kJ mol−1

I I 151

Cl Cl 243

(2)

(b) Iodine monochloride is a polar molecule which adds rapidly to double bonds in

a similar way to hydrogen chloride. This reaction can be used to determine the

degree of unsaturation in oils.

(i) Add the dipole to a molecule of iodine monochloride.

(1)

I Cl

(ii) Draw the mechanism for the addition of iodine monochloride to propene.

You should include all curly arrows and relevant lone pairs and dipoles.

(3)

(c) (i) To determine the extent of unsaturation of an oil, 0.250g of the oil was treated

with 25.00 cm3 of a 0.100 mol dm−3 ICl solution. Unreacted ICl reacted with

excess potassium iodide solution, forming iodine according to the equation:

ICl + KI → I2 + KCl

The amount of iodine produced was measured by reacting the mixture with a

solution of sodium thiosulfate, Na2S2O3.

The iodine released reacted with 32.65*P62307A01428*cm3of 0.100moldm−3sodium thiosulfate

solution in the mole ratio of 1 mol I2 : 2 mol Na2S2O3.

Calculate the number of moles of iodine monochloride which reacted with

0.250g of the oil.

(3)

(ii) Unsaturation in oils is measured using a scale called ‘Iodine number’.

This is the mass of iodine which will react with 100g of the oil.

Because iodine adds very slowly to double bonds, the reaction of iodine monochloride

is used instead.

Given that 1 mol of I2 is equivalent to 1 mol of ICl, use your answer in (c)(i) to

calculate the mass of iodine that would react with 100g of oil and hence identify

the unsaturated oil from the list of possible oils and their iodine numbers.

Oil Iodine number

cocoa butter 35–40

coconut oil 7–10

*P62307A01528*cod liver oil145–180

palm oil 44–51

peanut oil 84–106

(2)

*P62307A01628*

(iii) Give a reason why the reaction of iodine monochloride is significantly faster

than the reaction of iodine.

(1)

(Total for Question 5 = 12 marks)

Mark scheme

Show the mark scheme The mark scheme providing step-by-step answers and calculations for parts 5(a) through 5(c)(iii), including bond energy calculations, mechanism curly arrows and intermediates, titration mole calculations, iodine number derivation, and explanations regarding polarity and electrophilicity.

How to answer it

Iodine Monochloride: Structure, Mechanism & Calculations

What this question tests

This question assesses core AS Level Physical and Organic Chemistry concepts: calculating bond enthalpies using enthalpy change of reaction, drawing bond polarities and electrophilic addition mechanisms with curly arrows, performing multi-step titration stoichiometry (back-titrations), determining "iodine numbers" to identify fats/oils, and explaining reaction kinetics based on bond polarity.

Part (a): Bond Enthalpy Calculation

Calculate the bond energy of the I—Cl bond

✅ Correct Answer

+212 kJ mol⁻¹

Mark breakdown: (1) for bonds broken calculation (394); (1) for rearranging/calculating final I—Cl bond energy.

📐 Step-by-Step Calculation

  1. Energy required to break bonds (reactants):
    I—I (151) + Cl—Cl (243) = 394 kJ mol⁻¹
  2. Use formula for enthalpy change:
    ΔH = Σ(Bonds broken) - Σ(Bonds formed)
    -30 = 394 - 2(I—Cl)
  3. Rearrange for 2 × (I—Cl):
    2(I—Cl) = 394 + 30 = 424 kJ mol⁻¹
  4. Divide by 2 for one bond:
    I—Cl = 424 / 2 = +212 kJ mol⁻¹

❌ Common Errors

Forgetting to multiply the final value or divide by 2 correctly. Missing the sign rules when rearranging the thermochemical equation.

Part (b): Polarity and Reaction Mechanism

Dipole orientation and electrophilic addition to propene

✅ Correct Answers

(i) Dipole: Iodine carries partial positive charge ( δ+ ) and chlorine carries partial negative charge ( δ- ).

(ii) Mechanism: Curly arrow from C=C double bond to I(δ+); I—Cl bond breaks heterolytically with arrow to Cl; formation of secondary carbocation intermediate; curly arrow from Cl⁻ lone pair to electron-deficient carbon.

Marks: (i) 1 mark | (ii) 3 marks

💡 Key Knowledge

Electronegativity dictates polarity: Chlorine (3.0) is more electronegative than Iodine (2.5), making I the electrophile ( δ+ ).

🧠 Exam Technique

Ensure curly arrows start precisely from bonds/lone pairs and point accurately to atoms. The intermediate must show a secondary carbocation (positive charge on carbon 2).

Part (c)(i): Back-Titration Calculations

Calculate moles of ICl reacted with 0.250 g of oil

✅ Correct Answer

0.0008675 mol (or 8.68 × 10⁻⁴ mol )

Mark breakdown: (1) Moles of ICl added; (1) Moles of S₂O₃²⁻ / I₂ liberated; (1) Final subtraction for reacted ICl.

📐 Step-by-Step Calculation

  1. Moles of ICl added initially:
    (25.0 / 1000) × 0.100 = 0.00250 mol
  2. Moles of Na₂S₂O₃ used in titration:
    (32.65 / 1000) × 0.100 = 0.003265 mol
  3. Moles of unreacted ICl (via I₂ liberated):
    From stoichiometry (1 mol I₂ : 2 mol Na₂S₂O₃), unreacted ICl = Na₂S₂O₃ / 2 = 0.003265 / 2 = 0.0016325 mol
  4. Moles of ICl reacted with oil:
    Initial (0.00250) - Unreacted (0.0016325) = 0.0008675 mol

Part (c)(ii): Iodine Number and Oil Identification

Calculate mass of iodine per 100g of oil and identify the oil

✅ Correct Answer

Iodine number = 88.07 → peanut oil

Mark breakdown: (1) Calculation of equivalent mass of iodine; (1) Scaling to 100g and correct table match with Error Carried Forward (TE).

📐 Step-by-Step Calculation

  1. Mass of iodine equivalent to reacted ICl:
    Moles of ICl (0.0008675) × Mᵣ of I₂ (253.8) = 0.2202 g
  2. Scale up for 100 g of oil:
    (0.2202 g / 0.250 g) × 100 = 88.09 g (or using 254 gives 88.07 g )
  3. Identify oil:
    88.09 falls directly within the range for peanut oil (84–106).

Part (c)(iii): Reaction Kinetics Explanation

Why is ICl reaction significantly faster than with I₂?

✅ Correct Answer

Iodine monochloride has a permanent dipole (whereas I—I is non-polar), making the iodine atom more δ+ , a stronger electrophile, and more susceptible to attack by the electron-rich double bond.

Mark: 1 mark (Requires explicit comparison to I₂).

❌ Common Errors

Stating "ICl is polar" without comparing it to iodine (I₂). Examiners penalize answers that fail to state that I₂ is non-polar or has no permanent dipole.

Topics

Physical Chemistry · Organic Chemistry · Topic 8: Energetics I · Topic 2: Bonding and Structure · Topic 5: Formulae, Equations and Amounts of Substance · Topic 6: Organic Chemistry I

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.