DP Chemistry (last assessment 2024)

Question 20N.2.hl.TZ0.1

Date	November 2020	Marks available	[Maximum mark: 32]	Reference code	20N.2.hl.TZ0.1
Level	hl	Paper	2	Time zone	TZ0
Command term	Calculate, Comment, Deduce, Determine, Draw, Explain, Justify, Outline, Predict, State, Write	Question number	1	Adapted from	N/A

[Maximum mark: 32]

20N.2.hl.TZ0.1

Chlorine undergoes many reactions.

(a(i))

State the full electron configuration of the chlorine atom.

[1]

Markscheme

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}$ ✔

Do not accept condensed electron configuration.

Examiners report

Well answered question with 90% of candidates correctly identifying the complete electron configuration for chlorine.

(a(ii))

State, giving a reason, whether the chlorine atom or the chloride ion has a larger radius.

[1]

Markscheme

${Cl}^{-}$ AND more «electron–electron» repulsion ✔

Accept $C l^{-}$ AND has an extra electron.

Examiners report

Most candidates could correctly explain the relative sizes of chlorine atom and chloride ion.

(a(iii))

Outline why the chlorine atom has a smaller atomic radius than the sulfur atom.

[2]

Markscheme

$Cl$ has a greater nuclear charge/number of protons/ $Z_{eff}$ «causing a stronger pull on the outer electrons» ✔

same number of shells
OR
same «outer» energy level
OR
similar shielding ✔

Examiners report

Fairly well answered though some candidates missed M2 for not recognizing the same number of shells affected.

(a(iv))

The mass spectrum of chlorine is shown.

Outline the reason for the two peaks at $m / z = 35$ and $37$ .

[1]

Markscheme

«two major» isotopes «of atomic mass $35$ and $37$ » ✔

Examiners report

More than 80% could identify that the two peaks in the MS of chlorine are due to different isotopes.

(a(v))

Explain the presence and relative abundance of the peak at $m / z = 74$ .

[2]

Markscheme

«diatomic» molecule composed of «two» chlorine-37 atoms ✔

chlorine-37 is the least abundant «isotope»
OR
low probability of two ${}_{{}_{37}}{Cl}$ «isotopes» occurring in a molecule ✔

Examiners report

Not well answered. Some candidates were able to identify m/z 74 being due to the m/z of two Cl-37 atoms, however fewer candidates were able to explain the relative abundance of the isotope.

$2.67 g$ of manganese(IV) oxide was added to $200.0 {cm}^{3}$ of $2.00 mol {dm}^{- 3} HCl$ .

${MnO}_{2} (s) + 4 HCl (aq) \to {Cl}_{2} (g) + 2 H_{2} O (l) + {MnCl}_{2} (aq)$

(b(i))

Calculate the amount, in $mol$ , of manganese(IV) oxide added.

[1]

Markscheme

$« \frac{2.67 g}{86.94 g {mol}^{- 1}} = » 0.0307 « mol »$ ✔

Examiners report

Stoichiometric calculations were generally well done and over 90% could calculate mol from a given mass.

(b(ii))

Determine the limiting reactant, showing your calculations.

[2]

Markscheme

$« n_{HCl} = 2.00 mol {dm}^{- 3} \times 0.2000 {dm}^{3} » = 0.400 mol$ ✔

$« \frac{0.400}{4} = » 0.100 mol$ AND ${MnO}_{2}$ is the limiting reactant ✔

Accept other valid methods of determining the limiting reactant in M2.

Examiners report

90% of candidates earned full marks on this 2-mark question involving finding a limiting reactant.

(b(iii))

Determine the excess amount, in $mol$ , of the other reactant.

[1]

Markscheme

$« 0.0307 mol \times 4 = 0.123 mol »$

$« 0.400 mol - 0.123 mol = » 0.277 « mol »$ ✔

Examiners report

Surprisingly, quite a number of candidates struggled with the quantity of excess reactant despite correctly identifying limiting reactant previously.

(b(iv))

Calculate the volume of chlorine, in ${dm}^{3}$ , produced if the reaction is conducted at standard temperature and pressure (STP). Use section 2 of the data booklet.

[1]

Markscheme

$« 0.0307 mol \times 22.7 {dm}^{3} {mol}^{- 1} = » 0.697 « {dm}^{3} »$ ✔

Accept methods employing $p V = n R T$ .

Examiners report

Most candidates could find the volume of gas produced in a reaction under standard conditions.

(b(v))

State the oxidation state of manganese in ${MnO}_{2}$ and ${MnCl}_{2}$ .

[2]

Markscheme

${MnO}_{2} : + 4$ ✔

${MnCl}_{2} : + 2$ ✔

Examiners report

More than 90% could identify the oxidation number of manganese in both MnO₂ and MnCl₂.

(b(vi))

Deduce, referring to oxidation states, whether ${MnO}_{2}$ is an oxidizing or reducing agent.

[1]

Markscheme

oxidizing agent AND oxidation state of $Mn$ changes from $+ 4$ to $+ 2$ /decreases ✔

Examiners report

Most candidates stated that MnO₂ is an oxidizing agent in the reaction but many did not get the mark because there was no reference to oxidation states.

Chlorine gas reacts with water to produce hypochlorous acid and hydrochloric acid.

${Cl}_{2} (g) + H_{2} O (l) ⇌ HClO (aq) + HCl (aq)$

(c(i))

Hypochlorous acid is considered a weak acid. Outline what is meant by the term weak acid.

[1]

Markscheme

partially dissociates/ionizes «in water» ✔

Examiners report

Another well answered 1-mark question where candidates correctly identified a weak acid as an acid which partially dissociates in water.

(c(ii))

State the formula of the conjugate base of hypochlorous acid.

[1]

Markscheme

${ClO}^{-}$ ✔

Examiners report

Roughly ⅓ of the candidates failed to identify the conjugate base, perhaps distracted by the fact it was not contained in the equation given.

(c(iii))

Calculate the concentration of $H^{+} (aq)$ in a $HClO (aq)$ solution with a $pH = 3.61$ .

[1]

Markscheme

$« [H^{+}] = 10^{- 3.61} = » 2.5 \times 10^{- 4} « mol {dm}^{- 3} »$ ✔

Examiners report

Vast majority of candidates could calculate the concentration of H⁺ (aq) in a HClO (aq) solution with a pH =3.61.

(d(i))

State the type of reaction occurring when ethane reacts with chlorine to produce chloroethane.

[1]

Markscheme

«free radical» substitution/ $S_{R}$ ✔

Do not accept electrophilic or nucleophilic substitution.

Examiners report

Many identified the reaction of chlorine with ethane as free-radical substitution, or just substitution, with some erroneously stating nucleophilic or electrophilic substitution.

(d(ii))

Predict, giving a reason, whether ethane or chloroethane is more reactive.

[1]

Markscheme

chloroethane AND C–Cl bond is weaker/ $324 kJ {mol}^{- 1}$ than C–H bond/ $414 kJ {mol}^{- 1}$
OR
chloroethane AND contains a polar bond ✔

Accept “chloroethane AND polar”.

Examiners report

The underlying reasons for the relative reactivity of ethane and chloroethane were not very well known with a few giving erroneous reasons and some stating ethane more reactive.

(d(iii))

Explain the mechanism of the reaction between chloroethane and aqueous sodium hydroxide, $NaOH (aq)$ , using curly arrows to represent the movement of electron pairs.

[3]

Markscheme

curly arrow going from lone pair/negative charge on $O$ in ⁻OH to $C$ ✔

curly arrow showing $Cl$ leaving ✔

representation of transition state showing negative charge, square brackets and partial bonds ✔

Accept $O H^{-}$ with or without the lone pair.

Do not accept curly arrows originating on $H$ in $O H^{-}$ .

Accept curly arrows in the transition state.

Do not penalize if $H O$ and $C l$ are not at 180°.

Do not award M3 if $O H - C$ bond is represented.

Examiners report

Few earned full marks for the curly arrow mechanism of the reaction between sodium hydroxide and chloroethane. Mistakes being careless curly arrow drawing, inappropriate –OH notation, curly arrows from the hydrogen or from the carbon to the C–Cl bond, or a method that missed the transition state.

(d(iv))

Ethoxyethane (diethyl ether) can be used as a solvent for this conversion.
Draw the structural formula of ethoxyethane

[1]

Markscheme

/ ${CH}_{3} {CH}_{2} {OCH}_{2} {CH}_{3}$ ✔

Accept $(C H_{3} C H_{2})_{2} O$ .

Examiners report

Approximately 60% could draw ethoxyethane however many demonstrated little knowledge of structure of an ether molecule.

(d(v))

Deduce the number of signals and chemical shifts with splitting patterns in the ¹H NMR spectrum of ethoxyethane. Use section 27 of the data booklet.

[3]

Markscheme

2 «signals» ✔

0.9−1.0 AND triplet ✔

3.3−3.7 AND quartet ✔

Accept any values in the ranges.

Award [1] for two correct chemical shifts or two correct splitting patterns.

Examiners report

A poorly answered question with some getting full marks on this 1HNMR spectrum of ethoxyethane question. Very few could identify all 3 of number of signals, chemical shift, and splitting pattern.

${CCl}_{2} F_{2}$ is a common chlorofluorocarbon, $CFC$ .

(e(i))

Calculate the percentage by mass of chlorine in ${CCl}_{2} F_{2}$ .

[2]

Markscheme

$« M ({CCl}_{2} F_{2}) = » 120.91 « g {mol}^{- 1} »$ ✔

$\frac{2 \times 35.45 g {mol}^{- 1}}{120.91 g {mol}^{- 1}} \times 100 % = » 58.64 « % »$ ✔

Award [2] for correct final answer.

Examiners report

Another good example of candidates being well rehearsed in calculations with 90% earning 2/2 on this question of calculation percentage by mass composition.

(e(ii))

Comment on how international cooperation has contributed to the lowering of $CFC$ emissions responsible for ozone depletion.

[1]

Markscheme

Any of:

research «collaboration» for alternative technologies «to replace $CFC$ s»
OR
technologies «developed»/data could be shared
OR
political pressure/Montreal Protocol/governments passing legislations ✔

Do not accept just “collaboration”.

Do not accept any reference to $C F C$ as greenhouse gas or product of fossil fuel combustion.

Accept reference to specific measures, such as agreement on banning use/manufacture of $C F C$ s.

Examiners report

Somewhat disappointing answers on this question about how international cooperation has contributed to the lowering of CFC emissions. Many gave vague answers and some referred to carbon emissions and global warming.

(e(iii))

$CFC$ s produce chlorine radicals. Write two successive propagation steps to show how chlorine radicals catalyse the depletion of ozone.

[2]

Markscheme

$O_{3} + Cl \cdot \to O 2 + ClO \cdot$ ✔

$ClO \cdot + O \cdot \to O_{2} + Cl \cdot$
OR
$ClO \cdot + O_{3} \to Cl \cdot + 2 O_{2}$ ✔