DP Chemistry (last assessment 2024)

Question 20N.2.sl.TZ0.1

Date	November 2020	Marks available	[Maximum mark: 28]	Reference code	20N.2.sl.TZ0.1
Level	sl	Paper	2	Time zone	TZ0
Command term	Calculate, Comment, Deduce, Determine, Draw, Explain, Formulate, Justify, Outline, Predict, Show, State	Question number	1	Adapted from	N/A

[Maximum mark: 28]

20N.2.sl.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

Most candidates wrote the electron configuration of chlorine correctly.

(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

Only half of the candidates deduced that the chloride ion has a larger radius than the chlorine atom with a valid reason. Many candidates struggled with this question and decided that the extra electron in the chloride ion caused a greater attraction between the nucleus and the outer electrons.

(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

Only about a third of the candidates identified the extra proton in the chlorine nucleus as the cause of the smaller atomic radius when compared to the sulfur atom, and only the stronger candidates also compared the shielding or the number of shells in the two atoms. Many candidates had a poor understanding of factors affecting atomic radius and could not explain the difference.

(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

About 60% of the candidates recognized that the peaks at m/z 35 and 37 in the mass spectrum of chlorine are due to its isotopes. A few students wrote 'isomers' instead of '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

This was the lowest scoring question on the paper, that was also left blank by 10% of the candidates. About 20% of the candidates identified the peak at m/z = 74 to be due to a molecule made up of two 37Cl atoms. And only very few candidates commented that the low abundance of the peak was due to the low abundance of the 37Cl isotope. A common incorrect answer was that chlorine has an isotope of mass number 74.

$2.67 g$ 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

Most candidates were able to determine the number of moles of MnO₂ using the 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

It was pleasing that the majority of the candidates were able to determine the limiting reactant by using the stoichiometric ratio.

(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

Half of the candidates were able to determine the amount of excess reactant. Some candidates who determined the limiting reactant in the previous part correctly, forgot to use the stoichiometric ratio in this part, and ended up with incorrect answers.

(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

60% of the candidates determined the volume of chlorine produced correctly. Some candidates made mistakes in the units when using PV = nRT and had a power of 10 error.

(b(v))

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

[2]

Markscheme

${MnO}_{2} : + 4$ ✔

${MnCl}_{2} : + 2$ ✔

Examiners report

The majority of candidates were able to determine the oxidation states of Mn in the two compounds correctly.

(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

Less than half of the candidates were awarded the mark. Some did identify MnO2 as the oxidizing agent but did not give the explanation in terms of oxidation state as required in the question. Other candidates did not have an understanding of oxidizing and reducing agents.

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

A very well answered question - 80% of candidates understood what is meant by the term weak acid. Incorrect answers included 'acids that have high pH'.

(c(ii))

State the formula of the conjugate base of hypochlorous acid.

[1]

Markscheme

${ClO}^{-}$ ✔

Examiners report

Half of the candidates deduced the formula of the conjugate base of hypochlorous acid. Incorrect answers included H₂O and HCl.

(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

A well answered question. It was pleasing to see that 70% of the candidates were able to calculate [H⁺] from the given pH.

(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

More than half of the candidates identified the type of reaction between ethane and chlorine as a substitution reaction. A few candidates lost the marks for writing 'electrophilic substitution' or 'nucleophilic substitutions'.

(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

This was a challenging question that was answered correctly by only 30% of the candidates. A variety of incorrect answers were seen such as 'chlorine is a halogen and hence it is reactive', and 'ethane is more reactive because it is an alkane'. For students who answered correctly, the polarity was the most frequently given reason.

(d(iii))

Write the equation for the reaction of chloroethane with a dilute aqueous solution of sodium hydroxide.

[1]

Markscheme

${CH}_{3} {CH}_{2} Cl (l) + {OH}^{-} (aq) \to {CH}_{3} {CH}_{2} OH (aq) + {Cl}^{-} (aq)$
OR
${CH}_{3} {CH}_{2} Cl (l) + NaOH (aq) \to {CH}_{3} {CH}_{2} OH (aq) + NaCl (aq)$ ✔

Accept use of $C_{2} H_{5} C l$ and $C_{2} H_{5} O H / C_{2} H_{6} O$ in the equation.

Examiners report

Half of the candidates wrote the correct equation for the hydrolysis of chloroethane. Incorrect answers often included carbon dioxide and water as the products.

(d(iv))

Deduce the nucleophile for the reaction in d(iii).

[1]

Markscheme

hydroxide «ion»/ ${OH}^{-}$ ✔

Do not accept $N a O H$ .

Examiners report

This was a highly discriminating question. Only 30% of the candidates were able to identify the hydroxide ion as the nucleophile in the hydrolysis of chloroethane. Incorrect answers included NaOH where the ion was not specified. 14% of the candidates left this question blank.

(d(v))

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

Half of the candidates were able to give the structural formula of ethoxyethane. Incorrect answers included methoxymethane, ketones and esters.

(d(vi))

Deduce the number of signals and their chemical shifts in the ${}_{1}H NMR$ spectrum of ethoxyethane. Use section 27 of the data booklet.

[2]

Markscheme

$2$ «signals» ✔

$0.9 - 1.0 « ppm »$ AND $3.3 - 3.7 « ppm »$ ✔

Accept any values in the ranges.

Award [1 max] for two incorrect chemical shifts.

Examiners report

Nearly half of the candidates were able to identify the number of signals obtained in the 1H NMR spectrum of ethoxyethane, obtaining the first mark of this question. Many candidates were awarded the mark as 'error carried forward' from an incorrect structure of ethoxyethane. The second mark for this question required candidates to look up values of chemical shift from the data booklet. Nearly a third of the candidates were able to match the chemical environments of the hydrogen atoms in ethoxyethane to those listed in the data booklet successfully.

${CCl}_{2} F_{2}$ 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

This was the highest scoring question in the paper. The majority of candidates were able to calculate the percentage by mass of chlorine in CCl₂F₂. Mistakes included incorrect rounding and arithmetic errors.

(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

This nature of science question was well answered by half of the candidates. Some teachers commented that the wording was rather vague. Incorrect answers were mainly assuming that CFCs were related to the combustion of fuels and greenhouse gas emissions.