Chemistry 9701/22 — May/June 2021

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Halogen Compounds · Introduction to Organic Chemistry · Group 17 · Atoms, Molecules and Stoichiometry · Group 2 · Atomic Structure · +8 more

Q1Group 17Atoms, Molecules and StoichiometryGroup 2Atomic StructureFree sample

A Group 2 metal combines with bromine to form a crystalline solid, $\text{MBr}_2$ .

Excess aqueous $\text{AgNO}_3$ is added to a solution of $\text{MBr}_2$ and a precipitate forms. The mixture is filtered. The precipitate is dried and the mass of the precipitate is recorded.

(a)

State the formula and colour of the precipitate.

DifficultyEasy

Worked solution

Answer

Precipitate: AgBr, cream in colour.

Final answer

AgBr; cream

Detailed explanation

Background Concept

When aqueous silver nitrate is added to a solution containing halide ions, insoluble silver halides precipitate: AgCl is white, AgBr is cream, AgI is pale yellow. The colour of the precipitate identifies the halide ion present.

Understanding the Question

You must state both the formula and the colour of the precipitate formed when AgNO3(aq) is added to a solution containing bromide ions (from MBr2).

Approach

Recognise that MBr2 provides Br- ions, which react with Ag+ to give insoluble AgBr, and recall its colour.

Step-by-Step Reasoning

Bromide ions react with Ag+ ions: Ag+(aq) + Br-(aq) -> AgBr(s). Silver bromide precipitates as a cream solid. Both pieces of information carry one mark each, so both must be stated.

Key Takeaways

Learn the three silver halide colours in order of decreasing solubility: AgCl white, AgBr cream, AgI pale yellow.

Common Mistakes

Writing 'white' instead of 'cream', or giving only the colour without the formula. Both formula and colour are separately marked.

Things to Be Careful About

Cream (not white or yellow) is required for full credit; the precipitate is AgBr, not Ag2Br.

Techniques used

identify the silver halide precipitaterecall the colour of silver bromide

(b)

Complete the equation to represent the reaction between $\text{MBr}_2$ and $\text{AgNO}_3$ .

...... $\text{MBr}_2$ + ...... $\text{AgNO}_3 \rightarrow$ ...................................................................

DifficultyMedium-Easy

Worked solution

Answer

1\text{MBr}_2 + 2\text{AgNO}_3 \rightarrow 2\text{AgBr} + \text{M}(\text{NO}_3)_2

Final answer

1MBr2 + 2AgNO3 -> 2AgBr + M(NO3)2

Detailed explanation

Background Concept

In a double displacement (precipitation) reaction the ions simply swap partners: Ag+ pairs with Br- to form insoluble AgBr, and the metal cation M2+ pairs with NO3- to form soluble M(NO3)2 which stays in solution.

Understanding the Question

You must complete and balance the equation between MBr2 and AgNO3, filling in the coefficients and naming all products.

Approach

Each MBr2 contains two Br- ions, so two Ag+ ions are needed, hence 2AgNO3 giving 2AgBr. The remaining ions M2+ and two NO3- form M(NO3)2.

Step-by-Step Reasoning

Write the ionic swap: M2+ + 2Br- + 2Ag+ + 2NO3- -> 2AgBr(s) + M2+ + 2NO3-. Cancelling spectator ions leaves Ag+ + Br- -> AgBr. Reassembling the full equation gives 1MBr2 + 2AgNO3 -> 2AgBr + M(NO3)2, which is balanced in every element.

Key Takeaways

When one formula unit of a Group 2 halide reacts with AgNO3, the AgNO3 coefficient is always twice the number of halide ions per formula unit.

Common Mistakes

Writing AgBr2 or forgetting the coefficient 2 before AgNO3 and AgBr; writing MNO3 instead of M(NO3)2, which is unbalanced and wrong for a 2+ cation.

Things to Be Careful About

Check that nitrate appears twice in M(NO3)2; brackets are essential.

Techniques used

write and balance the precipitation equation

(c)

A $0.250\text{ g}$ sample of pure $\text{MBr}_2$ contains $8.415 \times 10^{-4}\text{ mol } \text{MBr}_2$ .

Calculate the relative formula mass, $M_r$ , of $\text{MBr}_2$ . Use this to identify $\text{M}$ .

Show your working.

DifficultyMedium-Easy

Worked solution

Working

M_r(\text{MBr}_2) = \frac{0.250}{8.415 \times 10^{-4}} = 297.1

A_r(\text{M}) = 297.1 - 2(79.9) = 137.3

Answer

$A_r \approx 137$ , so $\text{M}$ is barium (Ba).

Final answer

Mr = 297.1; Ar(M) = 137.3; M = barium

Detailed explanation

Background Concept

Moles = mass / Mr. Rearranging, Mr = mass / moles. For an ionic compound, the formula mass equals the sum of the atomic masses of its constituent atoms, so the metal's Ar can be found by subtracting the bromine contribution.

Understanding the Question

Given the mass (0.250 g) and the amount (8.415 x 10^-4 mol) of MBr2, find Mr, then Ar of M, then identify which Group 2 metal matches that Ar.

Approach

Step 1: Mr = mass / moles. Step 2: Ar(M) = Mr - 2 x Ar(Br) using Ar(Br) = 79.9. Step 3: match the Ar to the Group 2 elements (Mg 24.3, Ca 40.1, Sr 87.6, Ba 137.3, Ra 226).

Step-by-Step Reasoning

Mr = 0.250 / (8.415 x 10^-4) = 297.1 (1 d.p.). Bromine's Ar is 79.9, and there are two Br atoms, contributing 159.8. So Ar(M) = 297.1 - 159.8 = 137.3. Comparing with Group 2 atomic masses, 137.3 corresponds to barium.

Key Takeaways

The reverse of moles = mass/Mr lets you find an unknown Mr from experimental data; subtracting known atomic masses identifies an unknown element.

Common Mistakes

Subtracting only one bromine (forgetting the subscript 2), using Ar(Br) = 80 inconsistently, or identifying Sr (87.6) or Ca (40.1) due to arithmetic slips.

Things to Be Careful About

Carry enough significant figures through (at least 4) so the subtraction gives 137.3 unambiguously; quote Mr as 297.1 (or 297).

Techniques used

calculate moles from mass and Mrdeduce Ar from the formula massidentify the Group 2 element from its Ar

(d)

A sample of $\text{MBr}_2$ is dissolved in water. Chlorine gas is then bubbled into the solution.

(i)

Describe the observations for this reaction.

DifficultyEasy

Worked solution

Answer

The solution turns orange/brown as bromine is displaced.

Final answer

Solution turns orange/brown

Detailed explanation

Background Concept

Halogens displace less reactive halogens from their halides: reactivity decreases down Group 17 (Cl2 > Br2 > I2). Chlorine can oxidise bromide ions to bromine molecules.

Understanding the Question

Chlorine gas is bubbled into an aqueous solution of MBr2 (bromide ions). You must state what is observed.

Approach

Apply the reactivity order: Cl2 displaces Br2 from bromide solution. Bromine in water is orange/brown, so the colourless solution takes on this colour.

Step-by-Step Reasoning

Cl2(g) + 2Br-(aq) -> 2Cl-(aq) + Br2(aq). The displaced bromine colours the solution orange/brown. This single observation is the one mark.

Key Takeaways

Memorise the solution colours of the halogens: chlorine pale green (barely visible when dilute), bromine orange, iodine brown.

Common Mistakes

Saying a precipitate forms (bromine stays dissolved), or describing the colour of chlorine gas rather than the solution colour change.

Things to Be Careful About

The observation is a colour change in the solution/mixture — credit requires the orange or brown colour of bromine.

Techniques used

predict displacement reaction observationsrecall bromine solution colour

(ii)

Name the type of reaction that occurs when $\text{MBr}_2$ reacts with chlorine gas.

DifficultyEasy

Worked solution

Answer

Displacement (a redox reaction).

Final answer

Displacement

Detailed explanation

Background Concept

A more reactive halogen displaces a less reactive halogen from its salt; the more reactive halogen is reduced and the halide ion is oxidised, so displacement of halogens is also a redox reaction.

Understanding the Question

Name the type of reaction between chlorine and bromide ions.

Approach

Chlorine 'kicks out' bromine from MBr2 — this is displacement.

Step-by-Step Reasoning

Cl2 has displaced Br from the bromide salt; chlorine gains electrons (reduction) and bromide loses electrons (oxidation). The expected single term is 'displacement'; 'redox' may also be credited by some schemes, but 'displacement' is the mark-scheme answer.

Key Takeaways

Displacement reactions of halogens demonstrate the decreasing oxidising power down Group 17.

Common Mistakes

Writing 'disproportionation' (chlorine's oxidation state changes from 0 to -1, bromine's from -1 to 0 — different species change in different directions, so it is not disproportionation).

Things to Be Careful About

Give the single, specific term the mark scheme wants: displacement.

Techniques used

classify the reaction type

(e)

Compound $\mathbf{Y}$ is a pure insoluble solid which contains halide ions.

A single reagent is added directly to compound $\mathbf{Y}$ to determine the halide ion present.

Identify the reagent added. State the observation which would confirm that $\mathbf{Y}$ contains bromide ions.

reagent ......................................................................................................................................

observation ................................................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

Reagent: concentrated sulfuric acid.

Observation: brown vapour/fumes (of bromine) are released.

Final answer

Concentrated H2SO4; brown vapour/gas

Detailed explanation

Background Concept

Solid halide salts are commonly tested with concentrated H2SO4. NaCl gives misty HCl fumes; NaBr gives misty HBr fumes plus brown Br2 vapour (HBr reduces the H2SO4); NaI gives purple I2 vapour and rotten-egg H2S smell.

Understanding the Question

Compound Y is an insoluble solid containing halide ions; a single reagent added directly must reveal whether bromide is present. Both reagent and observation are required.

Approach

Concentrated sulfuric acid works on solid halides without prior dissolution, unlike AgNO3 which needs a solution. Bromide gives brown bromine vapour — the distinguishing observation.

Step-by-Step Reasoning

Adding conc. H2SO4 to a bromide salt: Br- is first protonated to HBr, then HBr reduces H2SO4, producing Br2 which appears as brown fumes. Chloride would give only steamy HCl fumes (no brown vapour), so brown fumes confirm bromide.

Key Takeaways

Conc. H2SO4 distinguishes solid halides by the reduction products: no colour change (Cl-), brown fumes (Br-), purple vapour/black solid (I-).

Common Mistakes

Naming AgNO3 (unsuitable — Y is insoluble, and AgNO3 gives colours not 'brown vapour'); writing 'brown solution' instead of brown vapour/gas.

Things to Be Careful About

The observation must be brown fumes/vapour released from the solid; both reagent and observation carry one mark each.

Techniques used

select the correct halide test reagent for a solidstate the confirmatory observation for bromide

(f)

Separate $1.0\text{ g}$ samples of three different magnesium salts are tested in order to identify the anion present in each sample.

(i)

Explain how the action of heat is used to identify which sample is:

$\text{MgCO}_3$
$\text{Mg(NO}_3)_2$
$\text{MgO}$ .

DifficultyMedium-Easy

Worked solution

Answer

Heat each 1.0 g sample strongly:
Mg(NO3)2 and MgCO3 both decompose, so both samples lose mass (final mass less than 1.0 g).
Mg(NO3)2 alone produces brown fumes of NO2 — this identifies the nitrate.
MgO shows no change / no reaction and no loss of mass — it is already fully oxidised, so this identifies the oxide.
The remaining sample that loses mass but gives no brown fumes is MgCO3 (it releases colourless CO2).

Final answer

Nitrate and carbonate lose mass; brown fumes identify the nitrate; MgO shows no change; remaining sample is MgCO3

Detailed explanation

Background Concept

Group 2 nitrates and carbonates thermally decompose while oxides are stable. Mg(NO3)2 -> MgO + 2NO2 + O2 (brown NO2 fumes); MgCO3 -> MgO + CO2 (colourless gas). MgO undergoes no reaction on heating.

Understanding the Question

You must explain how heating alone distinguishes the three 1.0 g samples, using observations available: mass change, colour of fumes, or nothing happening.

Approach

Use a decision tree: (1) does the sample lose mass? (2) are brown fumes produced? (3) nothing happens?

Step-by-Step Reasoning

Heating the nitrate gives MgO + 2NO2(g) + O2(g): mass falls below 1.0 g and dense brown NO2 fumes appear — uniquely identifying Mg(NO3)2. Heating the carbonate gives MgO + CO2(g): mass also falls, but no brown fumes — identifying MgCO3 by elimination. Heating MgO causes no change and no mass loss — identifying the oxide. The three mark points are: mass loss (nitrate and carbonate), brown fumes (nitrate only), no change (MgO).

Key Takeaways

Thermal stability increases down Group 2 for both nitrates and carbonates, but Mg salts at the top of the group decompose readily — a standard identification tool.

Common Mistakes

Saying the carbonate gives brown fumes (confusing CO2 with NO2); failing to state the mass-loss point that separates oxide from the decomposing salts; describing only one of the three samples.

Things to Be Careful About

All three distinguishing observations are needed for full marks: mass loss, brown fumes, no change.

Techniques used

compare mass loss on heatingrecall brown NO2 fumes from nitrate decomposition

(ii)

Complete the electron configuration of the magnesium cation present in these salts.

$1\text{s}^2$ .................................................................................................................................

DifficultyEasy

Worked solution

Answer

\text{Mg}^{2+}: 1\text{s}^2 2\text{s}^2 2\text{p}^6

Final answer

1s2 2s2 2p6

Detailed explanation

Background Concept

Magnesium has configuration 1s2 2s2 2p6 3s2 (12 electrons). In forming Mg2+, the two 3s electrons (the outermost shell) are removed first, leaving the 10-electron configuration of neon.

Understanding the Question

Complete the configuration of the Mg2+ cation starting from 1s2.

Approach

Remove two electrons from Mg's outer 3s subshell.

Step-by-Step Reasoning

Mg: 1s2 2s2 2p6 3s2; remove the two 3s electrons -> Mg2+: 1s2 2s2 2p6. This is the complete 10-electron noble-gas (neon) configuration.

Key Takeaways

Ions are formed by removing the highest-energy (outermost) electrons first; Mg2+ is isoelectronic with Ne.

Common Mistakes

Writing 3s2 (forgetting the electrons are lost), or writing the neutral Mg configuration.

Things to Be Careful About

The superscripts must sum to 10 for Mg2+ (2 + 2 + 6 = 10).

Techniques used

write the electron configuration of an ion

(g)

A sample of $\text{MgCO}_3(\text{s})$ is distinguished from a sample of $\text{Mg(OH)}_2(\text{s})$ by adding a small amount of each solid to $\text{HCl}(\text{aq})$ .

State one similarity and one difference in these two reactions.

similarity .....................................................................................................................................

difference ...................................................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

Similarity: both solids dissolve/react — the solid disappears (both react with HCl to form soluble MgCl2).

Difference: MgCO3 fizzes / effervesces (CO2 gas released), whereas Mg(OH)2 does not fizz.

Final answer

Similarity: solid disappears; Difference: MgCO3 fizzes, Mg(OH)2 does not

Detailed explanation

Background Concept

Both MgCO3 and Mg(OH)2 are basic compounds that react with hydrochloric acid to give soluble MgCl2 + H2O. However, the carbonate also produces CO2 gas: MgCO3 + 2HCl -> MgCl2 + H2O + CO2; Mg(OH)2 + 2HCl -> MgCl2 + 2H2O.

Understanding the Question

You need one observation common to both reactions and one observation that differs.

Approach

Common outcome: both ionic solids react and dissolve ( disappear) in the acid. Distinguishing outcome: gas evolution only for the carbonate.

Step-by-Step Reasoning

In both cases the reaction produces soluble magnesium chloride, so the solid disappears — the similarity. Only the carbonate releases CO2, seen as effervescence — the difference. Either 'MgCO3 fizzes' or 'Mg(OH)2 does not fizz' (ORA) satisfies the mark scheme.

Key Takeaways

Effervescence with dilute acid is the classic test for carbonate ions, and it distinguishes carbonates from other basic solids like hydroxides.

Common Mistakes

Giving 'both neutralise the acid' as the difference (that is a similarity); naming temperature change as the difference when the mark scheme wants the fizzing contrast.

Things to Be Careful About

Word the difference as a clear contrast: MgCO3 fizzes due to CO2, Mg(OH)2 does not.

Techniques used

compare reactions of carbonate and hydroxide with acid

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