Section A Answer all the questions in the spaces provided. (a) Write a balanced equation for the reaction of each of the following chlorides with water. phosphorus(V) chloride silicon(IV) chloride (b) When sulfur is heated under pressure with chlorine, the major product is $\text{SCl}_2$ ($\text{Cl-S-Cl}$). $$ \text{S}_8(\text{g}) + 8\text{Cl}_2(\text{g}) \rightarrow 8\text{SCl}_2(\text{g}) $$ Use data from the Data Booklet to calculate the enthalpy change, $\Delta H$, for…

(a) PCl5 + 4H2O → H3PO4 + 5HCl; SiCl4 + 2H2O → SiO2 + 4HCl · (b) +64 kJ mol^-1 · (c)(i) +2 · (c)(ii) Half the sulfur is oxidised from +2 to +4; half is reduced from +2 to 0 · (c)(iii) HCl · (c)(iv) 2SCl2 + 2H2O → S + SO2 + 4HCl · (c)(v) AgNO3: white precipitate; K2Cr2O7: solution turns green

Chemistry 9701/41 — October/November 2010

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

questions

100

marks

105

minutes

Topics Introduction to A Level Organic Chemistry · Equilibria · Electrochemistry · Transition Elements · Hydroxy Compounds · Carboxylic Acids and Derivatives · +6 more

Q1Group 2Chemical EnergeticsElectrochemistryFree sample

Section A

Answer all the questions in the spaces provided.

(a)

Write a balanced equation for the reaction of each of the following chlorides with water.

phosphorus(V) chloride

silicon(IV) chloride

DifficultyMedium-Easy

Worked solution

Answer

$\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}$

$\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}$

Final answer

PCl5 + 4H2O → H3PO4 + 5HCl; SiCl4 + 2H2O → SiO2 + 4HCl

Detailed explanation

Background Concept

Covalent chlorides of non-metals (such as $\text{PCl}_5$ and $\text{SiCl}_4$ ) undergo hydrolysis when added to water. The chlorine atoms are replaced by oxygen or hydroxide groups from water, forming the corresponding oxoacid (or oxide) and hydrogen chloride. The reaction is essentially a nucleophilic substitution where water attacks the electron-deficient central atom.

For $\text{PCl}_5$ , phosphorus is in oxidation state +5 and forms phosphoric acid, $\text{H}_3\text{PO}_4$ , with the chloride ions combining with hydrogen from water to form $\text{HCl}$ .

For $\text{SiCl}_4$ , silicon is in oxidation state +4 and forms silicon dioxide, $\text{SiO}_2$ (or hydrated forms such as $\text{H}_2\text{SiO}_3$ or $\text{Si(OH)}_4$ ), with $\text{HCl}$ also formed.

Understanding the Question

The question asks for balanced equations for the hydrolysis of two covalent chlorides: phosphorus(V) chloride and silicon(IV) chloride. The Roman numerals indicate the oxidation state of the central atom. The key is to identify the correct products: phosphoric acid for $\text{PCl}_5$ and silicon dioxide (or silicic acid) for $\text{SiCl}_4$ , with $\text{HCl}$ as the common by-product.

Approach

For each chloride, identify the oxidation state of the central atom. The hydrolysis products are determined by the oxidation state:

P(V) forms $\text{H}_3\text{PO}_4$ (phosphoric acid, P in +5)
Si(IV) forms $\text{SiO}_2$ (silicon dioxide, Si in +4)

Then balance the equation: balance P or Si, then Cl (as HCl), then H and O.

Step-by-Step Reasoning

For $\text{PCl}_5$ :

P is in +5 oxidation state. The oxoacid of P(V) is $\text{H}_3\text{PO}_4$ (phosphoric acid).
Each Cl becomes HCl. 5 Cl atoms $\rightarrow$ 5 HCl.
Balance the equation: $\text{PCl}_5 + x\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}$
Count H on the right: 3 (from $\text{H}_3\text{PO}_4$ ) + 5 (from HCl) = 8 H atoms.
Each $\text{H}_2\text{O}$ provides 2 H atoms, so $x = 4$ .
Check O: 4 $\text{H}_2\text{O}$ gives 4 O atoms; $\text{H}_3\text{PO}_4$ needs 4 O atoms. ✓
Balanced: $\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}$

For $\text{SiCl}_4$ :

Si is in +4 oxidation state. The oxide of Si(IV) is $\text{SiO}_2$ .
Each Cl becomes HCl. 4 Cl atoms $\rightarrow$ 4 HCl.
Balance: $\text{SiCl}_4 + x\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}$
Count H on the right: 4 (from HCl). Each $\text{H}_2\text{O}$ gives 2 H, so $x = 2$ .
Check O: 2 $\text{H}_2\text{O}$ gives 2 O atoms; $\text{SiO}_2$ needs 2 O atoms. ✓
Balanced: $\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}$

Key Takeaways

Covalent chlorides hydrolyse to give the oxoacid/oxide of the central element and HCl.
The oxidation state of the central atom determines the product.
Balancing requires careful counting of H and O atoms.

Common Mistakes

Writing $\text{PCl}_3$ hydrolysis products instead of $\text{PCl}_5$ (forgetting the oxidation state).
Forgetting to balance the HCl coefficient.
Writing $\text{SiO}_2 \cdot x\text{H}_2\text{O}$ instead of just $\text{SiO}_2$ (the mark scheme allows hydrated forms too).

Things to Be Careful About

The mark scheme allows $\text{H}_2\text{SiO}_3$ or $\text{Si(OH)}_4$ as alternatives to $\text{SiO}_2$ for $\text{SiCl}_4$ hydrolysis.
State symbols are not required in this question, but including them shows understanding.

Techniques used

balance hydrolysis equationswrite products of covalent chloride hydrolysis

(b)

When sulfur is heated under pressure with chlorine, the major product is $\text{SCl}_2$ ( $\text{Cl-S-Cl}$ ).

\text{S}_8(\text{g}) + 8\text{Cl}_2(\text{g}) \rightarrow 8\text{SCl}_2(\text{g})

Use data from the Data Booklet to calculate the enthalpy change, $\Delta H$ , for this reaction. The eight sulfur atoms in the $\text{S}_8$ molecule are all joined in a single ring by single bonds.

\Delta H = \text{.......................................... kJ mol}^{-1}

DifficultyMedium

Worked solution

Working

Bonds broken:

8 S−S bonds: $8 \times 264 = 2112\ \text{kJ mol}^{-1}$
8 Cl−Cl bonds: $8 \times 244 = 1952\ \text{kJ mol}^{-1}$
Total energy input: $2112 + 1952 = 4064\ \text{kJ mol}^{-1}$

Bonds formed:

16 S−Cl bonds: $16 \times 250 = 4000\ \text{kJ mol}^{-1}$

$\Delta H = 4064 - 4000 = +64\ \text{kJ mol}^{-1}$

Answer

$\Delta H = +64\ \text{kJ mol}^{-1}$

Final answer

+64 kJ mol^-1

Detailed explanation

Background Concept

The enthalpy change of a reaction can be estimated using average bond energies (bond enthalpies). The enthalpy change equals the energy required to break bonds in the reactants minus the energy released when bonds form in the products:

$\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds formed})$

Breaking bonds requires energy (endothermic, positive contribution); forming bonds releases energy (exothermic, negative contribution).

Understanding the Question

The reaction is $\text{S}_8(\text{g}) + 8\text{Cl}_2(\text{g}) \rightarrow 8\text{SCl}_2(\text{g})$ . The $\text{S}_8$ molecule is a ring of 8 sulfur atoms joined by single S−S bonds. The Data Booklet provides bond energies: S−S = 264 kJ mol⁻¹, Cl−Cl = 244 kJ mol⁻¹, S−Cl = 250 kJ mol⁻¹. We must calculate $\Delta H$ for this reaction.

Approach

Identify all bonds broken in reactants: 8 S−S bonds (in the $\text{S}_8$ ring) and 8 Cl−Cl bonds (in 8 $\text{Cl}_2$ molecules).
Identify all bonds formed in products: each $\text{SCl}_2$ has 2 S−Cl bonds, so 8 $\text{SCl}_2$ molecules have 16 S−Cl bonds.
Apply $\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds formed})$ .

Step-by-Step Reasoning

Bonds broken:

The $\text{S}_8$ ring contains 8 S−S single bonds (one between each adjacent pair of S atoms in the ring). Energy: $8 \times 264 = 2112\ \text{kJ mol}^{-1}$ .
8 $\text{Cl}_2$ molecules each contain one Cl−Cl bond. Energy: $8 \times 244 = 1952\ \text{kJ mol}^{-1}$ .
Total energy input: $2112 + 1952 = 4064\ \text{kJ mol}^{-1}$ .

Bonds formed:

Each $\text{SCl}_2$ molecule has 2 S−Cl bonds (Cl−S−Cl). With 8 $\text{SCl}_2$ molecules: $8 \times 2 = 16$ S−Cl bonds.
Energy released: $16 \times 250 = 4000\ \text{kJ mol}^{-1}$ .

Enthalpy change:
$\Delta H = 4064 - 4000 = +64\ \text{kJ mol}^{-1}$

The positive sign indicates the reaction is endothermic — more energy is absorbed breaking bonds than is released forming new bonds.

Key Takeaways

Bond energy calculation: $\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds formed})$ .
Count bonds carefully, especially in ring structures like $\text{S}_8$ — each S atom contributes one S−S bond, so 8 S atoms give 8 S−S bonds.
Each $\text{SCl}_2$ molecule has 2 S−Cl bonds.

Common Mistakes

Forgetting that $\text{S}_8$ has 8 S−S bonds (one per S atom in the ring), not 7 or 9.
Counting 8 S−Cl bonds instead of 16 (each $\text{SCl}_2$ has 2 S−Cl bonds).
Reversing the formula (using bonds formed − bonds broken), which gives the wrong sign.

Things to Be Careful About

The $\text{S}_8$ ring is a closed loop of 8 atoms, so there are 8 S−S bonds, not 7 (a chain of 8 atoms would have 7 bonds).
The units are kJ mol⁻¹ of reaction as written.
The positive sign is important — the reaction is endothermic.

Techniques used

calculate enthalpy change from bond energiescount bonds broken and formed in a ring structure

(c)

Under suitable conditions, $\text{SCl}_2$ reacts with water to produce a yellow precipitate of sulfur and a solution A. Solution A contains a mixture of $\text{SO}_2(\text{aq})$ and compound B.

(i)

What is the oxidation number of sulfur in $\text{SCl}_2$ ?

DifficultyEasy

Worked solution

Answer

Final answer

Detailed explanation

Background Concept

The oxidation number (oxidation state) is a formal charge assigned to an atom in a compound based on a set of rules. For a neutral molecule, the sum of all oxidation numbers is zero. For covalent compounds, the more electronegative element is assigned the negative oxidation number.

In $\text{SCl}_2$ , chlorine is more electronegative than sulfur. Each Cl atom has oxidation number −1. Since the molecule is neutral, the sum must be zero.

Understanding the Question

We need to find the oxidation number of sulfur in $\text{SCl}_2$ . This is a straightforward application of the oxidation number rules.

Approach

Use the rule that the sum of oxidation numbers in a neutral molecule is zero. Assign Cl = −1 and solve for S.

Step-by-Step Reasoning

Let the oxidation number of S be $x$ .

$x + 2(-1) = 0$
$x - 2 = 0$
$x = +2$

The oxidation number of sulfur in $\text{SCl}_2$ is +2.

Key Takeaways

In a neutral compound, oxidation numbers sum to zero.
The more electronegative element gets the negative oxidation number.
Cl is almost always −1 in compounds (except in $\text{Cl}_2$ where it is 0, or with more electronegative elements like O or F).

Common Mistakes

Assigning Cl a positive oxidation number (Cl is more electronegative than S, so it must be negative).
Forgetting that there are 2 Cl atoms in $\text{SCl}_2$ .

Things to Be Careful About

The oxidation number is written with the sign before the number: +2, not 2+.
The oxidation number is a formal concept, not a real charge on the atom.

Techniques used

determine oxidation number from electronegativity rules

(ii)

Work out how the oxidation number of sulfur changes during the reaction of $\text{SCl}_2$ with water.

DifficultyMedium-Easy

Worked solution

Answer

Half of the sulfur is oxidised from +2 to +4 (in $\text{SO}_2$ ), an increase of +2; the other half is reduced from +2 to 0 (elemental S), a decrease of −2.

Final answer

Half the sulfur is oxidised from +2 to +4; half is reduced from +2 to 0

Detailed explanation

Background Concept

Disproportionation is a redox reaction in which the same element in a single oxidation state is simultaneously oxidised and reduced. In this case, $\text{SCl}_2$ (S in +2) reacts with water to produce elemental sulfur (S in 0) and $\text{SO}_2$ (S in +4).

Understanding the Question

We need to track how the oxidation number of sulfur changes during the reaction of $\text{SCl}_2$ with water. The products are S (elemental, oxidation state 0) and $\text{SO}_2$ (S in +4).

Approach

Compare the oxidation number of S in the reactant ( $\text{SCl}_2$ , +2) with the oxidation numbers in the products (S, 0; and $\text{SO}_2$ , +4).

Step-by-Step Reasoning

In $\text{SCl}_2$ , S is +2.

In elemental S, S is 0. Change: +2 → 0, a decrease of 2 (reduction).

In $\text{SO}_2$ , O is −2 (each). For the molecule to be neutral: $x + 2(-2) = 0$ , so $x = +4$ . Change: +2 → +4, an increase of 2 (oxidation).

So half the sulfur is oxidised (by +2) and half is reduced (by −2). This is a disproportionation reaction.

The stoichiometry of the overall equation ( $2\text{SCl}_2 \rightarrow \text{S} + \text{SO}_2$ ) reflects the 1:1 ratio of oxidised to reduced sulfur.

Key Takeaways

Disproportionation: the same element is simultaneously oxidised and reduced.
The oxidation number changes are equal and opposite (+2 and −2).
The stoichiometry of the equation reflects the 1:1 ratio of oxidised to reduced sulfur.

Common Mistakes

Saying the sulfur is only oxidised or only reduced, without recognising the disproportionation.
Calculating the wrong oxidation number for S in $\text{SO}_2$ (forgetting O is −2 each).

Things to Be Careful About

The oxidation number of S in $\text{SO}_2$ is +4 (not +6, which would be $\text{SO}_3$ ).
The changes are +2 and −2, not +4 and −4.
The mark scheme gives 1 mark for the increase and 1 mark for the decrease — both must be stated.

Techniques used

track oxidation number changes in disproportionation

(iii)

Suggest the identity of compound B.

DifficultyMedium-Easy

Worked solution

Answer

$\text{HCl}$

Final answer

HCl

Detailed explanation

Background Concept

When $\text{SCl}_2$ reacts with water, the chlorine atoms are released as chloride ions. These combine with H⁺ from water to form HCl. The solution A contains $\text{SO}_2(\text{aq})$ and HCl(aq).

Understanding the Question

We need to identify compound B, which is present in solution A along with $\text{SO}_2(\text{aq})$ . The question asks us to "suggest" the identity, meaning we should deduce it from the chemistry of the reaction.

Approach

Consider what happens to the Cl atoms in $\text{SCl}_2$ when it reacts with water. The Cl atoms become Cl⁻ ions, which form HCl with H⁺ from water.

Step-by-Step Reasoning

In $\text{SCl}_2 + \text{H}_2\text{O}$ , the S−Cl bonds break. The Cl atoms acquire electrons to become Cl⁻ ions. Water provides H⁺, so HCl is formed. The HCl remains dissolved in the aqueous solution as HCl(aq) (hydrochloric acid).

Compound B is therefore HCl.

This is confirmed by the observation in part (v): $\text{AgNO}_3(\text{aq})$ gives a white precipitate (AgCl), confirming the presence of Cl⁻ ions.

Key Takeaways

Hydrolysis of chlorides produces HCl.
HCl(aq) provides Cl⁻ ions that can be detected with $\text{AgNO}_3$ .

Common Mistakes

Suggesting $\text{H}_2\text{S}$ or another sulfur compound for B (B is the chloride product, not a sulfur compound).
Not recognising that HCl is the aqueous product.

Things to Be Careful About

The mark scheme notes that B can be "read into (iv)" — i.e., if the equation in (iv) shows HCl, then B is HCl.
The identity of B is confirmed by the $\text{AgNO}_3$ test in part (v).

Techniques used

deduce compound identity from reaction chemistry

(iv)

Construct an equation for the reaction between $\text{SCl}_2$ and water.

DifficultyMedium-Easy

Worked solution

Answer

$2\text{SCl}_2 + 2\text{H}_2\text{O} \rightarrow \text{S} + \text{SO}_2 + 4\text{HCl}$

Final answer

2SCl2 + 2H2O → S + SO2 + 4HCl

Detailed explanation

Background Concept

The reaction of $\text{SCl}_2$ with water is a disproportionation reaction. $\text{SCl}_2$ (S in +2) is simultaneously oxidised to $\text{SO}_2$ (S in +4) and reduced to elemental S (S in 0). The Cl atoms become HCl.

Understanding the Question

We need to construct a balanced equation for the reaction between $\text{SCl}_2$ and water, given that the products are S (yellow precipitate), $\text{SO}_2(\text{aq})$ , and compound B (HCl).

Approach

Write the unbalanced equation: $\text{SCl}_2 + \text{H}_2\text{O} \rightarrow \text{S} + \text{SO}_2 + \text{HCl}$
Balance the sulfur atoms: 2 $\text{SCl}_2$ gives 1 S + 1 $\text{SO}_2$ (2 S atoms total).
Balance the Cl atoms: 2 $\text{SCl}_2$ gives 4 Cl, so 4 HCl.
Balance the H atoms: 4 HCl needs 4 H, so 2 $\text{H}_2\text{O}$ .
Balance the O atoms: 2 $\text{H}_2\text{O}$ gives 2 O; $\text{SO}_2$ needs 2 O. ✓

Step-by-Step Reasoning

Unbalanced: $\text{SCl}_2 + \text{H}_2\text{O} \rightarrow \text{S} + \text{SO}_2 + \text{HCl}$

Balance S: $2\text{SCl}_2 \rightarrow \text{S} + \text{SO}_2 + \text{HCl}$ (2 S atoms on each side)

Balance Cl: $2\text{SCl}_2 \rightarrow \text{S} + \text{SO}_2 + 4\text{HCl}$ (4 Cl atoms on each side)

Balance H: $2\text{SCl}_2 + 2\text{H}_2\text{O} \rightarrow \text{S} + \text{SO}_2 + 4\text{HCl}$ (4 H atoms on each side)

Balance O: $2\text{SCl}_2 + 2\text{H}_2\text{O} \rightarrow \text{S} + \text{SO}_2 + 4\text{HCl}$ (2 O atoms on each side)

Final balanced equation:
$2\text{SCl}_2 + 2\text{H}_2\text{O} \rightarrow \text{S} + \text{SO}_2 + 4\text{HCl}$

Check:

S: 2 on left, 1 + 1 = 2 on right ✓
Cl: 4 on left, 4 on right ✓
H: 4 on left (2×2), 4 on right (4×1) ✓
O: 2 on left (2×1), 2 on right ( $\text{SO}_2$ ) ✓

Key Takeaways

Disproportionation equations are balanced by treating the two products of the same element separately.
The stoichiometric coefficient 2 on $\text{SCl}_2$ reflects the 1:1 ratio of oxidised (to $\text{SO}_2$ ) and reduced (to S) sulfur.

Common Mistakes

Writing $\text{SCl}_2 + \text{H}_2\text{O} \rightarrow \text{S} + \text{SO}_2 + 2\text{HCl}$ (unbalanced — wrong coefficients).
Forgetting to balance the equation fully.

Things to Be Careful About

The equation must be balanced — this is a mark in the mark scheme.
State symbols are not required but show understanding.
The coefficient 2 on $\text{SCl}_2$ is essential — without it, the sulfur atoms don't balance.

Techniques used

balance a disproportionation redox equation

(v)

What would you observe when each of the following reagents is added to separate samples of solution A?

$\text{AgNO}_3(\text{aq})$

$\text{K}_2\text{Cr}_2\text{O}_7(\text{aq})$

DifficultyMedium-Easy

Worked solution

Answer

$\text{AgNO}_3(\text{aq})$ : white precipitate forms
$\text{K}_2\text{Cr}_2\text{O}_7(\text{aq})$ : the orange solution turns green

Final answer

AgNO3: white precipitate; K2Cr2O7: solution turns green

Detailed explanation

Background Concept

Solution A contains $\text{SO}_2(\text{aq})$ and HCl(aq).

$\text{AgNO}_3(\text{aq})$ test: Ag⁺ ions react with Cl⁻ ions to form a white precipitate of AgCl. This is the standard test for chloride ions.
$\text{K}_2\text{Cr}_2\text{O}_7(\text{aq})$ test: Potassium dichromate(VI) is an orange oxidising agent. In acidic conditions, it oxidises $\text{SO}_2$ (a reducing agent) to $\text{SO}_4^{2-}$ (sulfate), while the dichromate is reduced to Cr³⁺, which is green. The colour change from orange to green confirms the presence of a reducing agent such as $\text{SO}_2$ .

Understanding the Question

We need to state what is observed when $\text{AgNO}_3(\text{aq})$ and $\text{K}_2\text{Cr}_2\text{O}_7(\text{aq})$ are added to separate samples of solution A, which contains $\text{SO}_2(\text{aq})$ and HCl(aq).

Approach

Identify what each reagent tests for:

$\text{AgNO}_3$ tests for Cl⁻ ions (white precipitate of AgCl).
$\text{K}_2\text{Cr}_2\text{O}_7$ tests for reducing agents (orange → green colour change).

Step-by-Step Reasoning

$\text{AgNO}_3(\text{aq})$ : $\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})$ . AgCl is a white precipitate. Since solution A contains HCl(aq) (i.e., Cl⁻ ions), a white precipitate forms.
$\text{K}_2\text{Cr}_2\text{O}_7(\text{aq})$ : The orange dichromate ion, $\text{Cr}_2\text{O}_7^{2-}$ , is reduced by $\text{SO}_2$ to green Cr³⁺. The colour change from orange to green confirms $\text{SO}_2$ is present as a reducing agent.

Key Takeaways

$\text{AgNO}_3$ is the test for Cl⁻: white precipitate of AgCl.
$\text{K}_2\text{Cr}_2\text{O}_7$ is an oxidising agent that turns from orange to green when reduced by a reducing agent like $\text{SO}_2$ .

Common Mistakes

Forgetting the colour of the AgCl precipitate (white).
Saying the dichromate turns "green" without mentioning it was orange first.
Confusing the dichromate test with the iodide test (which gives brown iodine).

Things to Be Careful About

The mark scheme gives 1 mark for the white precipitate with $\text{AgNO}_3$ and 1 mark for the green colour with $\text{K}_2\text{Cr}_2\text{O}_7$ .
The green colour of Cr³⁺ is the key observation for the dichromate test.
The $\text{K}_2\text{Cr}_2\text{O}_7$ test requires acidic conditions in practice, though this is not explicitly stated in the question.

Techniques used

describe qualitative test observationsidentify chloride ions with AgNO3identify reducing agents with K2Cr2O7

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