(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 $\text{SCl}_2$ ($\text{Cl–S–Cl}$). $$…

(a) PCl5 + 4H2O -> H3PO4 + 5HCl; SiCl4 + 2H2O -> SiO2 + 4HCl · (b) +64 kJ mol^-1 · (c)(i) +2 · (c)(ii) Half of 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) White precipitate with AgNO3; orange solution turns green with K2Cr2O7.

Chemistry 9701/42 — 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 · Carboxylic Acids and Derivatives · Chemical Energetics · Electrochemistry · Transition Elements · Equilibria · +4 more

Q1Carboxylic Acids and DerivativesChemical EnergeticsElectrochemistryFree sample

(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

Non-metal chlorides are covalent compounds that hydrolyse when added to water. The chlorine atoms leave as hydrogen chloride, and the remaining element forms an oxide, hydroxide or oxoacid. Phosphorus(V) chloride contains phosphorus in the +5 oxidation state; with water it gives phosphoric(V) acid, $\text{H}_3\text{PO}_4$ . Silicon(IV) chloride gives silicon dioxide (or hydrated silica) and hydrogen chloride. These equations must be balanced.

Understanding the Question

The question asks for balanced equations for the hydrolysis of $\text{PCl}_5$ and $\text{SiCl}_4$ . No state symbols are required, but the atoms must balance.

Approach

Write the expected products, then balance by adjusting the coefficients of water and hydrogen chloride. For $\text{PCl}_5$ , the products are $\text{H}_3\text{PO}_4 + \text{HCl}$ ; for $\text{SiCl}_4$ , the products are $\text{SiO}_2 + \text{HCl}$ .

Step-by-Step Reasoning

For $\text{PCl}_5$ : start with $\text{PCl}_5 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + \text{HCl}$ . There are 5 chlorine atoms on the left, so 5 $\text{HCl}$ are needed on the right. The right side then has 3 H from $\text{H}_3\text{PO}_4$ plus 5 H from $\text{HCl}$ , giving 8 H in total, so 4 $\text{H}_2\text{O}$ are needed on the left. Oxygen: 4 O on the left and 4 O in $\text{H}_3\text{PO}_4$ . Balanced.

For $\text{SiCl}_4$ : start with $\text{SiCl}_4 + \text{H}_2\text{O} \rightarrow \text{SiO}_2 + \text{HCl}$ . Four $\text{HCl}$ are needed for the 4 chlorine atoms. Then hydrogen: 4 H on the right, so 2 $\text{H}_2\text{O}$ on the left. Oxygen: 2 O on the left and 2 O in $\text{SiO}_2$ . Balanced.

Key Takeaways

Hydrolysis of covalent chlorides is a balanced-atom exercise. The products are often an oxide or oxoacid plus hydrogen chloride.

Common Mistakes

Forgetting to balance the $\text{HCl}$ coefficient, using the wrong number of water molecules, or writing an unbalanced equation. Writing $\text{H}_2\text{SiO}_3$ instead of $\text{SiO}_2$ is also accepted, so it is not a mistake.

Things to Be Careful About

Check both hydrogen and oxygen balance. State symbols are not required here, but if included they should be correct.

Techniques used

balance hydrolysis equationsidentify products of hydrolysis of non-metal chlorides

(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: $8 \times \text{S-S} + 8 \times \text{Cl-Cl}$
Bonds formed: $16 \times \text{S-Cl}$

\Delta H = 8 \times 264 + 8 \times 244 - 16 \times 250 = +64\ \text{kJ mol}^{-1}

Answer

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

Final answer

+64 kJ mol^-1

Detailed explanation

Background Concept

Bond enthalpy (bond energy) is the energy needed to break one mole of a particular covalent bond in gaseous molecules. In a reaction, energy is absorbed to break reactant bonds and released when product bonds form. For gaseous reactants and products:

\Delta H \approx \sum (\text{bonds broken}) - \sum (\text{bonds formed})

The $\text{S}_8$ molecule is a ring of eight sulfur atoms joined by eight single S–S bonds. Each $\text{SCl}_2$ molecule has two S–Cl bonds.

Understanding the Question

Use the Data Booklet bond energies to calculate $\Delta H$ for:

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

You must count all bonds broken in the reactants and all bonds formed in the products.

Approach

Identify the bonds broken: 8 S–S bonds in $\text{S}_8$ and 8 Cl–Cl bonds in the 8 $\text{Cl}_2$ molecules. Identify the bonds formed: each $\text{SCl}_2$ has 2 S–Cl bonds, so 8 molecules give 16 S–Cl bonds. Then apply the bond-energy formula.

Step-by-Step Reasoning

Bonds broken:

8 \times 264 + 8 \times 244 = 2112 + 1952 = 4064\ \text{kJ}

Bonds formed:

16 \times 250 = 4000\ \text{kJ}

So:

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

The positive sign shows the reaction is endothermic: more energy is needed to break the reactant bonds than is released when the product bonds form.

Key Takeaways

Bond-energy calculations require careful bond counting. The enthalpy change is found from bonds broken minus bonds formed, and the sign matters.

Common Mistakes

Counting only 8 S–Cl bonds instead of 16; forgetting that the $\text{S}_8$ ring has 8 S–S bonds; using the wrong sign; or omitting the unit $\text{kJ mol}^{-1}$ .

Things to Be Careful About

Use the average bond energies from the Data Booklet. Keep the sign positive for an endothermic reaction. Always include the unit.

Techniques used

use average bond enthalpies to estimate enthalpy changecount bonds broken and bonds formedapply ΔH = bonds broken − bonds formed

(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

$+2$

Final answer

Detailed explanation

Background Concept

Oxidation number is a bookkeeping charge assigned to atoms in a compound. In covalent compounds, the more electronegative atom is assigned the negative oxidation number. Chlorine is more electronegative than sulfur, so in $\text{SCl}_2$ each Cl is $-1$ . The sum of oxidation numbers in a neutral molecule is zero.

Understanding the Question

Find the oxidation number of sulfur in $\text{SCl}_2$ .

Approach

Let $x$ be the oxidation number of sulfur. Since the molecule is neutral:

x + 2(-1) = 0

Step-by-Step Reasoning

x - 2 = 0 \Rightarrow x = +2

So sulfur is in the $+2$ oxidation state in $\text{SCl}_2$ .

Key Takeaways

Oxidation numbers are assigned using electronegativity and the rule that the sum of oxidation numbers in a neutral species is zero.

Common Mistakes

Assigning chlorine a positive oxidation number, or forgetting that the molecule is neutral.

Things to Be Careful About

Sulfur can have many oxidation states, including $+2$ , $+4$ and $+6$ . Here the correct value is $+2$ .

Techniques used

assign oxidation numbers using electronegativity rulesuse the neutrality of a molecule

(ii)

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

DifficultyMedium-Easy

Worked solution

Answer

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

Final answer

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

Detailed explanation

Background Concept

Disproportionation is a reaction in which the same element is simultaneously oxidised and reduced. This is recognised by following the oxidation number of that element into the different products.

Understanding the Question

The reaction of $\text{SCl}_2$ with water produces sulfur and $\text{SO}_2$ . You need to state how the oxidation number of sulfur changes when it forms these two products.

Approach

Find the oxidation number of sulfur in each product and compare it with the value $+2$ in $\text{SCl}_2$ .

Step-by-Step Reasoning

In $\text{SCl}_2$ , sulfur is $+2$ . In elemental sulfur, $\text{S}$ , the oxidation number is $0$ : this is a decrease of 2, so this sulfur is reduced. In $\text{SO}_2$ , oxygen is $-2$ each, so sulfur must be $+4$ : this is an increase of 2, so this sulfur is oxidised.

The balanced equation shows that two $\text{SCl}_2$ molecules give one $\text{S}$ and one $\text{SO}_2$ , so half of the sulfur atoms are reduced and half are oxidised.

Key Takeaways

A single element can be both oxidised and reduced in one reaction. Always compare oxidation numbers in reactants and products.

Common Mistakes

Only mentioning the oxidation to $+4$ and ignoring the reduction to $0$ ; confusing which product is oxidised and which is reduced.

Things to Be Careful About

Use the correct oxidation numbers: $\text{S}$ in $\text{SCl}_2$ is $+2$ , in $\text{S}$ it is $0$ , and in $\text{SO}_2$ it is $+4$ .

Techniques used

track oxidation number changesidentify disproportionation

(iii)

Suggest the identity of compound B.

DifficultyEasy

Worked solution

Answer

$\text{HCl}$

Final answer

HCl

Detailed explanation

Background Concept

When a chloride reacts with water, the chlorine atoms typically combine with hydrogen from water to form hydrogen chloride. The remaining atoms form the other products.

Understanding the Question

Solution A contains $\text{SO}_2(\text{aq})$ and compound B. You need to identify B from the atoms available in the reaction.

Approach

Write the balanced equation for the reaction of $\text{SCl}_2$ with water. The chlorine atoms must appear in a chlorine-containing product.

Step-by-Step Reasoning

The reaction is:

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

Chlorine appears only in $\text{HCl}$ , hydrogen comes from water, and oxygen appears in $\text{SO}_2$ . Therefore compound B is hydrogen chloride, $\text{HCl}$ .

Key Takeaways

Atom conservation is a powerful tool for identifying unknown products in a reaction.

Common Mistakes

Suggesting $\text{Cl}_2$ or $\text{HOCl}$ instead of $\text{HCl}$ .

Things to Be Careful About

In aqueous solution $\text{HCl}$ exists as $\text{H}^+$ and $\text{Cl}^-$ ions, but the molecular formula $\text{HCl}$ is the expected answer.

Techniques used

use atom conservation to deduce a productbalance hydrogen and chlorine atoms

(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

A redox equation can be balanced by writing separate oxidation and reduction half-equations and then combining them so that electrons cancel.

Understanding the Question

Construct a balanced equation for the reaction between $\text{SCl}_2$ and water, given that sulfur and $\text{SO}_2$ are products and $\text{HCl}$ is also formed.

Approach

Write the half-equations for the oxidation of $\text{SCl}_2$ to $\text{SO}_2$ and the reduction of $\text{SCl}_2$ to sulfur, then add them.

Step-by-Step Reasoning

Oxidation half-equation:

\text{SCl}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_2 + 4\text{H}^+ + 4\text{Cl}^- + 2\text{e}^-

Reduction half-equation:

\text{SCl}_2 + 2\text{e}^- \rightarrow \text{S} + 2\text{Cl}^-

Adding them cancels the electrons:

2\text{SCl}_2 + 2\text{H}_2\text{O} \rightarrow \text{S} + \text{SO}_2 + 4\text{H}^+ + 4\text{Cl}^-

Writing $\text{H}^+ + \text{Cl}^-$ as $\text{HCl}$ gives:

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

Check atoms: 2 S, 4 Cl, 4 H, 2 O on each side.

Key Takeaways

Half-equations must balance atoms and charge. Adding them with equal electron transfer gives the overall balanced equation.

Common Mistakes

Unbalanced chlorine or hydrogen; forgetting to cancel electrons; writing incorrect coefficients.

Things to Be Careful About

Ensure the final equation is balanced in all atoms. The coefficients 2, 2, 1, 1, 4 are required.

Techniques used

balance a redox equationcombine oxidation and reduction half-equations

(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

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

Final answer

White precipitate with AgNO3; orange solution turns green with K2Cr2O7.

Detailed explanation

Background Concept

Silver nitrate is used to test for halide ions. Silver chloride is a white precipitate. Acidified potassium dichromate is an oxidising agent; when it is reduced by a reducing agent such as $\text{SO}_2$ , the orange dichromate ion is converted to green $\text{Cr}^{3+}$ .

Understanding the Question

Solution A contains $\text{SO}_2(\text{aq})$ and $\text{HCl}$ . You need to predict the observations when $\text{AgNO}_3(\text{aq})$ and $\text{K}_2\text{Cr}_2\text{O}_7(\text{aq})$ are added to separate samples.

Approach

Identify the ions present: $\text{Cl}^-$ from $\text{HCl}$ and $\text{SO}_2$ as a reducing agent. Apply the known tests.

Step-by-Step Reasoning

With $\text{AgNO}_3$ :

\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}(\text{s})

Silver chloride is a white precipitate, so a white precipitate forms.

With $\text{K}_2\text{Cr}_2\text{O}_7$ : the orange dichromate ion, $\text{Cr}_2\text{O}_7^{2-}$ , is reduced by $\text{SO}_2$ to green $\text{Cr}^{3+}$ . The solution therefore turns green.

Key Takeaways

Qualitative tests link specific ions to characteristic observations: $\text{Cl}^-$ gives a white precipitate with $\text{AgNO}_3$ ; reducing agents turn acidified dichromate from orange to green.

Common Mistakes

Saying the silver chloride precipitate is cream or yellow (that would be $\text{AgBr}$ or $\text{AgI}$ ); saying the dichromate turns blue.

Things to Be Careful About

Dichromate is orange and $\text{Cr}^{3+}$ is green. The $\text{SO}_2$ is the reducing agent that causes the colour change.

Techniques used

identify chloride precipitate with silver nitrateidentify reducing agent with acidified dichromate

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