Chlorine, $\text{Cl}_2$, reacts with many elements and compounds to form chlorides. Table 2.1 shows information about some chlorides of Period 3 elements. (a) Complete Table 2.1. Table 2.1 Na Mg Si --- --- --- --- formula of chloride structure of chloride giant bonding of chloride covalent pH of solution formed on addition of chloride to water 6.2 (b) When $\text{Cl}_2$ reacts with cold $\text{NaOH}(\text{aq})$, $\text{Cl}_2$ is both oxidised and reduced. The products are…

(a) See completed table above. · (b)(i) Disproportionation · (b)(ii) NaClO · (b)(iii) 3Cl2 + 6NaOH -> NaClO3 + 5NaCl + 3H2O · (b)(iv) White precipitate; dissolves in dilute ammonia. · (c)(i) PCl5: trigonal bipyramidal; PCl6-: octahedral · (c)(ii) J: H2O; type: hydrolysis · (d)(i) See mechanism diagram below. · (d)(ii) NaOH in ethanol, heat · (d)(iii) Repeat unit: -[CH2-C(Cl)(CH3)]- with dashed bonds at ends.

9701/22 Feb/Mar 2024 — Chemistry Solved Past Paper

Q1MediumChemical BondingAtoms, Molecules and StoichiometryStates of MatterElectrochemistryHydrocarbonsCarbonyl Compounds

Bismuth is an element in Group 15 of the Periodic Table.

(a)

Bismuth has metallic bonding.

Draw a labelled diagram to show the metallic bonding in bismuth.

1M Similar

DifficultyEasy

Worked solution

Answer

A regular three-dimensional lattice of positive ions (cations) surrounded by a sea of delocalised electrons.

Final answer

Regular lattice of positive ions surrounded by delocalised electrons (see diagram)

Detailed explanation

Background Concept

Metallic bonding is the electrostatic attraction between a regular, three-dimensional lattice of positive metal ions (cations) and a 'sea' of delocalised electrons. The delocalised electrons are valence electrons that have been removed from their parent atoms and are free to move throughout the entire metallic structure. This model explains the typical properties of metals, such as high electrical and thermal conductivity, malleability, and ductility.

Understanding the Question

The question asks for a labelled diagram showing the metallic bonding in bismuth, a Group 15 element that exhibits metallic bonding. You need to represent the two key components: the positive ions and the delocalised electrons, and clearly label them.

Approach

Draw a regular, repeating arrangement of circles with a '+' sign inside to represent the positive metal ions. Scatter smaller circles with a '-' sign (or simply '-' signs) between the ions to represent the delocalised electrons. Add clear labels for 'positive ions' and 'delocalised electrons'.

Step-by-Step Reasoning

Positive ions: Metal atoms lose their valence electrons to form cations. In a solid metal, these cations are arranged in a regular, repeating 3D lattice. Draw rows or a grid of circles containing a '+' sign.
Delocalised electrons: The valence electrons are not bound to any specific ion; they are free to move throughout the structure. Represent these as smaller circles with a '-' sign, or just '-' signs, distributed in the spaces between the positive ions.
Labels: Clearly label the '+' circles as 'positive ions' (or 'metal cations') and the '-' symbols as 'delocalised electrons'.

Key Takeaways

Metallic bonding is always modelled as a regular lattice of positive ions in a sea of delocalised electrons. Diagrams must show both components clearly and be labelled.

Common Mistakes

Drawing covalent bonds between the metal ions (metals do not form directional covalent bonds in their bulk structure).
Forgetting to label the diagram.
Showing electrons as belonging to specific atoms rather than being delocalised throughout the lattice.

Things to Be Careful About

Ensure the arrangement of ions looks regular and repeating (like a grid or close-packed layers). The electrons must be shown between the ions, not attached to them. Labels are essential to score the mark.

Techniques used

draw a labelled diagram of metallic bondingidentify delocalised electrons

(b)

Bismuth reduces water to form bismuth oxide, $\text{Bi}_2\text{O}_3$ . A colourless gas that ignites with a squeaky pop also forms.

3M

(i)

Construct an equation for the reduction of water by bismuth.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

2\text{Bi}(\text{s}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{Bi}_2\text{O}_3(\text{s}) + 3\text{H}_2(\text{g})

Final answer

2Bi + 3H2O -> Bi2O3 + 3H2

Detailed explanation

Background Concept

When a metal reduces water, it typically forms a metal oxide (or hydroxide) and hydrogen gas. The metal is oxidised, and the hydrogen in water is reduced from an oxidation state of +1 to 0 in H2. Balancing such equations requires ensuring that the number of atoms of each element and the total charge are equal on both sides.

Understanding the Question

The question states that bismuth reduces water to form bismuth oxide (Bi2O3) and a colourless gas that ignites with a squeaky pop. The 'squeaky pop' test is the standard identification for hydrogen gas (H2). You need to write a balanced symbol equation for this reaction.

Approach

Write the unbalanced equation with the known reactants (Bi, H2O) and products (Bi2O3, H2).
Balance the bismuth atoms.
Balance the oxygen atoms by adjusting the coefficient of H2O.
Balance the hydrogen atoms by adjusting the coefficient of H2.
Add state symbols where appropriate (though the mark scheme does not strictly require them for this specific mark, it is good practice).

Step-by-Step Reasoning

Reactants and products: Bi + H2O -> Bi2O3 + H2
Balance Bi: There are 2 Bi atoms on the right, so put a 2 in front of Bi on the left: 2Bi + H2O -> Bi2O3 + H2
Balance O: There are 3 O atoms on the right (in Bi2O3), so put a 3 in front of H2O: 2Bi + 3H2O -> Bi2O3 + H2
Balance H: There are 6 H atoms on the left (3 x 2), so put a 3 in front of H2 on the right: 2Bi + 3H2O -> Bi2O3 + 3H2
Final check: 2 Bi, 3 O, 6 H on both sides. The equation is balanced.

Key Takeaways

Always identify the products from the description (e.g., 'squeaky pop' = H2) before balancing. Count atoms of each element systematically: balance metals first, then non-metals like oxygen, and finally hydrogen.

Common Mistakes

Forgetting to balance the equation (e.g., Bi + H2O -> Bi2O3 + H2).
Writing the wrong product for hydrogen (e.g., H instead of H2).
Incorrectly balancing oxygen and hydrogen simultaneously.

Things to Be Careful About

Ensure the subscripts in the formulae (Bi2O3, H2O, H2) are correct and not mixed up with coefficients. The mark scheme accepts the equation without state symbols, but including them (s, l, g) is good practice and shows chemical understanding.

Techniques used

balance a redox equationidentify products of metal-water reaction

(ii)

$\text{Bi}_2\text{O}_3$ is a yellow insoluble solid that melts at $1090\text{ K}$ . The molten compound conducts electricity.

Deduce the structure and bonding of $\text{Bi}_2\text{O}_3$ . Explain your answer.

2M Similar

DifficultyMedium

Worked solution

Answer

Structure and bonding: Giant ionic structure (or giant ionic lattice).
Explanation: The high melting point (1090 K) indicates strong electrostatic forces of attraction between oppositely charged ions in a giant lattice. The fact that it conducts electricity only when molten indicates the presence of mobile ions that are free to move and carry charge in the liquid state, which is characteristic of ionic compounds.

Final answer

Giant ionic structure; high melting point indicates strong ionic bonds; conducts when molten due to mobile ions

Detailed explanation

Background Concept

The physical properties of a substance are direct consequences of its structure and bonding.

Giant ionic lattices consist of a regular arrangement of alternating positive and negative ions held together by strong electrostatic forces. Breaking these forces requires a lot of energy, leading to high melting and boiling points.
Electrical conductivity in ionic compounds occurs only when the ions are free to move. In a solid lattice, ions are fixed in place, so they do not conduct. When molten (or dissolved in water), the lattice breaks down, and the ions become mobile, allowing them to carry electrical charge.

Understanding the Question

You are given two key properties of Bi2O3: it has a high melting point (1090 K) and it conducts electricity when molten. You must deduce the structure and bonding and explain how these properties support your deduction.

Approach

Use the high melting point to argue for a giant structure with strong bonds (giant ionic or giant covalent).
Use the electrical conductivity when molten to distinguish between ionic and giant covalent (giant covalent like diamond does not conduct when molten, ionic does).
Conclude it is a giant ionic structure and explicitly link each property to the bonding model.

Step-by-Step Reasoning

Deduce structure/bonding: The combination of a high melting point and conductivity when molten is the classic signature of a giant ionic structure.
Explain melting point: A melting point of 1090 K is very high. This means a large amount of energy is required to overcome the forces holding the structure together. In an ionic lattice, these are strong electrostatic forces of attraction between the positive and negative ions throughout the giant lattice.
Explain conductivity: Solid ionic compounds do not conduct electricity because the ions are fixed in the lattice. When melted, the lattice breaks down, and the ions become mobile (free to move). These mobile ions can carry charge, allowing the molten compound to conduct electricity. This rules out giant covalent (which doesn't conduct) and simple molecular (which has low melting points).

Key Takeaways

Always link the physical property directly to the structural feature: high melting point -> strong bonds in a giant lattice; molten conductivity -> mobile ions -> ionic bonding.

Common Mistakes

Saying 'it conducts because it has electrons' (ionic compounds conduct via ions, not delocalised electrons).
Stating 'strong bonds' without specifying electrostatic forces between ions.
Forgetting to mention that the lattice is giant.

Things to Be Careful About

The mark scheme specifically looks for 'giant' and 'ionic' as the structure/bonding deduction, and links the melting point to strong forces/lattice, and conductivity to molten/mobile ions. Ensure both halves of the explanation are present.

Techniques used

deduce structure and bonding from physical propertiesrelate melting point to lattice energyrelate electrical conductivity to mobile ions

(c)

$\text{Bi}_2\text{O}_3$ can be used to form $\text{NaBiO}_3$ , as shown in equation 1.

\text{equation 1} \quad \text{Na}_2\text{O} + \text{Bi}_2\text{O}_3 + \text{O}_2 \rightarrow 2\text{NaBiO}_3

2M

(i)

Deduce the oxidation number of Bi in $\text{Bi}_2\text{O}_3$ and in $\text{NaBiO}_3$ .

oxidation number of Bi:

in $\text{Bi}_2\text{O}_3$ ................................................. in $\text{NaBiO}_3$ ...........................................................

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

in $\text{Bi}_2\text{O}_3$ : +3
in $\text{NaBiO}_3$ : +5

Final answer

+3 in Bi2O3; +5 in NaBiO3

Detailed explanation

Background Concept

Oxidation number (or oxidation state) is the charge an atom would have if all bonds to atoms of different elements were 100% ionic. Rules for assigning oxidation numbers include: oxygen is usually -2 (except in peroxides), alkali metals (Group 1) are +1, and the sum of oxidation numbers in a neutral compound is zero.

Understanding the Question

You need to calculate the oxidation number of bismuth (Bi) in two different compounds: Bi2O3 and NaBiO3.

Approach

Apply the standard oxidation number rules to each compound, setting the sum of all oxidation numbers to zero.

Step-by-Step Reasoning

In Bi2O3:
- Oxygen has an oxidation number of -2.
- Let the oxidation number of Bi be $x$ .
- Sum: $2x + 3(-2) = 0 \Rightarrow 2x - 6 = 0 \Rightarrow 2x = 6 \Rightarrow x = +3$ .
- Oxidation number of Bi is +3.
In NaBiO3:
- Sodium (Group 1) has an oxidation number of +1.
- Oxygen has an oxidation number of -2.
- Let the oxidation number of Bi be $y$ .
- Sum: $1 + y + 3(-2) = 0 \Rightarrow 1 + y - 6 = 0 \Rightarrow y - 5 = 0 \Rightarrow y = +5$ .
- Oxidation number of Bi is +5.

Key Takeaways

Always use the known oxidation numbers of oxygen (-2) and Group 1 metals (+1) as anchors to solve for the unknown element. The sum must equal the overall charge of the species (0 for neutral compounds).

Common Mistakes

Forgetting to multiply the oxidation number of oxygen by its subscript (e.g., using -2 instead of -6 for O3).
Missing the contribution of sodium in NaBiO3 (treating it as if only Bi and O are present).

Things to Be Careful About

Ensure algebraic signs are correct. $2x - 6 = 0$ means $2x = +6$ , so $x = +3$ , not -3.

Techniques used

calculate oxidation numbers from formulae

(ii)

Identify the reducing agent in equation 1.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Bi2O3

Final answer

Bi2O3

Detailed explanation

Background Concept

Oxidation is the loss of electrons, resulting in an increase in oxidation number.
Reduction is the gain of electrons, resulting in a decrease in oxidation number.
The reducing agent is the substance that causes reduction in another species by losing electrons itself (i.e., it is oxidised).
The oxidising agent is the substance that causes oxidation in another species by gaining electrons itself (i.e., it is reduced).

Understanding the Question

In equation 1: $\text{Na}_2\text{O} + \text{Bi}_2\text{O}_3 + \text{O}_2 \rightarrow 2\text{NaBiO}_3$ , you need to identify the reducing agent.

Approach

Determine the oxidation number of the key elements (Bi and O) on both sides of the equation. The element that is oxidised (increases in oxidation number) belongs to the reducing agent.

Step-by-Step Reasoning

Track Bismuth (Bi):
- In reactant $\text{Bi}_2\text{O}_3$ , Bi is +3 (calculated in part c(i)).
- In product $\text{NaBiO}_3$ , Bi is +5 (calculated in part c(i)).
- Bi goes from +3 to +5. It has lost electrons, so Bi is oxidised.
Track Oxygen (O):
- In $\text{O}_2$ , O is 0.
- In products, O is -2.
- O goes from 0 to -2. It has gained electrons, so O is reduced.
Identify agents:
- Since Bi in $\text{Bi}_2\text{O}_3$ is oxidised, $\text{Bi}_2\text{O}_3$ acts as the reducing agent.
- Since O in $\text{O}_2$ is reduced, $\text{O}_2$ acts as the oxidising agent.

Key Takeaways

The reducing agent is the reactant containing the element that is oxidised (oxidation number increases). Always check the oxidation numbers of all elements that change.

Common Mistakes

Confusing the reducing agent with the element that is reduced. Remember: the reducing agent gets oxidised.
Forgetting to check the oxidation state of oxygen in O2 (which is 0, not -2).

Things to Be Careful About

The question asks for the 'reducing agent', which is the whole compound ( $\text{Bi}_2\text{O}_3$ ), not just the element (Bi). Writing 'Bi' would likely lose a mark.

Techniques used

identify reducing agent from oxidation number change

(d)

$\text{NaBiO}_3$ is an oxidising agent with similar properties to $\text{KMnO}_4$ .

Fig. 1.1 shows an example of the use of $\text{NaBiO}_3$ as an oxidising agent.

Fig. 1.1

7M

(i)

Explain the term oxidising agent.

1M Similar

(ii)

Compound X forms when methylbut-2-ene reacts with $\text{KMnO}_4$ .

State the essential conditions for this reaction.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Cold and dilute $\text{KMnO}_4$ (usually in alkaline/aqueous conditions).

Final answer

Cold and dilute

Detailed explanation

Background Concept

Alkenes react with acidified or alkaline potassium manganate(VII) ( $\text{KMnO}_4$ ) depending on the conditions:

Cold, dilute, alkaline $\text{KMnO}_4$ : Forms a 1,2-diol (glycol). The purple $\text{MnO}_4^-$ is reduced to brown $\text{MnO}_2$ (in alkaline) or colourless $\text{Mn}^{2+}$ (in acid, though acid is less common for diol formation to avoid cleavage).
Hot, concentrated, acidified $\text{KMnO}_4$ : Cleaves the C=C bond completely, forming ketones and/or carboxylic acids depending on the substituents.

Understanding the Question

Compound X is a diol (1,2-diol) formed from methylbut-2-ene. The question asks for the essential conditions for $\text{KMnO}_4$ to produce a diol rather than cleaving the molecule.

Approach

Recall the standard conditions for the dihydroxylation of alkenes using $\text{KMnO}_4$ .

Step-by-Step Reasoning

To stop the oxidation at the diol stage and prevent oxidative cleavage of the C=C bond, the $\text{KMnO}_4$ must be cold and dilute.
(Optional but good context: it is typically used in alkaline or neutral aqueous solution, where a brown precipitate of $\text{MnO}_2$ forms, but 'cold and dilute' is the key marking phrase for the conditions on the $\text{KMnO}_4$ itself).

Key Takeaways

Condition control is vital in organic chemistry. Cold/dilute $\text{KMnO}_4$ gives diols; hot/concentrated gives cleavage products.

Common Mistakes

Saying 'acidified' (this promotes cleavage).
Saying 'hot' (this promotes cleavage).
Forgetting 'dilute'.

Things to Be Careful About

The mark scheme specifically looks for 'cold' and 'dilute'. Mentioning 'alkaline' is often accepted or implied, but 'cold and dilute' are the primary discriminating factors.

Techniques used

identify conditions for alkene dihydroxylation

(iii)

Complete Table 1.1 to show what is observed when compounds Y and Z react separately with the named reagents.

Table 1.1

reagent	observation with Y	observation with Z
$\text{Na}_2\text{CO}_3(\text{aq})$	no reaction
alkaline $\text{I}_2(\text{aq})$
2,4-dinitrophenylhydrazine (2,4-DNPH)
Tollens' reagent

4M Similar

DifficultyMedium

Worked solution

Answer

reagent	observation with Y (propanone)	observation with Z (ethanal)
$\text{Na}_2\text{CO}_3(\text{aq})$	no reaction	no reaction
alkaline $\text{I}_2(\text{aq})$	yellow precipitate	yellow precipitate
2,4-dinitrophenylhydrazine (2,4-DNPH)	red / orange / yellow precipitate	red / orange / yellow precipitate
Tollens' reagent	no reaction	silver mirror

(Note: Y is propanone, a ketone with a $\text{CH}_3\text{CO}-$ group. Z is ethanal, an aldehyde with a $\text{CH}_3\text{CHO}-$ group.)

Final answer

See table above

Detailed explanation

Background Concept

Carbonyl compounds (aldehydes and ketones) can be distinguished and identified using specific chemical tests:

2,4-DNPH: Reacts with all carbonyl compounds (aldehydes and ketones) to form a yellow, orange, or red precipitate (a 2,4-dinitrophenylhydrazone). It tests for the presence of the C=O group.
Tollens' reagent (ammoniacal silver nitrate, $[\text{Ag}(\text{NH}_3)_2]^+$ ): Oxidises aldehydes to carboxylic acids, reducing $\text{Ag}^+$ to metallic silver, which forms a silver mirror on the test tube. Ketones do not react (no mirror).
Alkaline iodine (iodoform test, $\text{I}_2$ in $\text{NaOH}$ ): Tests for a methyl ketone ( $\text{CH}_3\text{CO}-$ group) or a methyl carbinol ( $\text{CH}_3\text{CH(OH)}-$ group). It produces a yellow precipitate of triiodomethane ( $\text{CHI}_3$ ). Ethanal ( $\text{CH}_3\text{CHO}$ ) also gives a positive test because it contains the $\text{CH}_3\text{C=O}$ fragment.
Sodium carbonate ( $\text{Na}_2\text{CO}_3$ ): Reacts with carboxylic acids to produce $\text{CO}_2$ gas. Aldehydes and ketones are not acidic enough to react, so there is no reaction.

Understanding the Question

Compound Y is propanone ( $\text{CH}_3\text{COCH}_3$ ), a ketone. It has a $\text{CH}_3\text{CO}-$ group.
Compound Z is ethanal ( $\text{CH}_3\text{CHO}$ ), an aldehyde. It has a $\text{CH}_3\text{CHO}-$ group.
You must predict the observation for each reagent with both Y and Z.

Approach

Evaluate each reagent against the structural features of Y and Z:

Does it have a C=O group? (Tests with 2,4-DNPH)
Is it an aldehyde or ketone? (Tests with Tollens')
Does it have a $\text{CH}_3\text{CO}-$ or $\text{CH}_3\text{CHO}$ group? (Tests with alkaline $\text{I}_2$ )
Is it acidic? (Tests with $\text{Na}_2\text{CO}_3$ )

Step-by-Step Reasoning

$\text{Na}_2\text{CO}_3(\text{aq})$ : Neither propanone nor ethanal is a carboxylic acid. Both are neutral. Observation: no reaction for both.
Alkaline $\text{I}_2(\text{aq})$ (Iodoform test):
- Y (propanone) has the $\text{CH}_3\text{CO}-$ group. Positive test: yellow precipitate ( $\text{CHI}_3$ ).
- Z (ethanal) has the $\text{CH}_3\text{CHO}$ group (which contains $\text{CH}_3\text{C=O}$ ). Positive test: yellow precipitate.
2,4-DNPH: Both Y and Z contain a C=O group. Positive test: red/orange/yellow precipitate for both.
Tollens' reagent:
- Y (ketone) cannot be oxidised. Negative test: no reaction.
- Z (aldehyde) is oxidised to ethanoic acid, reducing $\text{Ag}^+$ to Ag. Positive test: silver mirror.

Key Takeaways

Memorise the specific structural requirements for each carbonyl test: 2,4-DNPH for C=O, Tollens' for aldehydes, alkaline $\text{I}_2$ for methyl carbonyls ( $\text{CH}_3\text{CO}-$ or $\text{CH}_3\text{CHO}$ ).

Common Mistakes

Thinking Tollens' reagent reacts with ketones.
Forgetting that ethanal gives a positive iodoform test (many students only associate it with methyl ketones).
Confusing the colour of the 2,4-DNPH precipitate (must be red, orange, or yellow; 'white' or 'blue' is wrong).

Things to Be Careful About

Ensure the table is completed for both columns. A missing observation in one column loses a mark.
'Yellow precipitate' is the specific observation for the iodoform test; 'precipitate' alone is not enough.
'Silver mirror' is the specific observation for Tollens'; 'grey precipitate' or 'silver' alone might not score.

Techniques used

predict observations for carbonyl compound testsdistinguish aldehyde and ketone using chemical tests

(iv)

Construct an equation for the reaction of Z with $\text{NaBH}_4$ .

Use [H] to represent an atom of hydrogen from the reducing agent.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{CH}_3\text{CHO} + 2[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{OH}

Final answer

CH3CHO + 2[H] -> CH3CH2OH

Detailed explanation

Background Concept

Aldehydes can be reduced to primary alcohols using reducing agents like sodium borohydride ( $\text{NaBH}_4$ ) or lithium aluminium hydride ( $\text{LiAlH}_4$ ). In simplified equations, the reducing agent is often represented by $[\text{H}]$ , which denotes a hydrogen atom (or a hydride ion $\text{H}^-$ plus a proton $\text{H}^+$ ). The reduction of an aldehyde involves adding two hydrogen atoms across the C=O double bond: one to the carbon and one to the oxygen.

Understanding the Question

Compound Z is ethanal ( $\text{CH}_3\text{CHO}$ ). You need to write the equation for its reaction with $\text{NaBH}_4$ , using $[\text{H}]$ to represent the reducing agent.

Approach

Identify the product: reduction of an aldehyde gives a primary alcohol. Ethanal -> ethanol ( $\text{CH}_3\text{CH}_2\text{OH}$ ).
Balance the equation using $[\text{H}]$ . The C=O becomes CH-OH, requiring 2 hydrogen atoms.

Step-by-Step Reasoning

Reactant: $\text{CH}_3\text{CHO}$ (ethanal).
Product: $\text{CH}_3\text{CH}_2\text{OH}$ (ethanol).
Reduction: The C=O double bond is reduced to a C-O single bond, and the carbon gains an H, the oxygen gains an H. Total of 2 H atoms added.
Equation: $\text{CH}_3\text{CHO} + 2[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{OH}$ .

Key Takeaways

Aldehydes reduce to primary alcohodes; ketones reduce to secondary alcohols. The notation $2[\text{H}]$ represents the addition of two hydrogen atoms (reduction).

Common Mistakes

Writing the wrong product (e.g., carboxylic acid, which is oxidation, not reduction).
Using $[\text{H}]_2$ instead of $2[\text{H}]$ (the mark scheme specifically asks for $2[\text{H}]$ to represent atoms from the reducing agent).
Forgetting to balance the hydrogens.

Things to Be Careful About

The question explicitly says: 'Use [H] to represent an atom of hydrogen from the reducing agent.' Therefore, you must write $2[\text{H}]$ , not $\text{H}_2$ or $[\text{H}_2]$ .

Techniques used

write reduction equation using [H] notation

(e)

$\text{NaBiO}_3$ can be used to determine the concentration of $\text{Mn}^{2+}(\text{aq})$ . The ionic equation for the reaction is shown in equation 2.

\text{equation 2} \quad 2\text{Mn}^{2+} + 5\text{BiO}_3^- + 14\text{H}^+ \rightarrow 2\text{MnO}_4^- + 5\text{Bi}^{3+} + 7\text{H}_2\text{O}

A student uses the following procedure in an experiment.

Add $100.0\text{ cm}^3$ of a saturated solution of $\text{Mn}^{2+}(\text{aq})$ to a volumetric flask.
Add distilled water to the flask to make a $1.00\text{ dm}^3$ diluted solution.
Titrate a $25.00\text{ cm}^3$ sample of the diluted solution with $0.100\text{ mol dm}^{-3}$ $\text{NaBiO}_3(\text{aq})$ .

The $25.00\text{ cm}^3$ sample of the diluted solution of $\text{Mn}^{2+}(\text{aq})$ reacts completely with exactly $21.50\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ $\text{NaBiO}_3(\text{aq})$ .

Calculate the concentration, in $\text{mol dm}^{-3}$ , of $\text{Mn}^{2+}(\text{aq})$ in the saturated solution.

Show your working.

3M Similar

DifficultyMedium

Worked solution

Working

Moles of $\text{NaBiO}_3$ used:
$n(\text{NaBiO}_3) = \frac{21.50}{1000} \times 0.100 = 2.15 \times 10^{-3} \text{ mol}$
Moles of $\text{Mn}^{2+}$ in the $25.00 \text{ cm}^3$ sample (from equation 2, ratio $\text{Mn}^{2+} : \text{BiO}_3^- = 2 : 5$ ):
$n(\text{Mn}^{2+})_{\text{sample}} = 2.15 \times 10^{-3} \times \frac{2}{5} = 8.60 \times 10^{-4} \text{ mol}$
Moles of $\text{Mn}^{2+}$ in the $1.00 \text{ dm}^3$ diluted solution (dilution factor = $1000 / 25 = 40$ ):
$n(\text{Mn}^{2+})_{\text{diluted}} = 8.60 \times 10^{-4} \times 40 = 0.0344 \text{ mol}$
Concentration of $\text{Mn}^{2+}$ in the saturated solution (the $1.00 \text{ dm}^3$ diluted solution was made from $100.0 \text{ cm}^3$ of the saturated solution, so moles are conserved):
$c(\text{Mn}^{2+})_{\text{saturated}} = \frac{0.0344 \text{ mol}}{0.1000 \text{ dm}^3} = 0.344 \text{ mol dm}^{-3}$
(Alternatively: $c = \frac{0.0344 \times 1000}{100} = 0.344 \text{ mol dm}^{-3}$ )

Answer

$0.344 \text{ mol dm}^{-3}$

Final answer

0.344 mol dm-3

Detailed explanation

Background Concept

This is a titration calculation involving a redox reaction and a dilution step. The key principles are:

Titration: Use $n = c \times V$ (with $V$ in $\text{dm}^3$ ) to find moles of the titrant.
Stoichiometry: Use the molar ratio from the balanced equation to find moles of the analyte in the titrated sample.
Dilution: Moles are conserved during dilution ( $c_1V_1 = c_2V_2$ or simply scaling the moles by the dilution factor). The concentration of the original saturated solution can be found by dividing the total moles in the diluted volume by the original volume taken.

Understanding the Question

A $100.0 \text{ cm}^3$ sample of saturated $\text{Mn}^{2+}$ is diluted to $1.00 \text{ dm}^3$ ( $1000 \text{ cm}^3$ ).
A $25.00 \text{ cm}^3$ aliquot of this diluted solution is titrated with $0.100 \text{ mol dm}^{-3}$ $\text{NaBiO}_3$ .
The titre is $21.50 \text{ cm}^3$ .
Equation 2: $2\text{Mn}^{2+} + 5\text{BiO}_3^- + 14\text{H}^+ \rightarrow 2\text{MnO}_4^- + 5\text{Bi}^{3+} + 7\text{H}_2\text{O}$ (ratio $\text{Mn}^{2+} : \text{BiO}_3^- = 2 : 5$ ).
Find the concentration of $\text{Mn}^{2+}$ in the original saturated solution.

Approach

Calculate moles of $\text{BiO}_3^-$ used in the titration.
Use the 2:5 ratio to find moles of $\text{Mn}^{2+}$ in the $25.00 \text{ cm}^3$ aliquot.
Scale up to find moles of $\text{Mn}^{2+}$ in the full $1.00 \text{ dm}^3$ diluted solution (multiply by $1000/25 = 40$ ).
These moles came from the original $100.0 \text{ cm}^3$ of saturated solution. Calculate the concentration: $c = n / V$ (with $V = 0.1000 \text{ dm}^3$ ).

Step-by-Step Reasoning

Moles of titrant ( $\text{NaBiO}_3$ ):
$n = \frac{21.50 \text{ cm}^3}{1000} \text{ dm}^3 \times 0.100 \text{ mol dm}^{-3} = 2.15 \times 10^{-3} \text{ mol}$
Moles of $\text{Mn}^{2+}$ in $25.00 \text{ cm}^3$ sample:
From the equation, 5 moles of $\text{BiO}_3^-$ react with 2 moles of $\text{Mn}^{2+}$ .
$n(\text{Mn}^{2+}) = 2.15 \times 10^{-3} \times \frac{2}{5} = 8.60 \times 10^{-4} \text{ mol}$
Moles of $\text{Mn}^{2+}$ in the $1.00 \text{ dm}^3$ diluted solution:
The $25.00 \text{ cm}^3$ sample is $1/40$ of the total $1000 \text{ cm}^3$ .
$n_{\text{total}} = 8.60 \times 10^{-4} \text{ mol} \times \frac{1000}{25} = 8.60 \times 10^{-4} \times 40 = 0.0344 \text{ mol}$
Concentration in the saturated solution:
The $0.0344 \text{ mol}$ of $\text{Mn}^{2+}$ originally came from $100.0 \text{ cm}^3$ ( $0.1000 \text{ dm}^3$ ) of the saturated solution.
$c = \frac{0.0344 \text{ mol}}{0.1000 \text{ dm}^3} = 0.344 \text{ mol dm}^{-3}$
(Note: The mark scheme shows a slightly different grouping: $2.15 \times 10^{-3} \times (2/5) \times 40 = 0.0344$ , then $0.0344 \times (1000/100) = 0.344$ . Both are mathematically identical and correct.)

Key Takeaways

In dilution-titration problems, always track the moles. Calculate moles in the aliquot, scale to the total diluted volume, and then use the original volume to find the original concentration. Do not mix up the dilution factor (total volume / aliquot volume) with the concentration dilution factor.

Common Mistakes

Forgetting to convert $\text{cm}^3$ to $\text{dm}^3$ for the titre volume.
Using the wrong stoichiometric ratio (e.g., 5:2 instead of 2:5 for Mn:Bi).
Forgetting to account for the dilution step (calculating the concentration of the diluted solution instead of the saturated one).
Dividing by 1000 twice or missing a factor of 10 in the dilution calculation.

Things to Be Careful About

Significant figures: The data given (21.50, 0.100, 25.00, 100.0, 1.00) has 3 or 4 sig figs. The answer $0.344$ has 3 sig figs, which is appropriate.
Ensure state symbols or units are correct in intermediate steps if required, though for the final answer, 'mol dm^-3' is essential.
The equation uses $\text{BiO}_3^-$ , but the titrant is $\text{NaBiO}_3$ . Since NaBiO3 dissociates to give $\text{BiO}_3^-$ , 1 mole of NaBiO3 gives 1 mole of $\text{BiO}_3^-$ . This is a subtle point that doesn't change the math but is good to understand.

Techniques used

calculate moles from a titreuse stoichiometric ratio from ionic equationaccount for dilution factor

Q2Medium-HardChemical PeriodicityGroup 17Chemical BondingHydrocarbonsHalogen CompoundsPolymerisation

Chlorine, $\text{Cl}_2$ , reacts with many elements and compounds to form chlorides.

Table 2.1 shows information about some chlorides of Period 3 elements.

(a)

Complete Table 2.1.

Table 2.1

	Na	Mg	Si
formula of chloride
structure of chloride	giant
bonding of chloride			covalent
pH of solution formed on addition of chloride to water		6.2

3M Similar

DifficultyEasy

Worked solution

Answer

	Na	Mg	Si
formula of chloride	NaCl	MgCl $_2$	SiCl $_4$
structure of chloride	giant	giant	simple
bonding of chloride	ionic	ionic	covalent
pH of solution formed	7	6.2	1–4

Final answer

See completed table above.

Detailed explanation

Background Concept

The properties of Period 3 chlorides change across the period as the bonding changes from ionic (metals on the left) to covalent (non-metals on the right). Sodium chloride (NaCl) and magnesium chloride (MgCl $_2$ ) are giant ionic lattices. Silicon tetrachloride (SiCl $_4$ ) is a simple covalent molecule. When added to water, ionic chlorides of Group 1 and 2 (except Mg $^{2+}$ which is slightly acidic due to polarisation) dissolve to give neutral or near-neutral solutions. Simple covalent chlorides like SiCl $_4$ react vigorously with water (hydrolyse) to form acidic solutions (HCl and H $_2$ SiO $_3$ or SiO $_2$ · $n$ H $_2$ O), giving a low pH (1–4). MgCl $_2$ hydrolyses slightly to give a weakly acidic solution (pH ~6.2).

Understanding the Question

The question asks to complete a table comparing the chloride of sodium (Na), magnesium (Mg), and silicon (Si). We need to provide the formula, structure type, bonding type, and the pH of the aqueous solution formed. The table already gives 'giant' structure for NaCl, 'covalent' bonding for SiCl $_4$ , and pH 6.2 for MgCl $_2$ .

Approach

Formulae: Use group numbers to determine oxidation states: Na $^+$ , Mg $^{2+}$ , Si $^{4+}$ , Cl $^-$ . Formulas are NaCl, MgCl $_2$ , SiCl $_4$ .
Structure and Bonding: Na and Mg are metals; their chlorides are giant ionic lattices. Si is a metalloid/non-metal; SiCl $_4$ is a simple molecular structure with covalent bonding.
pH: NaCl is neutral (pH 7). MgCl $_2$ is given as 6.2. SiCl $_4$ hydrolyses to form HCl, which is a strong acid, so pH is 1–4.

Step-by-Step Reasoning

NaCl: Sodium is a Group 1 metal. Formula is NaCl. It forms a giant ionic lattice with ionic bonding. Dissolving in water gives Na $^+$ and Cl $^-$ , neither hydrolyses significantly, so pH = 7.
MgCl $_2$ : Magnesium is a Group 2 metal. Formula is MgCl $_2$ . Giant ionic lattice, ionic bonding. The small, highly charged Mg $^{2+}$ ion polarises water molecules slightly, releasing H $^+$ , giving a weakly acidic pH of 6.2 (given).
SiCl $_4$ : Silicon is a Group 4 element. Formula is SiCl $_4$ . It forms simple covalent molecules. In water, it hydrolyses: SiCl $_4$ (l) + 2H $_2$ O(l) → SiO $_2$ (s) + 4HCl(aq). The HCl makes the solution strongly acidic, pH 1–4.

Key Takeaways

Period 3 chlorides transition from ionic (NaCl, MgCl $_2$ ) to covalent (SiCl $_4$ ) across the period.
Ionic chlorides of reactive metals give neutral solutions; covalent chlorides hydrolyse to give acidic solutions.
Mg $^{2+}$ is an exception among Group 2 chlorides in giving a slightly acidic solution due to cation polarisation.

Common Mistakes

Writing SiCl $_2$ instead of SiCl $_4$ (silicon is Group 4, needs 4 bonds).
Describing SiCl $_4$ as having a 'giant' structure or 'ionic' bonding.
Giving pH = 7 for SiCl $_4$ (forgetting hydrolysis produces HCl).
Writing 'molecular' instead of 'simple' for the structure of SiCl $_4$ (though 'molecular' is often accepted, 'simple' is the precise term used in mark schemes to contrast with 'giant').

Things to Be Careful About

Ensure formulae are correctly sub-scripted (MgCl $_2$ , SiCl $_4$ ).
The pH for SiCl $_4$ is a range (1–4) because it depends on concentration, but must be clearly acidic.
Do not confuse the bonding in MgCl $_2$ (ionic) with SiCl $_4$ (covalent).

Techniques used

predict formulae of Period 3 chloridesdeduce structure and bonding from element typepredict pH of chloride hydrolysis solutions

(b)

When $\text{Cl}_2$ reacts with cold $\text{NaOH}(\text{aq})$ , $\text{Cl}_2$ is both oxidised and reduced. The products are $\text{NaCl}$ , water and G.

5M

(i)

State the type of redox reaction in which the same species is both oxidised and reduced.

1M Similar

(ii)

Identify G.

1M Similar

(iii)

Write an equation for the reaction between $\text{Cl}_2$ and hot $\text{NaOH}(\text{aq})$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

3\text{Cl}_2 + 6\text{NaOH} \rightarrow \text{NaClO}_3 + 5\text{NaCl} + 3\text{H}_2\text{O}

Final answer

3Cl2 + 6NaOH -> NaClO3 + 5NaCl + 3H2O

Detailed explanation

Background Concept

The products of chlorine disproportionation depend on temperature.

Cold: Cl $_2$ + 2NaOH → NaCl + NaClO + H $_2$ O (chlorate(I), Cl is +1)
Hot: 3Cl $_2$ + 6NaOH → NaClO $_3$ + 5NaCl + 3H $_2$ O (chlorate(V), Cl is +5)

The hot reaction is thermodynamically more favourable; chlorate(I) disproportionates further to chlorate(V) and chloride when heated.

Understanding the Question

Write the equation for Cl $_2$ reacting with hot NaOH(aq).

Approach

Recall the hot disproportionation products: chloride (Cl $^-$ ) and chlorate(V) (ClO $_3^-$ ). Balance the redox half-equations or use oxidation number changes.

Step-by-Step Reasoning

Reduction: Cl $_2$ + 2e $^-$ → 2Cl $^-$ (oxidation state 0 to -1)
Oxidation: Cl $_2$ + 6OH $^-$ → 2ClO $_3^-$ + 3H $_2$ O + 10e $^-$ (oxidation state 0 to +5)
Multiply reduction by 5 to balance electrons: 5Cl $_2$ + 10e $^-$ → 10Cl $^-$
Add: 6Cl $_2$ + 6OH $^-$ → 2ClO $_3^-$ + 10Cl $^-$ + 3H $_2$ O
Simplify (divide by 2): 3Cl $_2$ + 3OH $^-$ → ClO $_3^-$ + 5Cl $^-$ + 1.5H $_2$ O
Multiply by 2 to get whole numbers for NaOH/H $_2$ O: 3Cl $_2$ + 6NaOH → NaClO $_3$ + 5NaCl + 3H $_2$ O

Key Takeaways

Cold NaOH + Cl $_2$ → NaClO (chlorate(I)).
Hot NaOH + Cl $_2$ → NaClO $_3$ (chlorate(V)).
Always check the temperature condition in the question.

Common Mistakes

Writing the cold equation (NaClO product) instead of the hot one.
Incorrect balancing (e.g., missing the 5NaCl).
Forgetting state symbols (though often not strictly required unless specified, good practice: (g), (aq), (l)).

Things to Be Careful About

The equation is 3Cl $_2$ + 6NaOH → NaClO $_3$ + 5NaCl + 3H $_2$ O. Do not write 1:1 ratio.

Techniques used

write equation for hot chlorine disproportionation

(iv)

Describe fully what is observed when $\text{AgNO}_3(\text{aq})$ is added to the aqueous solution of the chloride of sodium, followed by dilute $\text{NH}_3(\text{aq})$ .

2M Similar

DifficultyEasy

Worked solution

Answer

A white precipitate is formed.
The precipitate dissolves in dilute ammonia.

Working

The aqueous solution of sodium chloride contains Cl $^-$ ions.
Test: Add AgNO $_3$ (aq) → AgCl(s) (white ppt).
Add dilute NH $_3$ (aq) → [Ag(NH $_3$ ) $_2$ ] $^+$ (aq) (soluble complex), ppt dissolves.

Final answer

White precipitate; dissolves in dilute ammonia.

Detailed explanation

Background Concept

The test for halide ions (Cl $^-$ , Br $^-$ , I $^-$ ) uses aqueous silver nitrate (acidified with dilute HNO $_3$ to remove carbonate/sulfite interference).

Cl $^-$ + Ag $^+$ → AgCl(s) [white precipitate]
Br $^-$ + Ag $^+$ → AgBr(s) [cream precipitate]
I $^-$ + Ag $^+$ → AgI(s) [yellow precipitate]

Solubility in ammonia:

AgCl dissolves in dilute NH $_3$ (aq).
AgBr dissolves in concentrated NH $_3$ (aq).
AgI is insoluble in ammonia.

Understanding the Question

Describe observations when AgNO $_3$ (aq) is added to aqueous NaCl (chloride solution), followed by dilute NH $_3$ (aq).

Approach

Recall the standard test for Cl $^-$ ions: reagent, observation with AgNO $_3$ , observation with dilute NH $_3$ .

Step-by-Step Reasoning

Addition of AgNO $_3$ (aq): The solution contains Na $^+$ and Cl $^-$ (from NaCl). Ag $^+$ reacts with Cl $^-$ to form silver chloride, AgCl, which is a white precipitate.
$\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})$
Addition of dilute NH $_3$ (aq): Silver chloride is soluble in dilute ammonia because it forms the diamminesilver(I) complex ion.
$\text{AgCl}(\text{s}) + 2\text{NH}_3(\text{aq}) \rightarrow [\text{Ag}(\text{NH}_3)_2]^+(\text{aq}) + \text{Cl}^-(\text{aq})$
The precipitate dissolves to give a colourless solution.

Key Takeaways

Cl $^-$ gives a white ppt with AgNO $_3$ that dissolves in dilute NH $_3$ .
Br $^-$ gives a cream ppt that dissolves in conc. NH $_3$ .
I $^-$ gives a yellow ppt that is insoluble in NH $_3$ .

Common Mistakes

Forgetting to mention the colour of the precipitate (must say 'white').
Saying the ppt 'reacts' with ammonia (it dissolves/complexes, 'reacts' is vague).
Confusing dilute and concentrated ammonia solubility rules.

Things to Be Careful About

The question asks to 'describe fully what is observed'. Give both observations: ppt formation and subsequent dissolution.
Specify 'dilute' ammonia as given in the question.

Techniques used

describe silver nitrate test for chloride ionsdescribe solubility in ammonia

(c)

An excess of $\text{Cl}_2$ reacts with phosphorus to form $\text{PCl}_5$ .

4M

(i)

$\text{PCl}_5$ is a simple molecule in the gas phase.

It also exists in a solid form as two ions, $\text{PCl}_4^+$ and $\text{PCl}_6^-$ .

Complete Table 2.2 to identify the shapes of each of these species.

Table 2.2

species	$\text{PCl}_5$	$\text{PCl}_4^+$	$\text{PCl}_6^-$
shape		tetrahedral

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

species	PCl $_5$	PCl $_4^+$	PCl $_6^-$
shape	trigonal bipyramidal	tetrahedral	octahedral

Final answer

PCl5: trigonal bipyramidal; PCl6-: octahedral

Detailed explanation

Background Concept

VSEPR (Valence Shell Electron Pair Repulsion) theory predicts molecular shapes based on the number of bonding pairs (BP) and lone pairs (LP) around the central atom. Electron pairs arrange themselves to minimise repulsion.

5 BP, 0 LP → Trigonal bipyramidal (e.g., PCl $_5$ )
4 BP, 0 LP → Tetrahedral (e.g., CH $_4$ , PCl $_4^+$ )
6 BP, 0 LP → Octahedral (e.g., SF $_6$ , PCl $_6^-$ )

Phosphorus is in Group 15, so it has 5 valence electrons. It can expand its octet using empty d-orbitals.

Understanding the Question

Complete the table with the shapes of PCl $_5$ , PCl $_4^+$ , and PCl $_6^-$ . PCl $_4^+$ is given as tetrahedral.

Approach

Determine the number of electron domains (bonding pairs + lone pairs) for the central P atom in each species.

Step-by-Step Reasoning

PCl $_5$ : P has 5 valence electrons. Forms 5 single bonds with 5 Cl atoms. 5 BP, 0 LP. Total 5 domains. Shape: trigonal bipyramidal.
PCl $_4^+$ : P has 5 valence electrons. Positive charge means 1 less electron → 4 valence electrons. Forms 4 bonds with Cl. 4 BP, 0 LP. Total 4 domains. Shape: tetrahedral (given).
PCl $_6^-$ : P has 5 valence electrons. Negative charge means 1 more electron → 6 valence electrons. Forms 6 bonds with Cl. 6 BP, 0 LP. Total 6 domains. Shape: octahedral.

Key Takeaways

P can form 5 or 6 bonds (expanded octet).
Cation removes an electron (PCl $_4^+$ has 4 bonds), anion adds an electron (PCl $_6^-$ has 6 bonds).
Shapes follow standard VSEPR: 5=trigonal bipyramidal, 6=octahedral.

Common Mistakes

Thinking PCl $_5$ has a lone pair (it doesn't, 5 bonds use all 5 valence electrons).
Confusing trigonal bipyramidal with square pyramidal (square pyramidal is 5 BP + 1 LP, e.g., BrF $_5$ ).

Things to Be Careful About

Ensure correct spelling: 'trigonal bipyramidal' and 'octahedral'.

Techniques used

apply VSEPR theory to PCl5, PCl4+, PCl6-count bonding pairs and lone pairs

(ii)

$\text{PCl}_5$ reacts with J to form $\text{H}_3\text{PO}_4$ .

Identify J and state the type of reaction.

J: ................................. type of reaction: .............................................................................

2M Similar

(d)

$\text{Cl}_2$ reacts readily with propene to form K, 1,2-dichloropropane.

K can be used to form L.

Fig. 2.1

6M

(i)

Complete Fig. 2.2 to show the mechanism for the reaction of $\text{Cl}_2$ with propene in reaction 1.

Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

Fig. 2.2

4M Similar

DifficultyMedium-Hard

Worked solution

Answer

Step 1:

Curly arrow from the C=C double bond to the closer Cl atom (labelled $\delta^+$ ).
Curly arrow from the Cl–Cl bond to the further Cl atom (labelled $\delta^-$ ).
Dipole on Cl $_2$ : closer Cl is $\delta^+$ , further Cl is $\delta^-$ .

Intermediate:

Carbocation formed on the central carbon (C2) of the propane chain.
Cl atom bonded to C1 (the end carbon).
Positive charge on C2.

Step 2:

Curly arrow from a lone pair on the Cl $^-$ ion to the positively charged carbon (C2).
Final product: 1,2-dichloropropane.

Working

See diagram description below for full details of arrows and charges.

Final answer

See mechanism diagram below.

Detailed explanation

Background Concept

Alkenes undergo electrophilic addition with halogens (Cl $_2$ , Br $_2$ ). The mechanism occurs in two steps:

Formation of the carbocation: The $\pi$ -electrons of the C=C bond attack the electrophilic halogen molecule. The halogen molecule becomes polarised (induced dipole) by the approaching electron-rich alkene. A curly arrow goes from the C=C bond to one halogen atom, and another from the halogen-halogen bond to the other atom (heterolytic fission). This forms a halogen-substituted carbocation and a halide ion.
Nucleophilic attack: The halide ion (nucleophile) attacks the carbocation (electrophile) using a lone pair to form the second C-X bond.

Markovnikov's Rule: In asymmetric alkenes, the hydrogen (or the electrophile in the first step if considering the intermediate stability) adds to the carbon with more hydrogens to form the more stable carbocation. For Cl $_2$ addition, the Cl $^+$ equivalent adds to the less substituted carbon to leave the positive charge on the more substituted (more stable) carbon.

Understanding the Question

Draw the mechanism for the reaction of propene (CH $_3$ CH=CH $_2$ ) with Cl $_2$ to form 1,2-dichloropropane. Include charges, dipoles, lone pairs, and curly arrows.

Approach

Draw propene and Cl $_2$ with an induced dipole.
Show the first step: arrow from C=C to Cl( $\delta^+$ ), arrow from Cl-Cl to Cl( $\delta^-$ ).
Draw the intermediate: secondary carbocation on C2, Cl on C1.
Show the second step: arrow from Cl $^-$ lone pair to C $^+$ .

Step-by-Step Reasoning

Reactants: Propene is CH $_3$ –CH=CH $_2$ . Cl $_2$ is non-polar but becomes polarised near the alkene: Cl $^{\delta+}$ –Cl $^{\delta-}$ .
Step 1 (Electrophilic attack):
- A curly arrow starts from the center of the C=C double bond and points to the Cl atom that is closer (the $\delta^+$ one).
- A second curly arrow starts from the Cl–Cl bond and points to the further Cl atom (the $\delta^-$ one), showing heterolytic fission.
- This breaks the $\pi$ -bond and the Cl–Cl bond.
Intermediate:
- The Cl atom is now bonded to the terminal carbon (C1, the CH $_2$ end).
- The central carbon (C2, the CH end) loses a bond and becomes a carbocation (C $^+$ ). This is a secondary carbocation, which is more stable than the primary one that would form if Cl added to C2.
- The other chlorine is now a chloride ion, Cl $^-$ , with 4 lone pairs and a negative charge.
Step 2 (Nucleophilic attack):
- A curly arrow starts from a lone pair on the Cl $^-$ ion and points to the positively charged carbon (C2).
- This forms the second C–Cl bond, giving 1,2-dichloropropane (CH $_3$ –CHCl–CH $_2$ Cl).

Key Takeaways

Electrophilic addition involves two steps: carbocation formation, then nucleophilic attack.
The induced dipole on the halogen is crucial; arrows must show the flow of electrons from electron-rich to electron-poor.
The carbocation forms on the more substituted carbon (Markovnikov-like stability).

Common Mistakes

Drawing arrows from the Cl atom to the C=C bond (electrons flow from nucleophile/electron-rich to electrophile/electron-poor).
Forgetting the dipole on Cl $_2$ ( $\delta^+$ and $\delta^-$ ).
Drawing the wrong carbocation (primary instead of secondary).
Forgetting the lone pairs and charge on the Cl $^-$ ion in the second step.
Drawing the final product as the intermediate (forgetting the second step arrow).

Things to Be Careful About

Curly arrows must start from a bond or lone pair (electron source) and end at an atom or bond (electron destination).
The intermediate must clearly show the positive charge on the correct carbon (C2, not C1 or C3).
State symbols are not required for mechanisms, but charges and dipoles are essential.

Techniques used

draw electrophilic addition mechanism for Cl2 and alkeneshow curly arrows, dipoles, and intermediates

(ii)

Identify the reagent and conditions for reaction 2.

1M Similar

(iii)

Draw one repeat unit of the addition polymer that forms from L.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The monomer L is 2-chloropropene: CH $_2$ =C(Cl)CH $_3$ .

The repeat unit is:

\begin{array}{c} \text{Cl} \quad \text{H} \\ | \quad \quad | \\ \cdots - \text{C} - \text{C} - \cdots \\ | \quad \quad | \\ \text{CH}_3 \quad \text{H} \end{array}

(With dashed bonds extending from the two central carbons to indicate the polymer chain)

Working

Monomer: CH $_2$ =C(Cl)(CH $_3$ )
Break the C=C double bond to form single bonds in the polymer chain.
Each carbon in the repeat unit retains its substituents.
Left C (from CH $_2$ ): bonded to 2 H's.
Right C (from C(Cl)(CH $_3$ )): bonded to Cl and CH $_3$ .

Final answer

Repeat unit: -[CH2-C(Cl)(CH3)]- with dashed bonds at ends.

Detailed explanation

Background Concept

In addition polymerisation, the $\pi$ -bond of the alkene monomer breaks, and the carbons form new $\sigma$ -bonds with adjacent monomers. The repeat unit is the smallest structural unit that repeats in the polymer chain.

To draw a repeat unit:

Draw the monomer with the C=C double bond vertical or horizontal.
Break the double bond to a single bond.
Extend bonds from both carbons with dashed lines (or just lines with dots/ellipses) to show continuation.
Keep all substituents attached to their respective carbons.

Understanding the Question

Draw one repeat unit of the polymer formed from L (2-chloropropene).
L is CH $_2$ =C(Cl)–CH $_3$ (the Cl and CH $_3$ are on the same carbon, C2).

Approach

Identify the monomer structure, break the double bond, and arrange substituents on the repeat unit carbons.

Step-by-Step Reasoning

Monomer L: 2-chloropropene. Structure: H $_2$ $_{2}$ C=C(Cl)(CH $_3$ $_{3}$ ).
- C1 (terminal): bonded to 2 H atoms.
- C2 (central): bonded to 1 Cl atom and 1 CH $_3$ group.
Polymerisation: The double bond opens up.
Repeat unit: A two-carbon chain (the backbone).
- One carbon (from C1) is bonded to 2 H atoms.
- The other carbon (from C2) is bonded to a Cl atom and a CH $_3$ group.
- Dashed bonds extend from both backbone carbons to indicate the repeating chain.

Structure:

  Cl   H
  |    |
--C -- C--
  |    |
 CH3   H

(Note: The orientation can vary, e.g., Cl and CH $_3$ on the same carbon, H's on the other. The key is the connectivity: one C has Cl and CH $_3$ , the other has two H's.)

Key Takeaways

Add polymers have the same atoms as the monomer (no loss of small molecules).
The repeat unit has single bonds in the backbone and dashed bonds at the ends.
Substituents stay on the same carbon as in the monomer.

Common Mistakes

Drawing the repeat unit with a double bond (it's an addition polymer, so backbone is single bonds).
Forgetting the dashed bonds at the ends (indicates it's a repeat unit, not a molecule).
Putting Cl and CH $_3$ on different carbons (they were on the same carbon in the monomer).
Writing 'CH $_3$ CHCl–CH $_2$ Cl' as the repeat unit (that's the monomer K, not L's polymer).

Things to Be Careful About

Ensure the repeat unit is enclosed in brackets or has clear dashed bonds.
The formula is often written as poly(2-chloropropene) or poly(propene chloride). The repeat unit is –[CH $_2$ –C(Cl)(CH $_3$ )]–.

Techniques used

draw repeat unit of addition polymer from alkene monomer

Q3Medium-EasyNitrogen and SulfurChemical BondingChemical EnergeticsReaction Kinetics

Nitrogen, $\text{N}_2$ , is generally an unreactive molecule but it does react under certain conditions.

(a)

Give two reasons to explain the lack of reactivity of nitrogen.

................................................................................................................................................
................................................................................................................................................

2M Similar

DifficultyEasy

Worked solution

Answer

Strong triple bond / high triple bond enthalpy (941 kJ mol⁻¹).
Non-polar molecule (no permanent dipole).

Final answer

Strong triple bond; non-polar molecule

Detailed explanation

Background Concept

Nitrogen gas ( $\text{N}_2$ ) is the most abundant gas in the atmosphere. Its chemical inertness under standard conditions is a fundamental concept in nitrogen chemistry. This stability arises from two main factors: the strength of the bond holding the atoms together and the electronic symmetry of the molecule.

Understanding the Question

The question asks for two reasons why $\text{N}_2$ is unreactive. This is a standard recall question testing knowledge of the physical properties of the nitrogen molecule that prevent it from undergoing reactions easily.

Approach

Identify the bond type in $\text{N}_2$ and calculate/remember its bond enthalpy. Determine the polarity of the molecule based on electronegativity differences.

Step-by-Step Reasoning

Bond Strength: Nitrogen atoms are joined by a triple bond ( $\text{N}\equiv\text{N}$ ). The bond enthalpy is very high, approximately 941 kJ mol⁻¹. Breaking this bond requires a large amount of energy, so reactions with $\text{N}_2$ have high activation energies and are slow at room temperature.
Polarity: Both atoms are identical, so the electronegativity difference is zero. The molecule is non-polar. This means there is no permanent dipole to attract nucleophiles (electron pair donors) or electrophiles (electron pair acceptors), which are the usual participants in many reaction mechanisms.

Key Takeaways

The unreactivity of $\text{N}_2$ is due to the high bond enthalpy of the triple bond and the lack of polarity (no dipoles to attract attacking species).

Common Mistakes

Stating "nitrogen is a noble gas" (it is not; it is a non-metal diatomic molecule).
Saying "it has no lone pairs" (nitrogen has a lone pair on each atom).
Vague answers like "it is stable" without explaining why (bond strength/polarity).

Things to Be Careful About

Ensure you mention "triple bond" or "bond enthalpy" specifically. "Strong bond" is often acceptable, but "triple bond" is more precise. For the second point, "non-polar" is the key term.

Techniques used

recall bond propertiesidentify molecular polarity

(b)

$\text{N}_2$ can react with oxygen in an internal combustion engine to form a mixture of $\text{NO}$ and $\text{NO}_2$ .

Fig. 3.1 shows a reaction scheme involving $\text{N}_2$ .

Fig. 3.1

3M

(i)

Write an equation to show the formation of a mixture of $\text{NO}$ and $\text{NO}_2$ in reaction 1.

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{N}_2 + 1\frac{1}{2}\text{O}_2 \rightarrow \text{NO} + \text{NO}_2$
(Or $2\text{N}_2 + 3\text{O}_2 \rightarrow 2\text{NO} + 2\text{NO}_2$ )

Final answer

N2 + 1.5O2 -> NO + NO2

Detailed explanation

Background Concept

In internal combustion engines, the high temperatures (around 2000-3000 K) provide enough energy to break the strong $\text{N}\equiv\text{N}$ bond and $\text{O}=\text{O}$ bond, allowing nitrogen and oxygen from the air to react. This forms nitrogen oxides, primarily $\text{NO}$ and $\text{NO}_2$ .

Understanding the Question

Reaction 1 shows $\text{N}_2$ reacting with $\text{O}_2$ to form a mixture of $\text{NO}$ and $\text{NO}_2$ . We need to write a balanced equation representing this process.

Approach

Write the reactants ( $\text{N}_2$ , $\text{O}_2$ ) and products ( $\text{NO}$ , $\text{NO}_2$ ). Balance the atoms. The mark scheme accepts fractional coefficients if the simplest ratio is maintained, or integer coefficients.

Step-by-Step Reasoning

Reactants: $\text{N}_2$ and $\text{O}_2$ . Products: $\text{NO}$ and $\text{NO}_2$ .
Equation: $\text{N}_2 + \text{O}_2 \rightarrow \text{NO} + \text{NO}_2$ .
Balancing N: 2 on left, 2 on right (1 in NO, 1 in NO2). Balanced.
Balancing O: 2 on left. On right, 1 (in NO) + 2 (in NO2) = 3. So we need 1.5 $\text{O}_2$ .
Equation: $\text{N}_2 + 1.5\text{O}_2 \rightarrow \text{NO} + \text{NO}_2$ .
Multiplying by 2 to remove fractions: $2\text{N}_2 + 3\text{O}_2 \rightarrow 2\text{NO} + 2\text{NO}_2$ .

Key Takeaways

High temperatures in engines allow $\text{N}_2$ and $\text{O}_2$ to react, forming a mixture of oxides.

Common Mistakes

Writing $\text{N}_2 + \text{O}_2 \rightarrow 2\text{NO}$ (this is the formation of NO only, not the mixture specified).
Incorrect balancing.

Things to Be Careful About

The question asks for a mixture, so both $\text{NO}$ and $\text{NO}_2$ must appear in the products.

Techniques used

write balanced equation for gas formation

(ii)

Give the formulae of the products of reaction 2.

1M Similar

(iii)

State one environmental consequence of reaction 3.

1M Similar

(c)

The Haber process involves the reaction of $\text{N}_2$ and $\text{H}_2$ to form ammonia, $\text{NH}_3$ .

A catalyst is used, which allows the process to be carried out at a lower temperature and pressure.

\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \quad \Delta H = -92 \text{ kJ mol}^{-1}

5M

(i)

Use the information in (c) to complete Table 3.1.

Table 3.1

compound	enthalpy change of formation, $\Delta H_f$ / $\text{kJ mol}^{-1}$
$\text{N}_2$
$\text{H}_2$
$\text{NH}_3$

2M Similar

DifficultyEasy

Worked solution

Answer

compound	enthalpy change of formation, $\Delta H_f$ / kJ mol⁻¹
$\text{N}_2$	0
$\text{H}_2$	0
$\text{NH}_3$	–46

Working:
$\Delta H_{\text{reaction}} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$
$-92 = [2 \times \Delta H_f(\text{NH}_3)] - [0 + 0]$
$\Delta H_f(\text{NH}_3) = -92 / 2 = -46 \text{ kJ mol}^{-1}$

Final answer

N2: 0, H2: 0, NH3: -46

Detailed explanation

Background Concept

The standard enthalpy change of formation ( $\Delta H_f^\ominus$ ) is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions. By definition, $\Delta H_f$ for any element in its standard state is zero.

Understanding the Question

Complete the table with $\Delta H_f$ values for $\text{N}_2$ , $\text{H}_2$ , and $\text{NH}_3$ . We are given the overall reaction enthalpy $\Delta H = -92 \text{ kJ mol}^{-1}$ for $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ .

Approach

Identify $\Delta H_f$ for elements ( $\text{N}_2$ , $\text{H}_2$ ) -> 0.
Use the equation $\Delta H_{\text{reaction}} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$ to find $\Delta H_f$ for $\text{NH}_3$ .

Step-by-Step Reasoning

$\text{N}_2(\text{g})$ and $\text{H}_2(\text{g})$ are elements in standard states, so their $\Delta H_f$ is 0 kJ mol⁻¹.
For the reaction $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ , $\Delta H = -92 \text{ kJ mol}^{-1}$ .
$-92 = [2 \times \Delta H_f(\text{NH}_3)] - [1(0) + 3(0)]$ .
$2 \times \Delta H_f(\text{NH}_3) = -92$ .
$\Delta H_f(\text{NH}_3) = -46 \text{ kJ mol}^{-1}$ .

Key Takeaways

$\Delta H_f$ of elements is zero. Use Hess's law / enthalpy cycle to find formation enthalpies from reaction enthalpies.

Common Mistakes

Forgetting that $\Delta H_f$ is per mole of product formed. The reaction forms 2 moles of $\text{NH}_3$ , so divide by 2.
Writing 0 for $\text{NH}_3$ .

Things to Be Careful About

Signs are crucial. The reaction is exothermic (-92), so formation of ammonia must be negative.

Techniques used

apply definition of enthalpy of formation

(ii)

Explain how the presence of a catalyst affects the reaction.

1M Similar

(iii)

State and explain the effect, if any, on the rate of the Haber process as the pressure is lowered.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The rate is lowered. Lower pressure means the gas particles are more spread out (fewer particles per unit volume), so there is a lower frequency of successful collisions between reactant molecules.

Final answer

Rate lowered due to lower frequency of successful collisions

Detailed explanation

Background Concept

For gaseous reactions, pressure is proportional to concentration ( $P = cRT$ ). Increasing pressure increases the number of particles per unit volume (concentration). According to collision theory, a higher concentration leads to a higher frequency of collisions, and thus a higher frequency of successful collisions (collisions with energy $\geq E_a$ and correct orientation), increasing the rate.

Understanding the Question

State and explain the effect of lowering the pressure on the rate of the Haber process.

Approach

State the effect on rate (lower).
Explain why: Pressure down -> Volume up (or particles spread out) -> Concentration down -> Collision frequency down -> Rate down.

Step-by-Step Reasoning

Lowering pressure reduces the concentration of the gas reactants ( $\text{N}_2$ and $\text{H}_2$ ).
With fewer particles in the same volume (or particles further apart), the frequency of collisions between reactant molecules decreases.
Consequently, the frequency of successful collisions decreases, so the rate of reaction is lowered.

Key Takeaways

Pressure affects rate for gases by changing concentration. Lower pressure = lower rate.

Common Mistakes

Confusing rate with equilibrium position. (Lower pressure shifts equilibrium to the left (fewer moles of gas), but the question asks about rate). Even if equilibrium shifts left, the rate is slower because concentrations are lower.
Saying "lower pressure means molecules move slower" (temperature is constant, so average kinetic energy/speed is constant).

Things to Be Careful About

The question asks about rate, not equilibrium yield. Be precise: rate is lowered because collision frequency is lower.

Techniques used

explain pressure effect on rate

(d)

The $\text{N}_2\text{F}_2$ molecule has a double covalent bond between its nitrogen atoms. This consists of a $\sigma$ and a $\pi$ bond.

5M

(i)

Complete Fig. 3.2 to show the dot-and-cross diagram for $\text{N}_2\text{F}_2$ .

Show outer electrons only.

Fig. 3.2

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

(See diagram below)

F–N single bond (1 bonding pair).
N=N double bond (2 bonding pairs).
Each F atom has 3 lone pairs (6 electrons).
Each N atom has 1 lone pair (2 electrons).

Final answer

Dot-and-cross diagram: F-N=N-F with correct lone pairs

Detailed explanation

Background Concept

$\text{N}_2\text{F}_2$ (diazene difluoride) is analogous to ethene ( $\text{C}_2\text{H}_4$ ) but with N and F. The structure is F–N=N–F. Each nitrogen forms a double bond with the other nitrogen and a single bond with a fluorine. Nitrogen is in Group 5, so it has 5 valence electrons. Fluorine is in Group 7, so it has 7 valence electrons.

Understanding the Question

Complete the dot-and-cross diagram for $\text{N}_2\text{F}_2$ , showing outer electrons only. The skeleton is F-N-N-F.

Approach

Determine bonding: N=N double bond, F-N single bonds.
Count valence electrons: $2 \times 5 (\text{N}) + 2 \times 7 (\text{F}) = 24$ electrons.
Distribute electrons: Bonding pairs use $2+4+2 = 8$ electrons. Remaining 16 electrons go to lone pairs.
F needs 3 lone pairs (6 e- each). N needs 1 lone pair (2 e- each) to complete octet.

Step-by-Step Reasoning

Bonding:
- N–N: Double bond. Use 4 electrons (2 from each N). In dot-and-cross, show 4 crosses (or 2 dots + 2 crosses) in the overlap region.
- F–N: Single bond. Use 2 electrons (1 from F, 1 from N). Show 1 dot and 1 cross in overlap.
Lone Pairs:
- Each F has 7 valence e-. 1 used in bond. 6 left -> 3 lone pairs. (Show as 6 dots).
- Each N has 5 valence e-. 3 used in bonds (1 for F-N, 2 for N=N). 2 left -> 1 lone pair. (Show as 2 crosses).

Key Takeaways

Dot-and-cross diagrams must show bonding pairs in overlaps and lone pairs on atoms. Octets must be satisfied.

Common Mistakes

Drawing N-N single bond (incorrect structure).
Missing lone pairs on Nitrogen (N has 1 lone pair here).
Wrong number of electrons in overlaps.

Things to Be Careful About

The question says "show outer electrons only". Inner shells of F (2s2 2p5 -> 7 outer) are shown. N (2s2 2p3 -> 5 outer) are shown. Ensure distinct symbols (dots and crosses) for the two different atoms contributing to bonds.

Techniques used

draw dot-and-cross diagram for covalent molecule

(ii)

Deduce the hybridisation of the N atoms in $\text{N}_2\text{F}_2$ .

1M Similar

(iii)

Draw a diagram of the $\pi$ bond between the N atoms in $\text{N}_2\text{F}_2$ and describe how it forms.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Diagram: Two lobes of electron density (orbitals) located above and below the plane of the atoms (the internuclear axis), with a node between them.

Description: Formed by the side-on overlap of unhybridised 2p orbitals (one from each nitrogen atom) that are perpendicular to the plane of the molecule.

Final answer

Side-on overlap of 2p orbitals forming pi bond above and below axis

Detailed explanation

Background Concept

A double bond consists of one sigma ( $\sigma$ ) bond and one pi ( $\pi$ ) bond. The sigma bond is formed by head-on overlap of orbitals (sp²-sp² in this case). The pi bond is formed by side-on overlap of unhybridised p-orbitals. The p-orbitals must be parallel and perpendicular to the plane of the sigma bond framework.

Understanding the Question

Draw a diagram of the pi bond between the N atoms and describe how it forms.

Approach

Draw the pi orbital: two lobes (electron density clouds) above and below the internuclear axis (the line connecting the nuclei). Often drawn as two ovals or clouds, one shaded, one unshaded, or just outlines.
Describe: Unhybridised 2p orbitals on each N atom overlap side-on.

Step-by-Step Reasoning

Diagram: The internuclear axis is horizontal. The pi orbital consists of two lobes, one above and one below this axis. There is a nodal plane containing the nuclei where electron density is zero.
Description: Each nitrogen atom has an unhybridised 2p orbital (perpendicular to the sp² plane). These two 2p orbitals are parallel. They overlap side-on (laterally) to form the pi bond. The electron density is concentrated above and below the plane of the atoms.

Key Takeaways

Pi bond = side-on overlap of p-orbitals. Electron density above and below the internuclear axis.

Common Mistakes

Drawing the pi bond as being in the plane (that's a sigma bond or wrong orientation).
Saying "overlap of sp² orbitals" (sp² form sigma bonds).
Not mentioning "side-on" or "lateral" overlap.

Things to Be Careful About

The diagram should clearly show the lobes above and below the axis, not on the axis. The description must mention 2p orbitals and side-on overlap.

Techniques used

describe pi bond formation

Q4MediumIntroduction to Organic ChemistryHydrocarbonsHalogen CompoundsNitrogen CompoundsAnalytical TechniquesCarboxylic Acids and Derivatives

Compound S is used in food flavourings. A possible synthesis of S is shown in Fig. 4.1.

Fig. 4.1

(a)

P, Q, R and S show stereoisomerism.

Complete Table 4.1 by identifying with a tick ( $\checkmark$ ) the type of stereoisomerism that each molecule shows.

The type of stereoisomerism shown by Q is given.

Table 4.1

	P	Q	R	S
geometrical isomerism
optical isomerism		$\checkmark$

2M Similar

DifficultyMedium

Worked solution

Answer

	P	Q	R	S
geometrical isomerism	$\checkmark$
optical isomerism		$\checkmark$	$\checkmark$	$\checkmark$

Final answer

P: geometrical only; Q: optical only; R: optical only; S: optical only.

Detailed explanation

Background Concept

Stereoisomers have the same structural formula but differ in the spatial arrangement of atoms. Two types are tested at AS level. Geometrical (cis–trans / E–Z) isomerism arises from restricted rotation about a carbon–carbon double bond; it is only possible when each carbon of the C=C carries two different groups. Optical isomerism arises from a chiral centre — a carbon atom bonded to four different groups — giving two non-superimposable mirror-image forms (enantiomers).

Understanding the Question

The reaction scheme (Fig. 4.1) gives: P = but-2-ene ( $\text{CH}_3\text{CH=CHCH}_3$ ); Q = 2-bromobutane ( $\text{CH}_3\text{CHBrCH}_2\text{CH}_3$ ); R = 2-methylbutanenitrile; S = 2-methylbutanoic acid. For each, tick the row(s) of stereoisomerism it shows. Q is already given as optical.

Approach

For each molecule ask two independent questions: (i) does it have a C=C where both carbons bear two different groups? → geometrical. (ii) does it have a carbon with four different groups? → optical.

Step-by-Step Reasoning

P (but-2-ene, $\text{CH}_3\text{CH=CHCH}_3$ ): the C=C has $\text{CH}_3$ and H on each carbon — two different groups on both carbons → geometrical ✓. No carbon carries four different groups → optical ✗.
Q (2-bromobutane): no C=C → geometrical ✗. The C-2 bears H, Br, $\text{CH}_3$ , $\text{CH}_2\text{CH}_3$ (four different) → optical ✓ (given).
R (2-methylbutanenitrile): the carbon bearing CN also bears H, $\text{CH}_3$ and $\text{CH}_2\text{CH}_3$ — four different groups → optical ✓; no C=C → geometrical ✗.
S (2-methylbutanoic acid): the carbon bearing COOH also bears H, $\text{CH}_3$ and $\text{CH}_2\text{CH}_3$ — four different groups → optical ✓; no C=C → geometrical ✗.

Key Takeaways

Geometrical needs a double bond with two different groups on each carbon; optical needs one carbon with four different groups. The two are independent — a molecule can show one, the other, both, or neither.

Common Mistakes

Ticking geometrical for Q, R or S (no C=C present). Ticking optical for P (but-2-ene has no chiral carbon). Assuming the C=C in but-2-ene cannot give geometrical isomerism because the two ends look similar — it still qualifies since each carbon has H and CH $_3$ .

Things to Be Careful About

Check both carbons of the double bond for geometrical isomerism, and check that the four groups on the candidate chiral carbon are genuinely all different (e.g. in R and S the ethyl and methyl groups are distinct).

Techniques used

identify geometrical isomerism from restricted rotation and two different groups on each carbon of the double bondidentify optical isomerism from a chiral carbon bonded to four different groups

(b)

3M

(i)

Give the structural formula of Q.

1M Similar

(ii)

Name the mechanism in reaction 2.

1M Similar

(iii)

Complete the equation for reaction 3. R is represented as $\text{C}_4\text{H}_9\text{CN}$ .

\text{C}_4\text{H}_9\text{CN} + \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots

1M Similar

DifficultyMedium

Worked solution

Working

Acid hydrolysis of a nitrile gives a carboxylic acid and ammonium ion:

\text{C}_4\text{H}_9\text{CN} + 2\text{H}_2\text{O} + \text{H}^+ \rightarrow \text{C}_4\text{H}_9\text{COOH} + \text{NH}_4^+

Answer

$\text{C}_4\text{H}_9\text{CN} + 2\text{H}_2\text{O} + \text{H}^+ \rightarrow \text{C}_4\text{H}_9\text{COOH} + \text{NH}_4^+$

Final answer

C4H9CN + 2H2O + H+ -> C4H9COOH + NH4+

Detailed explanation

Background Concept

Nitriles ( $\text{R–CN}$ ) are hydrolysed by heating with aqueous acid ( $\text{H}^+$ ) to give a carboxylic acid ( $\text{R–COOH}$ ) and an ammonium ion. The C≡N triple bond is converted to C=O plus the nitrogen leaves as $\text{NH}_4^+$ ; two water molecules supply the oxygen and hydrogens, and one $\text{H}^+$ protonates the nitrogen.

Understanding the Question

Reaction 3 hydrolyses R ( $\text{C}_4\text{H}_9\text{CN}$ ) with $\text{H}^+(aq)$ to give S ( $\text{C}_4\text{H}_9\text{COOH}$ ). Complete and balance the equation.

Approach

Write reactants (nitrile + water + H $^+$ ) and products (carboxylic acid + ammonium), then balance atoms and charge.

Step-by-Step Reasoning

Start: $\text{C}_4\text{H}_9\text{CN} \rightarrow \text{C}_4\text{H}_9\text{COOH}$ . The product gains one O and the N becomes $\text{NH}_4^+$ . Add $2\text{H}_2\text{O}$ to supply the O and H, and $\text{H}^+$ to give nitrogen its four hydrogens and balance charge. Check: left H = 9+4+1 = 14; right H = 9+1+4 = 14 ✓. O: left 2, right 2 ✓. N: 1 each ✓. Charge: left +1, right +1 ✓. Balanced equation:

\text{C}_4\text{H}_9\text{CN} + 2\text{H}_2\text{O} + \text{H}^+ \rightarrow \text{C}_4\text{H}_9\text{COOH} + \text{NH}_4^+

Key Takeaways

Acid hydrolysis of a nitrile: nitrile + 2H $_2$ O + H $^+$ → carboxylic acid + NH $_4^+$ .

Common Mistakes

Writing $\text{NH}_3$ instead of $\text{NH}_4^+$ in acid; omitting water or H $^+$ ; wrong coefficient of water.

Things to Be Careful About

Ensure the charge balances (+1 each side) and that the alkyl group $\text{C}_4\text{H}_9$ is unchanged.

Techniques used

balance the acid hydrolysis of a nitrile to a carboxylic acidinclude water and H+ and the ammonium product

(c)

Compounds S and T react to form organic compound U, which has a single functional group.

\text{S} + \text{T} \rightarrow \text{U} + \text{H}_2\text{O}

Table 4.2 shows some data from the mass spectrum of U.

6M

(i)

Use the data from Table 4.2 to show that U contains 7 carbon atoms.

Show your working.

Table 4.2

peak	relative abundance
$\text{M}^+$	7.2
$[\text{M}+1]^+$	0.55

2M Similar

DifficultyMedium

Worked solution

Working

The [M+1] peak arises from $^{13}\text{C}$ , which is about 1.1% abundant relative to $^{12}\text{C}$ .

n = \frac{\text{abundance of } [M+1]^+}{\text{abundance of } M^+ \times 1.1} = \frac{0.55}{7.2 \times 1.1} = \frac{0.55}{7.92} = 6.94 \approx 7

Answer

7 carbon atoms

Final answer

7

Detailed explanation

Background Concept

In a mass spectrum the molecular ion peak $M^+$ corresponds to the molecule with all $^{12}\text{C}$ . The small $[M+1]^+$ peak is due to molecules containing one $^{13}\text{C}$ atom. Since $^{13}\text{C}$ is about 1.1% as abundant as $^{12}\text{C}$ , each carbon atom contributes ~1.1% to the $[M+1]^+/M^+$ ratio. Thus number of C = (ratio of M+1 to M) / 1.1%.

Understanding the Question

Given $M^+$ = 7.2 and $[M+1]^+$ = 0.55 (relative abundances), show U contains 7 carbons.

Approach

Compute the ratio $0.55/7.2$ , express as a percentage, and divide by 1.1.

Step-by-Step Reasoning

Ratio = $0.55/7.2 = 0.0764$ , i.e. 7.64%. Number of C = $7.64/1.1 = 6.94 \approx 7$ . Equivalently $n = 0.55/(7.2 \times 1.1) = 6.94$ . Rounding to the nearest whole number gives 7 carbon atoms. (Error carried forward from the ratio is acceptable.)

Key Takeaways

Number of carbon atoms = (abundance of [M+1] / abundance of M) / 0.011.

Common Mistakes

Forgetting to divide by 1.1 (or 0.011); using the wrong peak ratio; reporting 6.94 without rounding to 7.

Things to Be Careful About

Use the relative abundance ratio, not the absolute values alone, and round to a whole number of carbons.

Techniques used

use the relative abundance of the M+1 peak to find the number of carbon atomsdivide the M+1/M ratio by the 1.1% natural abundance of carbon-13

(ii)

Fig. 4.2 shows the infrared spectrum of U.

Fig. 4.2

Table 4.3 shows characteristic infrared absorption ranges.

Table 4.3

bond	functional groups containing the bond	characteristic infrared absorption range (in wavenumbers) / $\text{cm}^{-1}$
C–O	hydroxy, ester	1040–1300
C=C	aromatic compound, alkene	1500–1680
C=O	amide carbonyl, carboxyl ester	1640–1690 1670–1740 1710–1750
C $\equiv$ N	nitrile	2200–2250
C–H	alkane	2850–2950
N–H	amine, amide	3300–3500
O–H	carboxyl hydroxy	2500–3000 3200–3650

Use Fig. 4.2 and Table 4.3 to identify the functional group present in U.

Explain your answer fully.

functional group: .................................................................................................................

explanation: ........................................................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Functional group: ester

Explanation: absorption at about $1740\ \text{cm}^{-1}$ (within $1710$ – $1750\ \text{cm}^{-1}$ ) indicates a C=O (carbonyl) stretch of an ester, and absorption at about $1200\ \text{cm}^{-1}$ (within $1040$ – $1300\ \text{cm}^{-1}$ ) indicates a C–O stretch. There is no broad O–H absorption, so it is not a carboxylic acid.

Final answer

ester

Detailed explanation

Background Concept

IR spectroscopy identifies functional groups by the wavenumbers at which bonds absorb. A C=O stretch appears strongly in the 1640–1750 cm $^{-1}$ region; the exact sub-range distinguishes amide, carbonyl/carboxyl and ester. A C–O single-bond stretch (in esters and alcohols) appears 1040–1300 cm $^{-1}$ . An O–H stretch of a carboxylic acid is very broad (2500–3000 cm $^{-1}$ ); an alcohol O–H is broad (3200–3650 cm $^{-1}$ ).

Understanding the Question

Using Fig. 4.2 and Table 4.3, identify the single functional group in U and explain fully.

Approach

Read the major absorptions from the spectrum and match them to bonds; combine the evidence.

Step-by-Step Reasoning

The spectrum shows a strong sharp peak near 1740 cm $^{-1}$ → C=O stretch (ester range 1710–1750 cm $^{-1}$ ). A strong peak near 1200 cm $^{-1}$ → C–O stretch (1040–1300 cm $^{-1}$ ). A C=O plus a C–O together, with no broad O–H band, uniquely indicate an ester. (A carboxylic acid would also show the broad O–H; an amide would show N–H near 3300–3500 cm $^{-1}$ .)

Key Takeaways

Ester = C=O (~1740 cm $^{-1}$ ) + C–O (~1200 cm $^{-1}$ ) with no O–H.

Common Mistakes

Identifying only the carbonyl and calling it a ketone/aldehyde (ignoring C–O); missing the absence of O–H; citing only one absorption for the explanation mark.

Things to Be Careful About

Quote both wavenumber ranges from the table and link each to the correct bond; the explanation mark needs both pieces of evidence.

Techniques used

match IR absorptions to bond stretches in the data tablecombine C=O and C-O evidence to identify an ester

(iii)

T also has a single functional group.

Use the information in (c)(i) and your answer to (c)(ii) to identify T and U.

Draw the structures of T and U in the boxes.

Fig. 4.3

2M Similar

DifficultyMedium

Worked solution

Answer

T = ethanol, $\text{CH}_3\text{CH}_2\text{OH}$

U = ethyl 2-methylbutanoate, $\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{COOCH}_2\text{CH}_3$

Final answer

T = ethanol (CH3CH2OH); U = ethyl 2-methylbutanoate (CH3CH2CH(CH3)COOCH2CH3)

Detailed explanation

Background Concept

An ester forms by condensation of a carboxylic acid with an alcohol, losing water: $\text{RCOOH} + \text{R'OH} \rightarrow \text{RCOOR'} + \text{H}_2\text{O}$ . The acyl part ( $\text{RCO–}$ ) comes from the acid and the –OR' part from the alcohol. Counting carbons across the ester lets you work out how many come from each partner.

Understanding the Question

S is 2-methylbutanoic acid ( $\text{C}_4\text{H}_9\text{COOH}$ , 5 carbons). S + T → U + H $_2$ O, U is an ester with 7 carbons. Identify T and U and draw both.

Approach

U has 7 carbons; the acid contributes 5, so the alcohol contributes 2 → T is ethanol. Combine S's acyl group with the ethoxy group to draw U.

Step-by-Step Reasoning

From (c)(i) U has 7 C. S = 2-methylbutanoic acid has 5 C. The ester's alkoxy group must therefore have 7 − 5 = 2 C → from ethanol ( $\text{CH}_3\text{CH}_2\text{OH}$ ), which has a single functional group (–OH) as required. Hence T = ethanol. The ester U is ethyl 2-methylbutanoate: $\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{COOCH}_2\text{CH}_3$ , retaining the chiral methyl branch at the 2-position of the acyl chain. Draw T as HO–CH $_2$ –CH $_3$ and U as the corresponding ester (see diagrams).

Key Takeaways

In ester formation the carbon count of the alcohol = total ester carbons − acid carbons; the –O– in the ester links the two fragments.

Common Mistakes

Assigning the 2 carbons to the acid side; drawing U as an ether or ketone; placing the ester oxygen on the wrong side of the C=O; losing the methyl branch of S.

Things to Be Careful About

Show all atoms and bonds (displayed formula) for full marks; the C=O and the single-bonded O must be on the same carbon; keep the chiral branch.

Techniques used

deduce the ester from acid S plus alcohol Tsplit the ester into its acyl (from the acid) and alkoxy (from the alcohol) partsdraw displayed structures of the alcohol and ester

Chemistry 9701/22 — February/March 2024

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways