The shapes of four different complexes, P, Q, R and S, are shown in Table 5.1. The symbol J represents an atom or ion of a transition element. The symbol L is used to represent a monodentate ligand. (a) Label one bond angle on each of complexes P, Q, R and S, and identify the size of the angle in degrees. (b) Identify the shapes of complexes P, Q, R and S. P:…

(a) P: 109.5°, Q: 90° (or 180°), R: 90° (or 180°), S: 180° · (b) P: tetrahedral, Q: square planar, R: octahedral, S: linear · (c) Q and R · (d) P and R

A reaction scheme is shown in Fig. 7.1. The reagents needed for reaction 2 and reaction 3 are stated. Reaction 5 takes place when $\text{C}_2\text{H}_5\text{NH}_2$ is mixed with compound V. No special conditions are required. (a) Identify compound U which contains only three elements. (b) Describe the reagents and conditions for reaction 1. (c) Identify compound V. (d) Complete the equation for reaction 3. $$ \text{CH}_3\text{COOH} + \text{SOCl}_2 \rightarrow…

(a) CH3CN · (b) dilute acid and heat · (c) CH3COCl · (d) CH3COCl + HCl + SO2 · (e) C2H5Br · (f) heat in ethanol under pressure · (g) LiAlH4 · (h) hydrolysis: 1; addition: 2; reduction: 2, 6; substitution: 3, 4, 5 · (i) C2H5NHC2H5 > NH3 > C2H5NHCOCH3

9701/43 Oct/Nov 2024 — Chemistry Solved Past Paper

Q1MediumEquilibriaGroup 2

(a)

Disodium phosphate, $(\text{Na}^+)_2(\text{HPO}_4^{2-})$ , reacts with an acid to form monosodium phosphate, $\text{Na}^+(\text{H}_2\text{PO}_4^-)$ .

6M

(i)

Identify the ions that are a conjugate acid–base pair in this reaction, using the formulae of the species involved.

conjugate acid: ....................................................................

conjugate base: ....................................................................

1M Similar

DifficultyEasy

Worked solution

Answer

conjugate acid: $\text{H}_2\text{PO}_4^-$

conjugate base: $\text{HPO}_4^{2-}$

Final answer

Conjugate acid: H2PO4^-; conjugate base: HPO4^2-

Detailed explanation

Background Concept

A conjugate acid–base pair consists of two species that differ by exactly one proton ( $\text{H}^+$ ). The acid is the species that can donate a proton; the base is the species that can accept a proton. When a base accepts a proton, it becomes its conjugate acid; when an acid donates a proton, it becomes its conjugate base. The two species in a conjugate pair therefore differ in both the number of hydrogen atoms and the overall charge.

Understanding the Question

The stem tells us that disodium phosphate, $(\text{Na}^+)_2(\text{HPO}_4^{2-})$ , reacts with an acid to form monosodium phosphate, $\text{Na}^+(\text{H}_2\text{PO}_4^-)$ . The question asks us to identify the two phosphate species that form a conjugate acid–base pair. The sodium ions are spectator ions and play no role in the acid–base chemistry.

Approach

Compare the formulae of the two phosphate-containing ions. The species with more hydrogen atoms is the acid; the species with fewer hydrogen atoms is the base. They form a conjugate pair if they differ by one $\text{H}^+$ .

Step-by-Step Reasoning

The two phosphate ions are $\text{HPO}_4^{2-}$ and $\text{H}_2\text{PO}_4^-$ . The reaction is:
$\text{HPO}_4^{2-} + \text{H}^+ \rightarrow \text{H}_2\text{PO}_4^-$

$\text{HPO}_4^{2-}$ accepts a proton to become $\text{H}_2\text{PO}_4^-$ . Therefore:

$\text{HPO}_4^{2-}$ is the base (proton acceptor)
$\text{H}_2\text{PO}_4^-$ is its conjugate acid (formed when the base gains a proton)

They differ by exactly one $\text{H}^+$ , so they constitute a conjugate acid–base pair. The mark scheme requires BOTH species to be identified correctly for the single mark.

Key Takeaways

A conjugate acid–base pair differs by one proton.
The acid member has one more hydrogen (and one less negative charge) than the base member.
Spectator ions such as $\text{Na}^+$ are not part of the acid–base pair.

Common Mistakes

Swapping the labels: the acid has MORE hydrogens, not fewer.
Including $\text{Na}^+$ in the pair — the sodium ions are spectators.
Writing $\text{PO}_4^{3-}$ instead of $\text{HPO}_4^{2-}$ — the question specifies the actual species present.

Things to Be Careful About

The mark is awarded only if BOTH species are correct — it is an all-or-nothing mark.
Give the ionic formulae exactly as shown in the stem: $\text{H}_2\text{PO}_4^-$ and $\text{HPO}_4^{2-}$ .

Techniques used

identify conjugate acid-base paircompare formulae differing by one proton

(ii)

Define buffer solution.

2M Similar

DifficultyEasy

Worked solution

Answer

A buffer solution is a solution that resists changes in pH / allows only small changes in pH when small amounts of acid or alkali (base) are added.

Final answer

A solution that resists changes in pH when small amounts of acid or alkali are added.

Detailed explanation

Background Concept

A buffer solution is a system that maintains a nearly constant pH despite the addition of small amounts of acid or base. It works because it contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable concentrations. When $\text{H}^+$ is added, the conjugate base neutralises it; when $\text{OH}^-$ is added, the weak acid neutralises it. The pH does change slightly, but only by a very small amount.

Understanding the Question

This is a definition question worth 2 marks. The mark scheme splits the definition into two distinct ideas: (1) the effect — resists changes in pH, and (2) the condition — when small amounts of acid or alkali are added. Both ideas must be present to earn both marks.

Approach

Recall the standard textbook definition of a buffer solution. Make sure both the effect and the condition are stated.

Step-by-Step Reasoning

The two marks are awarded as follows:

Effect: allows only small changes in pH / resists changes in pH.
Condition: when small amounts of acid or alkali/base ( $\text{H}^+$ or $\text{OH}^-$ ) are added.

A complete answer combines both: "A buffer solution resists changes in pH when small amounts of acid or alkali are added."

Key Takeaways

A buffer does not keep pH perfectly constant — it resists change.
The qualifier "small amounts" is essential — adding a large excess of acid or base will overwhelm the buffer.
Buffers contain a weak acid and its conjugate base (or vice versa).

Common Mistakes

Writing "maintains a constant pH" — the mark scheme wants "resists changes" or "allows only small changes".
Omitting the condition "when small amounts of acid or alkali are added" — this loses the second mark.
Saying "neutralises acids and bases" without mentioning the pH-resisting effect.

Things to Be Careful About

The mark scheme explicitly requires both parts. A one-sentence answer that covers both ideas scores full marks.
The terms "alkali" and "base" are both accepted; " $\text{OH}^-$ " is also acceptable.

Techniques used

recall the definition of a buffer solution

(iii)

Write two equations to show how a mixture of $(\text{Na}^+)_2(\text{HPO}_4^{2-})$ and $\text{Na}^+(\text{H}_2\text{PO}_4^-)$ can act as a buffer solution.

equation 1: .........................................................................................................................

equation 2: .........................................................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

equation 1: $\text{HPO}_4^{2-} + \text{H}^+ \rightarrow \text{H}_2\text{PO}_4^-$

equation 2: $\text{H}_2\text{PO}_4^- + \text{OH}^- \rightarrow \text{HPO}_4^{2-} + \text{H}_2\text{O}$

Final answer

HPO4^2- + H+ -> H2PO4^- and H2PO4^- + OH- -> HPO4^2- + H2O

Detailed explanation

Background Concept

A buffer contains both members of a conjugate acid–base pair. In this case, the pair is $\text{H}_2\text{PO}_4^-$ (weak acid) and $\text{HPO}_4^{2-}$ (its conjugate base). The buffer resists pH change through two complementary reactions:

When acid ( $\text{H}^+$ ) is added, the conjugate base $\text{HPO}_4^{2-}$ accepts the proton.
When alkali ( $\text{OH}^-$ ) is added, the weak acid $\text{H}_2\text{PO}_4^-$ donates a proton to neutralise it.

Understanding the Question

We must write two equations showing how the mixture of $(\text{Na}^+)_2(\text{HPO}_4^{2-})$ and $\text{Na}^+(\text{H}_2\text{PO}_4^-)$ acts as a buffer. One equation shows the response to added acid; the other shows the response to added alkali. Each equation is worth 1 mark.

Approach

Identify which species reacts with $\text{H}^+$ and which reacts with $\text{OH}^-$ . The base ( $\text{HPO}_4^{2-}$ ) reacts with $\text{H}^+$ ; the acid ( $\text{H}_2\text{PO}_4^-$ ) reacts with $\text{OH}^-$ . Write balanced ionic equations.

Step-by-Step Reasoning

Equation 1 — response to added acid:
$\text{HPO}_4^{2-} + \text{H}^+ \rightarrow \text{H}_2\text{PO}_4^-$
The hydrogen phosphate ion accepts the added proton, converting to dihydrogen phosphate. This removes the added $\text{H}^+$ from solution, so the pH barely changes.

Equation 2 — response to added alkali:
$\text{H}_2\text{PO}_4^- + \text{OH}^- \rightarrow \text{HPO}_4^{2-} + \text{H}_2\text{O}$
The dihydrogen phosphate ion donates a proton to the hydroxide ion, forming hydrogen phosphate and water. This removes the added $\text{OH}^-$ .

The mark scheme also accepts the full compound equations:
$\text{Na}_2\text{HPO}_4 + \text{H}^+ \rightarrow \text{NaH}_2\text{PO}_4 + \text{Na}^+$
$\text{NaH}_2\text{PO}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{HPO}_4 + \text{H}_2\text{O}$

The ionic equations are cleaner and always acceptable.

Key Takeaways

The conjugate base mops up added $\text{H}^+$ ; the weak acid mops up added $\text{OH}^-$ .
Both buffer components must be present for the buffer to work in both directions.
The equations show the interconversion between the two components of the buffer pair.

Common Mistakes

Writing the equations in the wrong direction (showing the components being regenerated rather than consumed).
Forgetting $\text{H}_2\text{O}$ as a product in the hydroxide reaction.
Using $\text{PO}_4^{3-}$ or $\text{H}_3\text{PO}_4$ instead of the two specified species.
Writing unbalanced equations.

Things to Be Careful About

The $\text{H}^+$ in equation 1 represents any acid added; the $\text{OH}^-$ in equation 2 represents any alkali added.
Both equations must be balanced — charge and atoms.
The mark scheme gives 1 mark per correct equation.

Techniques used

write equations for buffer action with added acidwrite equations for buffer action with added alkali

(iv)

Identify one inorganic ion that acts as a buffer in blood.

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{HCO}_3^-$ (hydrogen carbonate)

Final answer

HCO3^- (hydrogen carbonate)

Detailed explanation

Background Concept

Blood plasma is maintained at a pH of about 7.4 by buffer systems. The most important is the carbonic acid / hydrogen carbonate buffer:
$\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$
The hydrogen carbonate ion, $\text{HCO}_3^-$ , acts as the base member of this pair, neutralising excess acid. The phosphate buffer ( $\text{H}_2\text{PO}_4^-/\text{HPO}_4^{2-}$ ) also exists in blood but is less significant.

Understanding the Question

This is a one-mark recall question asking for one inorganic ion that acts as a buffer in blood. The answer is the hydrogen carbonate ion, $\text{HCO}_3^-$ .

Approach

Recall the main buffer system in blood: the carbonic acid / hydrogen carbonate system. The ion is $\text{HCO}_3^-$ .

Step-by-Step Reasoning

The hydrogen carbonate ion, $\text{HCO}_3^-$ , is the key buffer ion in blood. It works alongside carbonic acid, $\text{H}_2\text{CO}_3$ :

When blood becomes too acidic, $\text{HCO}_3^-$ reacts with $\text{H}^+$ to form $\text{H}_2\text{CO}_3$ .
When blood becomes too alkaline, $\text{H}_2\text{CO}_3$ dissociates, releasing $\text{H}^+$ .

Key Takeaways

$\text{HCO}_3^-$ (hydrogen carbonate) is the principal buffer ion in blood.
It is part of the $\text{H}_2\text{CO}_3/\text{HCO}_3^-$ buffer system.

Common Mistakes

Writing $\text{H}_2\text{CO}_3$ instead of $\text{HCO}_3^-$ — the question asks for the ION, not the acid.
Writing $\text{CO}_3^{2-}$ — this is not the blood buffer ion.
Writing a full compound like $\text{NaHCO}_3$ — the question asks for the ion.

Things to Be Careful About

The question specifies "inorganic ion" — $\text{HCO}_3^-$ is the correct answer.
Either the formula or the name "hydrogen carbonate" is accepted.

Techniques used

recall the blood buffer ion

(b)

Compound E is the hydroxide of a Group 2 element. Compound E is a strong alkali.

$2.63\text{ g}$ of E is dissolved in water to make $250\text{ cm}^3$ of solution F. Solution F has a pH of 13.09 at $298\text{ K}$ .

4M

(i)

Show that the concentration of hydroxide ions in solution F is $0.123\text{ mol dm}^{-3}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

$[\text{H}^+] = 10^{-13.09} = 8.1 \times 10^{-14}\text{ mol dm}^{-3}$

$K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}$

$[\text{OH}^-] = \dfrac{1 \times 10^{-14}}{8.1 \times 10^{-14}} = 0.123\text{ mol dm}^{-3}$

Answer

$[\text{OH}^-] = 0.123\text{ mol dm}^{-3}$ (shown)

Final answer

[OH-] = 0.123 mol dm^-3

Detailed explanation

Background Concept

The pH of a solution is defined as $\text{pH} = -\log_{10}[\text{H}^+]$ . The ionic product of water at 298 K is $K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}\text{ mol}^2\text{ dm}^{-6}$ . These two relationships allow conversion between pH and $[\text{OH}^-]$ . An alternative route uses $\text{pOH} = 14 - \text{pH}$ and $[\text{OH}^-] = 10^{-\text{pOH}}$ .

Understanding the Question

Solution F has pH 13.09 at 298 K. We must show that $[\text{OH}^-] = 0.123\text{ mol dm}^{-3}$ . This is a "show that" question — the final value is given, so the working must demonstrate how it is obtained.

Approach

Method 1 (using $K_w$ ):

Calculate $[\text{H}^+]$ from pH.
Use $K_w = [\text{H}^+][\text{OH}^-]$ to find $[\text{OH}^-]$ .

Method 2 (using pOH):

$\text{pOH} = 14 - \text{pH}$ .
$[\text{OH}^-] = 10^{-\text{pOH}}$ .

Both methods are accepted by the mark scheme.

Step-by-Step Reasoning

Method 1 — using $K_w$ :

$[\text{H}^+] = 10^{-13.09} = 8.1 \times 10^{-14}\text{ mol dm}^{-3}$

Since $K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}$ :

$[\text{OH}^-] = \frac{1 \times 10^{-14}}{8.1 \times 10^{-14}} = 0.123\text{ mol dm}^{-3}$

Method 2 — using pOH:

$\text{pOH} = 14 - 13.09 = 0.91$

$[\text{OH}^-] = 10^{-0.91} = 0.123\text{ mol dm}^{-3}$

Both routes give the same result, confirming the value given in the question.

Key Takeaways

$\text{pH} + \text{pOH} = 14$ at 298 K.
$K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}$ at 298 K.
Either the $K_w$ route or the pOH route can be used to convert between pH and $[\text{OH}^-]$ .

Common Mistakes

Using $\text{pH} = -\log[\text{OH}^-]$ instead of $\text{pOH} = -\log[\text{OH}^-]$ .
Forgetting that $K_w = 1 \times 10^{-14}$ only at 298 K.
Arithmetic errors in evaluating $10^{-13.09}$ or $10^{-0.91}$ .

Things to Be Careful About

The mark scheme awards 1 mark for showing $10^{-13.09}$ or $[\text{H}^+] = 8.1 \times 10^{-14}$ , and 1 mark for the division by $K_w$ (or for showing pOH = 0.91 and $10^{-0.91}$ ).
Give the answer to at least 2 significant figures.

Techniques used

convert pH to hydrogen ion concentrationuse Kw to calculate hydroxide ion concentration

(ii)

Explain why the concentration of compound E in solution F is $0.0615\text{ mol dm}^{-3}$ .

1M Similar

DifficultyEasy

Worked solution

Answer

Each formula unit of E, X(OH) $_2$ , contains two hydroxide ions, so $[\text{E}] = \frac{[\text{OH}^-]}{2} = \frac{0.123}{2} = 0.0615\text{ mol dm}^{-3}$ .

Final answer

Each formula unit of X(OH)2 contains 2 OH- ions, so [E] = [OH-]/2 = 0.0615 mol dm^-3.

Detailed explanation

Background Concept

Group 2 elements form hydroxides with the general formula X(OH) $_2$ , where X is the Group 2 metal. When such a hydroxide dissolves completely (as a strong alkali does), each formula unit dissociates into one $\text{X}^{2+}$ ion and two $\text{OH}^-$ ions:
$\text{X(OH)}_2(\text{s}) \rightarrow \text{X}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})$

Understanding the Question

We have established that $[\text{OH}^-] = 0.123\text{ mol dm}^{-3}$ in solution F. The question asks us to explain why the concentration of compound E itself is exactly half of this value, $0.0615\text{ mol dm}^{-3}$ .

Approach

Recognise that E is a Group 2 hydroxide with formula X(OH) $_2$ . The stoichiometry of dissociation gives a 1:2 ratio between E and $\text{OH}^-$ .

Step-by-Step Reasoning

E is a Group 2 hydroxide, so its formula is X(OH) $_2$ . On complete dissociation:
$\text{X(OH)}_2 \rightarrow \text{X}^{2+} + 2\text{OH}^-$

Each mole of X(OH) $_2$ produces 2 moles of $\text{OH}^-$ ions. Therefore:
$[\text{X(OH)}_2] = \frac{[\text{OH}^-]}{2} = \frac{0.123}{2} = 0.0615\text{ mol dm}^{-3}$

Key Takeaways

Group 2 hydroxides have the formula X(OH) $_2$ , with two $\text{OH}^-$ per formula unit.
For a fully dissociated Group 2 hydroxide, $[\text{OH}^-] = 2 \times [\text{X(OH)}_2]$ .

Common Mistakes

Forgetting the factor of 2 — this is the entire point of the question.
Assuming E is a Group 1 hydroxide (XOH) which would give a 1:1 ratio.
Confusing the concentration of the compound with the concentration of its ions.

Things to Be Careful About

The stem tells us E is a "hydroxide of a Group 2 element" and a "strong alkali" — both clues point to X(OH) $_2$ fully dissociating.
The answer must explain WHY the factor of 2 applies, not just state the result.

Techniques used

relate hydroxide ion concentration to concentration of Group 2 hydroxide

(iii)

Use the concentration given in (ii) to identify compound E.

compound E: ................................................................................

1M Similar

DifficultyMedium-Easy

Worked solution

Working

moles of E $= 0.0615 \times 0.250 = 0.0154\text{ mol}$

molar mass of E $= \dfrac{2.63}{0.0154} = 171\text{ g mol}^{-1}$

E = X(OH) $_2$ , so RAM of X $= 171 - 34 = 137$

Answer

compound E is barium hydroxide, $\text{Ba(OH)}_2$

Final answer

Ba(OH)2 (barium hydroxide)

Detailed explanation

Background Concept

To identify an unknown compound from solution data, we use the relationships:

moles = concentration $\times$ volume (volume in dm $^3$ )
molar mass = mass / moles

For a Group 2 hydroxide X(OH) $_2$ , the molar mass is the sum of the atomic mass of X plus two hydroxide groups: $M_r = A_r(\text{X}) + 2 \times 17 = A_r(\text{X}) + 34$ .

Understanding the Question

We know the mass of E (2.63 g), the volume of solution F (250 cm $^3$ = 0.250 dm $^3$ ), and the concentration of E (0.0615 mol dm $^{-3}$ , from part (ii)). We must use these to identify which Group 2 hydroxide E is.

Approach

Calculate the number of moles of E in solution F.
Calculate the molar mass of E.
Subtract the mass of two OH groups to find the atomic mass of the metal, then identify it.

Step-by-Step Reasoning

Step 1 — moles of E:

moles $= 0.0615 \times 0.250 = 0.0154\text{ mol}$

Step 2 — molar mass of E:

$M_r = \frac{2.63}{0.0154} = 171\text{ g mol}^{-1}$

Step 3 — identify the metal:

E = X(OH) $_2$ , so:
$A_r(\text{X}) = 171 - 2(16.0 + 1.0) = 171 - 34 = 137$

Barium has atomic mass approximately 137. Therefore E is barium hydroxide, $\text{Ba(OH)}_2$ .

Key Takeaways

moles = concentration $\times$ volume (volume must be in dm $^3$ ).
molar mass = mass / moles.
For X(OH) $_2$ , subtract 34 from the molar mass to find the atomic mass of the metal.

Common Mistakes

Using volume in cm $^3$ instead of dm $^3$ (250 cm $^3$ = 0.250 dm $^3$ ).
Forgetting to subtract the mass of two OH groups (34) to find the metal's atomic mass.
Confusing barium (Ba, $\approx$ 137) with other Group 2 elements such as strontium (Sr, $\approx$ 88).

Things to Be Careful About

The mark scheme allows two routes: calculating $M_r$ of E directly (171) OR calculating the RAM of the metal (137). Either route leads to Ba(OH) $_2$ .
The mark is awarded for the identification "barium hydroxide / Ba(OH) $_2$ " — the calculation must support it.

Techniques used

calculate moles from concentration and volumecalculate molar mass from mass and molesidentify Group 2 element from molar mass

(c)

Compound E is much more soluble than magnesium hydroxide.

A saturated solution of magnesium hydroxide in water has a concentration of $1.40 \times 10^{-4}\text{ mol dm}^{-3}$ at $298\text{ K}$ .

Calculate the solubility product, $K_{\text{sp}}$ , of magnesium hydroxide. Include units.

K_{\text{sp}} = \text{.............................. units ..............................}

3M Similar

DifficultyMedium-Easy

Worked solution

Working

$\text{Mg(OH)}_2(\text{s}) \rightleftharpoons \text{Mg}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})$

$[\text{Mg}^{2+}] = 1.40 \times 10^{-4}\text{ mol dm}^{-3}$

$[\text{OH}^-] = 2 \times 1.40 \times 10^{-4} = 2.80 \times 10^{-4}\text{ mol dm}^{-3}$

$K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^-]^2 = 1.40 \times 10^{-4} \times (2.80 \times 10^{-4})^2 = 1.10 \times 10^{-11}$

Answer

$K_{\text{sp}} = 1.10 \times 10^{-11}\text{ mol}^3\text{ dm}^{-9}$

Final answer

1.10 x 10^-11 mol^3 dm^-9

Detailed explanation

Background Concept

The solubility product, $K_{\text{sp}}$ , is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. For magnesium hydroxide:
$\text{Mg(OH)}_2(\text{s}) \rightleftharpoons \text{Mg}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})$
The equilibrium expression is:
$K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^-]^2$

The concentration of $\text{Mg}^{2+}$ equals the solubility (the amount that dissolves per dm $^3$ ), and the concentration of $\text{OH}^-$ is twice the solubility because each formula unit produces two hydroxide ions.

Understanding the Question

We are given the concentration of a saturated solution of $\text{Mg(OH)}_2$ as $1.40 \times 10^{-4}\text{ mol dm}^{-3}$ . This is the solubility, which equals $[\text{Mg}^{2+}]$ . We must calculate $K_{\text{sp}}$ and include units. The question is worth 3 marks: expression, value, and units.

Approach

Write the dissociation equilibrium and the $K_{\text{sp}}$ expression.
Determine $[\text{Mg}^{2+}]$ and $[\text{OH}^-]$ from the given solubility.
Substitute into the expression and calculate.
Determine the units from the expression.

Step-by-Step Reasoning

Step 1 — Write the $K_{\text{sp}}$ expression:
$K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^-]^2$

Step 2 — Determine ion concentrations:

The solubility is $1.40 \times 10^{-4}\text{ mol dm}^{-3}$ , meaning that this amount of $\text{Mg(OH)}_2$ dissolves per dm $^3$ of saturated solution.

$[\text{Mg}^{2+}] = 1.40 \times 10^{-4}\text{ mol dm}^{-3}$ (one $\text{Mg}^{2+}$ per formula unit)

$[\text{OH}^-] = 2 \times 1.40 \times 10^{-4} = 2.80 \times 10^{-4}\text{ mol dm}^{-3}$ (two $\text{OH}^-$ per formula unit)

Step 3 — Substitute and calculate:
$K_{\text{sp}} = 1.40 \times 10^{-4} \times (2.80 \times 10^{-4})^2$
$= 1.40 \times 10^{-4} \times 7.84 \times 10^{-8}$
$= 1.10 \times 10^{-11}$

Step 4 — Determine units:

Units of $K_{\text{sp}}$ come from the expression: $[\text{Mg}^{2+}]$ has units mol dm $^{-3}$ and $[\text{OH}^-]^2$ has units (mol dm $^{-3}$ ) $^2$ = mol $^2$ dm $^{-6}$ . Multiplying:
$\text{mol dm}^{-3} \times \text{mol}^2\text{ dm}^{-6} = \text{mol}^3\text{ dm}^{-9}$

Key Takeaways

$K_{\text{sp}} = [\text{cation}][\text{anion}]^{\text{stoichiometric coefficient}}$ .
For a hydroxide X(OH) $_2$ , $[\text{OH}^-] = 2 \times$ solubility.
Units of $K_{\text{sp}}$ depend on the stoichiometry of the dissolution — always derive them from the expression.

Common Mistakes

Forgetting to square $[\text{OH}^-]$ in the $K_{\text{sp}}$ expression.
Using $[\text{OH}^-] = 1.40 \times 10^{-4}$ instead of $2.80 \times 10^{-4}$ .
Giving units as mol dm $^{-3}$ instead of mol $^3$ dm $^{-9}$ .
Rounding the final answer to 1 significant figure when the data allows 3.

Things to Be Careful About

The correct answer is $1.10 \times 10^{-11}\text{ mol}^3\text{ dm}^{-9}$ (minimum 2 significant figures).
The mark scheme awards: 1 mark for the expression, 1 mark for the value, 1 mark for the units.
ecf applies: if the expression is correct but the arithmetic is wrong, the units mark can still be earned.

Techniques used

write the Ksp expression for Mg(OH)2calculate ion concentrations from solubilitydetermine units of Ksp

(d)

Explain why compound E is much more soluble than magnesium hydroxide.

3M Similar

DifficultyMedium

Worked solution

Answer

Down Group 2, both lattice enthalpy and hydration enthalpy become less exothermic (less negative). However, the lattice enthalpy decreases more than the hydration enthalpy, so the enthalpy of solution becomes more exothermic (more negative) down the group. Therefore, barium hydroxide dissolves more readily than magnesium hydroxide.

Final answer

Down Group 2, lattice enthalpy decreases more than hydration enthalpy, so enthalpy of solution becomes more exothermic, making Ba(OH)2 more soluble than Mg(OH)2.

Detailed explanation

Background Concept

The solubility of an ionic compound is governed by the enthalpy of solution, which is the sum of the lattice enthalpy and the hydration enthalpy:
$\Delta H_{\text{sol}} = \Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$

Both $\Delta H_{\text{latt}}$ and $\Delta H_{\text{hyd}}$ are exothermic (negative) for ionic compounds. Lattice enthalpy is the energy released when gaseous ions come together to form the solid lattice; hydration enthalpy is the energy released when gaseous ions are surrounded by water molecules.

Down Group 2, the cations get larger (Mg $^{2+}$ < Ca $^{2+}$ < Sr $^{2+}$ < Ba $^{2+}$ ). Larger ions have lower charge density, so both the lattice enthalpy and the hydration enthalpy become less exothermic (less negative).

Understanding the Question

We must explain why barium hydroxide (compound E, identified in part (b)) is much more soluble than magnesium hydroxide. This is a Group 2 solubility trend question, worth 3 marks. The three marks correspond to: (1) both enthalpies become less exothermic, (2) lattice enthalpy changes more than hydration enthalpy, (3) enthalpy of solution becomes more exothermic.

Approach

Use the enthalpy of solution cycle: $\Delta H_{\text{sol}} = \Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$ . Analyse how each term changes down the group and, crucially, compare the RATES of change. The relative magnitudes determine the trend in $\Delta H_{\text{sol}}$ .

Step-by-Step Reasoning

Mark 1 — both become less exothermic:

Down Group 2 (from Mg to Ba), the cation radius increases. Both the lattice enthalpy and the hydration enthalpy become less exothermic (less negative).

Mark 2 — lattice enthalpy changes more:

The lattice enthalpy decreases more rapidly than the hydration enthalpy. This is because the lattice enthalpy depends on the sum of the cation and anion radii (both contribute to the interionic distance), while the hydration enthalpy depends mainly on the cation radius interacting with water. The larger change in lattice enthalpy dominates.

Mark 3 — enthalpy of solution becomes more exothermic:

Since $\Delta H_{\text{sol}} = \Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$ , and $\Delta H_{\text{latt}}$ becomes less negative faster than $\Delta H_{\text{hyd}}$ , the sum becomes more negative (more exothermic) down the group. A more exothermic enthalpy of solution makes dissolution more thermodynamically favourable, so solubility increases.

Therefore, barium hydroxide (further down the group than magnesium) has a more exothermic enthalpy of solution and is much more soluble.

Key Takeaways

$\Delta H_{\text{sol}} = \Delta H_{\text{latt}} + \Delta H_{\text{hyd}}$ .
Down Group 2, both $\Delta H_{\text{latt}}$ and $\Delta H_{\text{hyd}}$ become less exothermic, but $\Delta H_{\text{latt}}$ changes more.
The net effect is that $\Delta H_{\text{sol}}$ becomes more exothermic down the group, increasing solubility.
This explains why the solubility of Group 2 hydroxides increases down the group (opposite to the sulfates).

Common Mistakes

Saying "both become less exothermic, so solubility decreases" — this ignores the relative rates of change, which is the crucial point.
Not mentioning that lattice enthalpy changes more than hydration enthalpy — this is a required mark.
Confusing the hydroxide trend (solubility increases down the group) with the sulfate trend (solubility decreases down the group).
Using entropy arguments instead of enthalpy — the mark scheme specifically requires the enthalpy of solution reasoning.

Things to Be Careful About

The mark scheme awards 3 independent marks (marked independently — each can be earned separately).
The explanation must specifically address the RELATIVE magnitudes of the changes, not just state that both change.
Use precise terminology: "lattice enthalpy" (or "lattice energy"), "hydration enthalpy", and "enthalpy of solution".

Techniques used

compare lattice enthalpy trends down Group 2compare hydration enthalpy trends down Group 2relate enthalpy of solution to solubility

Q2MediumChemical Energetics

(a)

Predict and explain the variation in enthalpy change of hydration for the ions $\text{F}^-$ , $\text{Cl}^-$ , $\text{Br}^-$ and $\text{I}^-$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The enthalpy change of hydration becomes less negative (or less exothermic) down the group from F⁻ to I⁻. This is due to the increase in ionic radius/size, which results in decreased attraction to water molecules (weaker ion-dipole forces).

Final answer

The enthalpy change of hydration becomes less negative down the group due to increased ionic radius, resulting in weaker ion-dipole attractions to water.

Detailed explanation

Background Concept

The enthalpy change of hydration, $\Delta H_{\text{hyd}}$ , is the enthalpy change when one mole of gaseous ions is fully surrounded by water molecules to form infinitely dilute solution. It is always exothermic (negative) because ion-dipole attractions form between the ions and the polar water molecules.

Understanding the Question

The question asks to predict and explain the trend in $\Delta H_{\text{hyd}}$ for the halide ions F⁻, Cl⁻, Br⁻, and I⁻. We need to identify how the value changes as we move down Group 7 and provide a physical explanation for this change.

Approach

Consider the factors that determine the strength of ion-dipole interactions: the charge on the ion and the size (ionic radius) of the ion. Since all these ions have a 1⁻ charge, the difference must be due to their size. As we move down the group, the ionic radius increases.

Step-by-Step Reasoning

Trend prediction: As the ionic radius increases down the group (F⁻ < Cl⁻ < Br⁻ < I⁻), the charge density of the ion decreases. A lower charge density means the ion is less effective at attracting the partially positive hydrogen atoms of water molecules. Therefore, less energy is released when the ions are hydrated. The enthalpy change of hydration becomes less negative (less exothermic) down the group.
Explanation: The larger the ion, the greater the distance between the centre of the negative charge and the water dipoles. According to Coulomb's law, the force of attraction decreases with increasing distance. Thus, the ion-dipole forces between the larger halide ions and water are weaker, releasing less energy.

Key Takeaways

Hydration enthalpy becomes less exothermic for larger ions with the same charge because of weaker ion-dipole attractions due to increased ionic radius.

Common Mistakes

Stating that the enthalpy change becomes 'less negative' without explaining why.
Using the term 'atomic radius' instead of 'ionic radius'.
Saying 'weaker bonds' instead of 'weaker ion-dipole forces' or 'decreased attraction'.

Things to Be Careful About

Always specify that the trend is 'down the group' or 'from F⁻ to I⁻'.
Ensure you mention 'ionic radius' or 'ionic size', not just 'size'.
Use precise terminology: 'ion-dipole forces' or 'attraction to water', not 'bonding'.

Techniques used

predict trend in hydration enthalpyexplain trend using ionic radius and ion-dipole forces

(b)

Fig. 2.1 shows an incomplete energy cycle involving calcium fluoride, $\text{CaF}_2$ .

Fig. 2.1

6M

(i)

Complete line D. Include state symbols.

1M Similar

DifficultyEasy

Worked solution

Answer

\text{CaF}_2\text{(aq)}

or

\text{Ca}^{2+}\text{(aq)} + 2\text{F}^-\text{(aq)}

Final answer

CaF2(aq)

Detailed explanation

Background Concept

An energy cycle (or Born-Haber type cycle) for the solution of an ionic compound relates the enthalpy of formation, lattice enthalpy, hydration enthalpies, and enthalpy of solution. Line D represents the final state after the solution process (process 4) is complete.

Understanding the Question

We are given an incomplete energy cycle for CaF₂. Line A is the elements in their standard states, Line B is the gaseous ions, Line C is the solid ionic compound, and Line D is the unknown final state. Process 4 is the enthalpy change of solution, which takes the solid ionic compound to the aqueous state.

Approach

The enthalpy change of solution, $\Delta H_{\text{sol}}$ , is the enthalpy change when one mole of an ionic solid dissolves in water to form an infinitely dilute solution. The final state must therefore be the aqueous ions or the dissolved compound.

Step-by-Step Reasoning

Process 4 goes from Line C ( $\text{CaF}_2\text{(s)}$ ) to Line D. Since process 4 is the enthalpy change of solution, Line D must represent the aqueous state. This can be written as $\text{CaF}_2\text{(aq)}$ or, more explicitly, as the separated hydrated ions: $\text{Ca}^{2+}\text{(aq)} + 2\text{F}^-\text{(aq)}$ . Both are acceptable and must include state symbols.

Key Takeaways

In an energy cycle for solution, the final line (aqueous state) represents the ions dissolved in water, either as a formula with (aq) or as separated hydrated ions.

Common Mistakes

Forgetting the state symbol (aq).
Writing the aqueous ions without the correct stoichiometry (e.g., $\text{Ca}^{2+}\text{(aq)} + \text{F}^-\text{(aq)}$ ).

Things to Be Careful About

Always include state symbols in energy cycle diagrams. (aq) is required for the aqueous state.

Techniques used

complete energy cycle

(ii)

The value of the enthalpy change for process 1 can be calculated using the values of five other enthalpy changes which are not referred to in Fig. 2.1.

\text{process 1: } \text{Ca(s)} + \text{F}_2\text{(g)} \rightarrow \text{Ca}^{2+}\text{(g)} + 2\text{F}^-\text{(g)}

Identify these five other enthalpy changes, using either names or symbols.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Atomisation energy of Ca ( $\Delta H_{\text{at}}(\text{Ca})$ )
Atomisation energy of F₂ ( $\Delta H_{\text{at}}(\text{F}_2)$ ) or F–F bond energy
First ionisation energy of Ca ( $\text{IE}_1$ or $\Delta H_{\text{i1}}$ )
Second ionisation energy of Ca ( $\text{IE}_2$ or $\Delta H_{\text{i2}}$ )
First electron affinity of F ( $\text{EA}$ or $\Delta H_{\text{ea}}$ )

Final answer

Atomisation energy of Ca, atomisation energy of F2 (or F-F bond energy), first ionisation energy of Ca, second ionisation energy of Ca, first electron affinity of F.

Detailed explanation

Background Concept

Process 1 in the cycle is:

\text{Ca(s)} + \text{F}_2\text{(g)} \rightarrow \text{Ca}^{2+}\text{(g)} + 2\text{F}^-\text{(g)}

To break this down into standard enthalpy changes, we must consider the steps to convert the elements in their standard states to gaseous ions.

Understanding the Question

We need to identify the five specific enthalpy changes that, when summed, give the enthalpy change for process 1. These are the standard steps used in a Born-Haber cycle to form gaseous ions from elements.

Approach

Break down the overall process into sub-processes:

Convert Ca(s) to Ca(g) and then to Ca²⁺(g).
Convert F₂(g) to F(g) and then to F⁻(g).

Step-by-Step Reasoning

Atomisation of Ca: $\text{Ca(s)} \rightarrow \text{Ca(g)}$ requires the enthalpy of atomisation of calcium, $\Delta H_{\text{at}}(\text{Ca})$ .
Atomisation of F₂: $\frac{1}{2}\text{F}_2\text{(g)} \rightarrow \text{F(g)}$ requires half the enthalpy of atomisation of fluorine, or we can consider $\text{F}_2\text{(g)} \rightarrow 2\text{F(g)}$ which is the bond dissociation energy (or atomisation energy of F₂), $\Delta H_{\text{at}}(\text{F}_2)$ .
First ionisation of Ca: $\text{Ca(g)} \rightarrow \text{Ca}^+(\text{g}) + \text{e}^-$ requires the first ionisation energy, $\text{IE}_1$ or $\Delta H_{\text{i1}}$ .
Second ionisation of Ca: $\text{Ca}^+(\text{g}) \rightarrow \text{Ca}^{2+}\text{(g)} + \text{e}^-$ requires the second ionisation energy, $\text{IE}_2$ or $\Delta H_{\text{i2}}$ .
Electron affinity of F: $\text{F(g)} + \text{e}^- \rightarrow \text{F}^-(\text{g})$ requires the first electron affinity, $\text{EA}$ or $\Delta H_{\text{ea}}$ . Since we need $2\text{F}^-$ , this step occurs twice, but the question asks for the five types of enthalpy changes.

Key Takeaways

The formation of gaseous ions from elements involves atomisation (for both metal and non-metal), successive ionisation energies (for the metal), and electron affinity (for the non-metal).

Common Mistakes

Confusing atomisation of F₂ with bond dissociation energy (both are acceptable, but be precise).
Forgetting that Ca requires both first and second ionisation energies.
Including lattice energy or hydration enthalpy, which are not part of process 1.

Things to Be Careful About

Ensure you list five distinct types of enthalpy changes.
Use correct terminology: 'atomisation energy', 'ionisation energy', 'electron affinity'.

Techniques used

identify enthalpy changes in Born-Haber cycle

(iii)

Define lattice energy, $\Delta H_{\text{latt}}$ .

2M Similar

(iv)

Complete the expression to give the mathematical relationship between $\Delta H_{\text{latt}}$ of calcium fluoride and the enthalpy changes for processes 1 and 3.

\Delta H_{\text{latt}} = \text{................................................................................}

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\Delta H_{\text{latt}} = \Delta H_{\text{f}}(\text{CaF}_2\text{(s)}) - \Delta H_{\text{f}}(\text{Ca}^{2+}\text{(g)}) - 2\Delta H_{\text{f}}(\text{F}^-\text{(g)})

or simply:

\Delta H_{\text{latt}} = \text{process 3} - \text{process 1}

Final answer

ΔH_latt = ΔH_f(CaF2(s)) - ΔH_f(Ca2+(g)) - 2ΔH_f(F-(g))

Detailed explanation

Background Concept

Hess's law states that the total enthalpy change for a reaction is independent of the route taken. In an energy cycle, the sum of enthalpy changes around a closed loop must be zero, or the enthalpy change of one path must equal the enthalpy change of an alternative path between the same start and end points.

Understanding the Question

We need to find a relationship between $\Delta H_{\text{latt}}$ (which is the enthalpy change from gaseous ions to solid, i.e., Line B to Line C) and processes 1 and 3.

Approach

Identify the paths from Line B to Line C:

Direct path: Line B to Line C, which is the formation of the lattice, $\Delta H_{\text{latt}}$ .
Alternative path: Line B to Line A (reverse of process 1) then Line A to Line C (process 3).

Step-by-Step Reasoning

Path 1: $\Delta H_{\text{latt}}$ (from $\text{Ca}^{2+}\text{(g)} + 2\text{F}^-\text{(g)}$ to $\text{CaF}_2\text{(s)}$ )
Path 2: Reverse of process 1 + process 3
Reverse of process 1: $\text{Ca}^{2+}\text{(g)} + 2\text{F}^-\text{(g)} \rightarrow \text{Ca(s)} + \text{F}_2\text{(g)}$ , enthalpy change = $-\Delta H_{\text{process 1}}$
Process 3: $\text{Ca(s)} + \text{F}_2\text{(g)} \rightarrow \text{CaF}_2\text{(s)}$ , enthalpy change = $\Delta H_{\text{process 3}} = \Delta H_{\text{f}}(\text{CaF}_2\text{(s)})$

Equating the paths:

\Delta H_{\text{latt}} = \Delta H_{\text{process 3}} - \Delta H_{\text{process 1}}

Substituting the expressions for process 1 and process 3:

\Delta H_{\text{latt}} = \Delta H_{\text{f}}(\text{CaF}_2\text{(s)}) - [\Delta H_{\text{f}}(\text{Ca}^{2+}\text{(g)}) + 2\Delta H_{\text{f}}(\text{F}^-\text{(g)})]

\Delta H_{\text{latt}} = \Delta H_{\text{f}}(\text{CaF}_2\text{(s)}) - \Delta H_{\text{f}}(\text{Ca}^{2+}\text{(g)}) - 2\Delta H_{\text{f}}(\text{F}^-\text{(g)})

Key Takeaways

Use Hess's law to equate different paths around an energy cycle. Be careful with signs when reversing processes.

Common Mistakes

Adding process 1 instead of subtracting it.
Forgetting the stoichiometric coefficient (2) for the fluoride ion formation enthalpy.

Things to Be Careful About

Ensure the direction of the lattice enthalpy matches the definition (formation from gaseous ions to solid is negative; if the cycle defines it as dissociation, the sign would be positive. Here, process 3 - process 1 gives the formation enthalpy).

Techniques used

derive mathematical relationship from energy cycle

(c)

Use data from Table 2.1 to calculate a value for the hydration energy, $\Delta H_{\text{hyd}}$ , of fluoride ions, $\text{F}^-\text{(g)}$ .

Table 2.1

	value / $\text{kJ mol}^{-1}$
enthalpy change of solution of calcium fluoride, $\text{CaF}_2\text{(s)}$	+13
overall enthalpy change of process 1 in Fig. 2.1	+1395
enthalpy change of formation of calcium fluoride	-1214
enthalpy change of hydration of $\text{Ca}^{2+}\text{(g)}$	-1650

\Delta H_{\text{hyd}} \text{ F}^-\text{(g)} = \text{.............................. kJ mol}^{-1}

2M Similar

DifficultyMedium

Worked solution

Working

From the energy cycle, the sum of enthalpy changes along one path equals the sum along the alternative path:

\Delta H_{\text{process 1}} + \Delta H_{\text{process 2}} = \Delta H_{\text{process 3}} + \Delta H_{\text{process 4}}

Substitute the known values:

1395 + (-1650 + 2\Delta H_{\text{hyd}}\text{ F}^-) = -1214 + 13

-255 + 2\Delta H_{\text{hyd}}\text{ F}^- = -1201

2\Delta H_{\text{hyd}}\text{ F}^- = -1201 + 255

2\Delta H_{\text{hyd}}\text{ F}^- = -946

\Delta H_{\text{hyd}}\text{ F}^- = -473 \text{ kJ mol}^{-1}

Answer

-473

Final answer

-473

Detailed explanation

Background Concept

The energy cycle relates the enthalpy changes of different processes. According to Hess's law, the total enthalpy change is path-independent. We can write an equation equating the sum of enthalpy changes along one route to the sum along another route.

Understanding the Question

We are given four values from Table 2.1 and need to calculate $\Delta H_{\text{hyd}}$ for F⁻(g). The cycle has four processes:

Process 1: $\text{Ca(s)} + \text{F}_2\text{(g)} \rightarrow \text{Ca}^{2+}\text{(g)} + 2\text{F}^-\text{(g)}$ , $\Delta H = +1395 \text{ kJ mol}^{-1}$
Process 2: $\text{Ca}^{2+}\text{(g)} + 2\text{F}^-\text{(g)} \rightarrow \text{Ca}^{2+}\text{(aq)} + 2\text{F}^-\text{(aq)}$ , $\Delta H = -1650 + 2x$ (where $x = \Delta H_{\text{hyd}}\text{ F}^-$ )
Process 3: $\text{Ca(s)} + \text{F}_2\text{(g)} \rightarrow \text{CaF}_2\text{(s)}$ , $\Delta H = -1214 \text{ kJ mol}^{-1}$
Process 4: $\text{CaF}_2\text{(s)} \rightarrow \text{Ca}^{2+}\text{(aq)} + 2\text{F}^-\text{(aq)}$ , $\Delta H = +13 \text{ kJ mol}^{-1}$

Approach

Set up the equation: Process 1 + Process 2 = Process 3 + Process 4. Solve for $x$ .

Step-by-Step Reasoning

Write the Hess's law equation:

1395 + (-1650 + 2x) = -1214 + 13

Simplify the left side:

-255 + 2x = -1201

Rearrange to solve for $2x$ :

2x = -1201 + 255 = -946

Solve for $x$ :

x = \frac{-946}{2} = -473 \text{ kJ mol}^{-1}

Key Takeaways

Carefully set up the Hess's law equation from the cycle. Pay attention to signs and stoichiometric coefficients (the factor of 2 for fluoride hydration).

Common Mistakes

Getting the signs wrong (e.g., adding 1650 instead of subtracting).
Forgetting to multiply $\Delta H_{\text{hyd}}\text{ F}^-$ by 2.
Arithmetic errors in solving the algebraic equation.

Things to Be Careful About

Ensure all values are in the correct units (kJ mol⁻¹).
The final answer should be negative, as hydration is exothermic.

Techniques used

apply Hess's law to energy cyclecalculate unknown enthalpy

(d)

Define entropy.

1M Similar

(e)

At $298\text{ K}$ , the Gibbs free energy change, $\Delta G$ , for the solution of compound T is $+6.00\text{ kJ mol}^{-1}$ .

The enthalpy change of solution, $\Delta H_{\text{sol}}$ , of compound T is $+30.0\text{ kJ mol}^{-1}$ at $298\text{ K}$ .

Calculate the value of the entropy change, $\Delta S$ , for the solution of compound T at $298\text{ K}$ .

\Delta S = \text{.............................. J K}^{-1}\text{ mol}^{-1}

2M Similar

DifficultyMedium-Easy

Worked solution

Working

The Gibbs free energy equation is:

\Delta G = \Delta H - T\Delta S

Rearrange to solve for $\Delta S$ :

\Delta S = \frac{\Delta H - \Delta G}{T}

Given:
$\Delta G = +6.00 \text{ kJ mol}^{-1} = 6000 \text{ J mol}^{-1}$
$\Delta H = +30.0 \text{ kJ mol}^{-1} = 30000 \text{ J mol}^{-1}$
$T = 298 \text{ K}$

Substitute the values:

\Delta S = \frac{30000 - 6000}{298}

\Delta S = \frac{24000}{298}

\Delta S = +80.537... \text{ J K}^{-1}\text{ mol}^{-1}

Rounding to 3 significant figures:

\Delta S = +80.5 \text{ J K}^{-1}\text{ mol}^{-1}

Answer

+80.5

Final answer

+80.5

Detailed explanation

Background Concept

The Gibbs free energy change, $\Delta G$ , determines the feasibility of a process at constant temperature and pressure. The equation is:

\Delta G = \Delta H - T\Delta S

where $\Delta H$ is the enthalpy change, $T$ is the temperature in Kelvin, and $\Delta S$ is the entropy change. For a process to be feasible (spontaneous), $\Delta G$ must be negative.

Understanding the Question

We are given $\Delta G$ , $\Delta H$ , and $T$ for the solution of compound T, and we need to calculate $\Delta S$ . Note the units: $\Delta G$ and $\Delta H$ are in kJ mol⁻¹, but $\Delta S$ is required in J K⁻¹ mol⁻¹.

Approach

Rearrange the Gibbs equation to solve for $\Delta S$ , ensuring consistent units (convert kJ to J).

Step-by-Step Reasoning

Convert $\Delta G$ and $\Delta H$ to J mol⁻¹:
$\Delta G = 6.00 \times 1000 = 6000 \text{ J mol}^{-1}$
$\Delta H = 30.0 \times 1000 = 30000 \text{ J mol}^{-1}$
Rearrange the equation:
$\Delta S = \frac{\Delta H - \Delta G}{T}$
Substitute:
$\Delta S = \frac{30000 - 6000}{298} = \frac{24000}{298}$
Calculate:
$\Delta S = 80.537... \text{ J K}^{-1}\text{ mol}^{-1}$
Round to 3 significant figures (as the data is given to 3 sf):
$\Delta S = +80.5 \text{ J K}^{-1}\text{ mol}^{-1}$

Key Takeaways

Always check units when using the Gibbs equation. Enthalpy and free energy are often in kJ, but entropy is in J. Convert before calculating.

Common Mistakes

Forgetting to convert kJ to J, resulting in an answer that is off by a factor of 1000.
Using the wrong sign for $\Delta G$ or $\Delta H$ .
Incorrect algebraic rearrangement.

Things to Be Careful About

The question asks for the answer in J K⁻¹ mol⁻¹, not kJ K⁻¹ mol⁻¹.
Maintain appropriate significant figures (3 sf based on the given data).

Techniques used

apply Gibbs free energy equationcalculate entropy change

(f)

Predict whether compound T becomes more or less soluble as the water is heated from $298\text{ K}$ to $360\text{ K}$ . Explain your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Compound T becomes more soluble as the water is heated.

Explanation: The entropy change, $\Delta S$ , is positive. As temperature ( $T$ ) increases, the term $T\Delta S$ becomes more positive, making $-T\Delta S$ more negative. This causes $\Delta G$ to become more negative (or less positive), making the solution more feasible and thus increasing solubility.

Final answer

Compound T becomes more soluble as the temperature increases because the positive entropy change means that -TΔS becomes more negative, making ΔG more negative.

Detailed explanation

Background Concept

The solubility of a compound is related to the feasibility of its dissolution process. A process is feasible (spontaneous) when $\Delta G < 0$ . The Gibbs equation is:

\Delta G = \Delta H - T\Delta S

If $\Delta H$ is positive (endothermic dissolution) and $\Delta S$ is positive (increase in disorder), then at higher temperatures, the $-T\Delta S$ term becomes larger and more negative, eventually making $\Delta G$ negative.

Understanding the Question

We need to predict whether compound T becomes more or less soluble when heated from 298 K to 360 K, and explain why. We know from part (e) that $\Delta S$ is positive (+80.5 J K⁻¹ mol⁻¹) and $\Delta H$ is positive (+30.0 kJ mol⁻¹).

Approach

Analyze how the terms in the Gibbs equation change with increasing temperature, given the signs of $\Delta H$ and $\Delta S$ .

Step-by-Step Reasoning

At 298 K, $\Delta G = +6.00 \text{ kJ mol}^{-1}$ , which is positive, meaning the solution is not feasible (or only slightly soluble).
As $T$ increases to 360 K, the term $T\Delta S$ increases (since $\Delta S$ is positive).
Therefore, $-T\Delta S$ becomes more negative.
Since $\Delta H$ is constant (approximately), the overall $\Delta G = \Delta H - T\Delta S$ becomes more negative.
A more negative $\Delta G$ means the dissolution process becomes more feasible, so the compound becomes more soluble.

Key Takeaways

For endothermic dissolutions with a positive entropy change, solubility increases with temperature.

Common Mistakes

Saying 'solubility increases because $\Delta H$ is positive' without mentioning the role of $T\Delta S$ .
Confusing the effect of temperature on $\Delta G$ .
Not explicitly stating that $\Delta S$ is positive.

Things to Be Careful About

Ensure the explanation clearly links the positive $\Delta S$ to the temperature dependence of $\Delta G$ .
Use the correct terminology: 'more feasible', 'more negative $\Delta G$ '.

Techniques used

predict effect of temperature on solubility using Gibbs equation

Q3Medium-HardReaction KineticsTransition ElementsElectrochemistry

A and B react together to give product AB.

\text{A} + \text{B} \rightarrow \text{AB}

When the concentrations of A and B are both $0.0100\text{ mol dm}^{-3}$ , the rate of formation of AB is $7.62 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}$ . When the concentrations of A and B are both $0.0200\text{ mol dm}^{-3}$ , the rate of formation of AB is $3.05 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}$ .

(a)

5M

(i)

Complete the three possible rate equations that are consistent with these data.

rate = .................................................................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Doubling both $[\text{A}]$ and $[\text{B}]$ increases the rate by a factor of 4, so the overall order is 2. The orders can be 1 in each reactant, 2 in A only, or 2 in B only.

Answer

\text{rate} = k[\text{A}][\text{B}]

\text{rate} = k[\text{A}]^2

\text{rate} = k[\text{B}]^2

Final answer

rate = k[A][B]; rate = k[A]^2; rate = k[B]^2

Detailed explanation

Background Concept

A rate equation shows how the rate of a reaction depends on the concentrations of the reactants. For $\text{A} + \text{B} \rightarrow \text{AB}$ , the general form is $\text{rate} = k[\text{A}]^m[\text{B}]^n$ , where $m$ and $n$ are the orders with respect to A and B. The overall order is $m + n$ . When concentrations are changed, the rate changes by a factor equal to the concentration factor raised to the order for each reactant.

Understanding the Question

Two experiments are given. In the second experiment both $[\text{A}]$ and $[\text{B}]$ are doubled from $0.0100\text{ mol dm}^{-3}$ to $0.0200\text{ mol dm}^{-3}$ . We need to write three rate equations that are all consistent with the observed rate change. This is a deduction from data, not a calculation of $k$ .

Approach

Compare the two rates. Since both concentrations are doubled at the same time, the rate factor is $2^{m+n}$ . Calculate the rate ratio and set it equal to $2^{m+n}$ . Then list the integer combinations of $m$ and $n$ that give that overall order.

Step-by-Step Reasoning

The first rate is $7.62 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}$ and the second is $3.05 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}$ .

Rate ratio $= \frac{3.05 \times 10^{-3}}{7.62 \times 10^{-4}} = 4.00$ .

Both concentrations are doubled, so rate ratio $= 2^m \times 2^n = 2^{m+n}$ . Since $4 = 2^2$ , $m + n = 2$ .

Possible non-negative integer orders are:

$m = 1$ , $n = 1$ , giving $\text{rate} = k[\text{A}][\text{B}]$
$m = 2$ , $n = 0$ , giving $\text{rate} = k[\text{A}]^2$
$m = 0$ , $n = 2$ , giving $\text{rate} = k[\text{B}]^2$

These are the three required rate equations.

Key Takeaways

Rate data can fix the overall order of a reaction, but when both concentrations are changed together it cannot distinguish which reactant is responsible. Several rate equations may fit the same data.

Common Mistakes

Assuming the orders are equal to the stoichiometric coefficients in $\text{A} + \text{B} \rightarrow \text{AB}$ .
Forgetting that a zero-order reactant can be omitted from the rate equation.
Trying to deduce individual orders when only simultaneous changes are given.

Things to Be Careful About

Use the rate ratio, not the difference in rates.
Include $k$ in every rate equation.
The three equations must be written separately as the question asks for three possible equations.

Techniques used

deduce overall order from rate datawrite possible rate equations

(ii)

Choose one of the rate equations you have written in (i), and calculate the value of the rate constant, $k$ . Include the units of $k$ .

k = \text{.............................. units ..............................}

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $\text{rate} = k[\text{A}][\text{B}]$ with the first set of data:

k = \frac{7.62 \times 10^{-4}}{0.0100 \times 0.0100} = 7.62

Units:

\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}

Answer

$k = 7.62\text{ mol}^{-1}\text{ dm}^3\text{ s}^{-1}$

Final answer

7.62 mol^-1 dm^3 s^-1

Detailed explanation

Background Concept

The rate constant $k$ is the proportionality constant in a rate equation. Its value depends on temperature and catalyst, and its units depend on the overall order of the reaction. For a second-order reaction such as $\text{rate} = k[\text{A}][\text{B}]$ , the units of $k$ are $\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}$ .

Understanding the Question

We must choose one of the rate equations written in part (i), substitute one set of concentrations and the corresponding rate, and calculate $k$ with correct units. The mark scheme allows any of the three equations because they all give the same numeric $k$ here.

Approach

Use the first experiment with $[\text{A}] = [\text{B}] = 0.0100\text{ mol dm}^{-3}$ and rate $= 7.62 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}$ . Substitute into $\text{rate} = k[\text{A}][\text{B}]$ and solve for $k$ .

Step-by-Step Reasoning

$k = \frac{7.62 \times 10^{-4}}{0.0100 \times 0.0100} = 7.62$

The units are obtained from:
$\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}$

So $k = 7.62\text{ mol}^{-1}\text{ dm}^3\text{ s}^{-1}$ . Using the second experiment would give $3.05 \times 10^{-3} / 0.0200^2 = 7.625$ , the same value to 3 significant figures.

Key Takeaways

The units of a rate constant are determined by the overall order of the reaction. For second order, the concentration terms in the denominator are squared, giving $\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}$ .

Common Mistakes

Forgetting the units of $k$ .
Using the wrong concentration power in the denominator.
Quoting too few significant figures; the mark scheme requires at least 3 significant figures.

Things to Be Careful About

Use the chosen rate equation consistently.
The numeric value is the same for all three equations here because $[\text{A}] = [\text{B}]$ in the data.
Include the unit in the final answer.

Techniques used

calculate rate constant from rate equationdetermine units of rate constant

(iii)

Explain why it is not possible to calculate a value for the half-life, $t_{\frac{1}{2}}$ , of this reaction using the value of the rate constant $k$ calculated in (ii) and the equation $k = 0.693 / t_{\frac{1}{2}}$ .

1M Similar

(b)

Catalysts may be homogeneous or heterogeneous.

6M

(i)

Identify two metals that act as heterogeneous catalysts in the removal of $\text{NO}_2$ from the exhaust gases of car engines.

............................................................... and ..............................................................

1M Similar

(ii)

Iron acts as a heterogeneous catalyst in the Haber process.

Describe the mode of action of this iron catalyst.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{N}_2$ and $\text{H}_2$ adsorb onto the iron surface, forming bonds with surface atoms.
Adsorption weakens the $\text{N}\equiv\text{N}$ and H–H bonds.
$\text{NH}_3$ molecules form and desorb from the surface.

Final answer

Adsorption, bond weakening, desorption.

Detailed explanation

Background Concept

A heterogeneous catalyst is in a different phase from the reactants. In the Haber process, solid iron catalyses the reaction between gaseous $\text{N}_2$ and $\text{H}_2$ . The reaction takes place on the iron surface.

Understanding the Question

Describe the mode of action of the iron catalyst. The mark scheme rewards three key ideas: adsorption, bond weakening and desorption.

Approach

Think of the surface-catalysed process as a sequence: reactant molecules stick to the surface, their bonds are weakened, reaction occurs, and products leave the surface.

Step-by-Step Reasoning

$\text{N}_2$ and $\text{H}_2$ molecules adsorb onto the iron surface, forming bonds between the reactant atoms and surface iron atoms.
Adsorption weakens the covalent bonds within the reactants, especially the strong $\text{N}\equiv\text{N}$ triple bond, lowering the activation energy.
Reaction occurs on the surface, and the $\text{NH}_3$ molecules formed desorb, freeing the surface for more reactant molecules.

Key Takeaways

Heterogeneous catalysis involves adsorption, weakening of reactant bonds, reaction on the surface, and desorption of products.

Common Mistakes

Describing the catalyst as taking part in the reaction and being used up.
Omitting one of the three key stages; the mark scheme gives 2 marks for all three, 1 mark for any two.

Things to Be Careful About

Use the terms adsorption and desorption precisely.
Mention bond weakening, not just 'the catalyst provides a surface'.
Do not say the catalyst is consumed.

Techniques used

describe adsorption of reactants on catalyst surfaceexplain bond weakeningdescribe desorption of products

(iii)

$\text{Fe}^{2+}$ ions act as a homogeneous catalyst in the reaction between $\text{I}^-\text{(aq)}$ and $\text{S}_2\text{O}_8^{2-}\text{(aq)}$ .

Write equations for the two reactions that occur when $\text{Fe}^{2+}\text{(aq)}$ is added to a mixture of $\text{I}^-\text{(aq)}$ and $\text{S}_2\text{O}_8^{2-}\text{(aq)}$ .

equation 1: $\text{S}_2\text{O}_8^{2-} + \text{........................................................................................}$

equation 2: .......................................................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{S}_2\text{O}_8^{2-} + 2\text{Fe}^{2+} \rightarrow 2\text{SO}_4^{2-} + 2\text{Fe}^{3+}

2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2

Final answer

S2O8^2- + 2Fe^2+ -> 2SO4^2- + 2Fe^3+; 2Fe^3+ + 2I^- -> 2Fe^2+ + I2

Detailed explanation

Background Concept

A homogeneous catalyst is in the same phase as the reactants. $\text{Fe}^{2+}$ ions catalyse the redox reaction between $\text{I}^-$ and $\text{S}_2\text{O}_8^{2-}$ by providing an alternative pathway with a lower activation energy. The catalyst is oxidised in one step and regenerated in another.

Understanding the Question

Write the two reactions that occur when $\text{Fe}^{2+}$ is added to a mixture of $\text{I}^-$ and $\text{S}_2\text{O}_8^{2-}$ . The catalyst must be regenerated, so $\text{Fe}^{2+}$ is first oxidised to $\text{Fe}^{3+}$ , then $\text{Fe}^{3+}$ oxidises $\text{I}^-$ back to $\text{Fe}^{2+}$ .

Approach

Identify the oxidising and reducing agents. $\text{S}_2\text{O}_8^{2-}$ is a strong oxidising agent and oxidises $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ . The $\text{Fe}^{3+}$ formed then oxidises $\text{I}^-$ to $\text{I}_2$ , regenerating $\text{Fe}^{2+}$ .

Step-by-Step Reasoning

Equation 1: $\text{S}_2\text{O}_8^{2-} + 2\text{Fe}^{2+} \rightarrow 2\text{SO}_4^{2-} + 2\text{Fe}^{3+}$ .
This shows $\text{S}_2\text{O}_8^{2-}$ being reduced to sulfate and $\text{Fe}^{2+}$ being oxidised to $\text{Fe}^{3+}$ .

Equation 2: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$ .
This shows $\text{Fe}^{3+}$ being reduced back to $\text{Fe}^{2+}$ and iodide being oxidised to iodine.

Adding the two equations gives the overall reaction $\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2$ , with $\text{Fe}^{2+}$ regenerated.

Key Takeaways

A homogeneous catalyst is regenerated during the reaction. Writing the two steps of the catalytic cycle shows how the catalyst is oxidised and then reduced.

Common Mistakes

Writing the overall reaction instead of the two catalysed steps.
Not balancing charges and atoms in the equations.
Showing $\text{Fe}^{2+}$ being consumed without being regenerated.

Things to Be Careful About

Both equations are needed for the 2 marks.
The equations may be written with any whole-number multiples.
Check that atoms and charges balance.

Techniques used

write redox equations for catalytic cyclebalance charges and atoms

(iv)

Explain the difference between a homogeneous catalyst and a heterogeneous catalyst.

1M Similar

(c)

$\text{Fe}^{2+}$ ions can be oxidised to $\text{Fe}^{3+}$ ions under alkaline conditions by suitable oxidising agents.

5M

(i)

Iron is a transition element. Explain why iron forms stable compounds in both the +2 and the +3 oxidation states.

1M Similar

(ii)

The half-equation for the reduction of $\text{Fe}^{3+}$ under alkaline conditions, and its $E^\ominus$ value, are shown.

\text{Fe(OH)}_3 + \text{e}^- \rightleftharpoons \text{Fe(OH)}_2 + \text{OH}^- \quad E^\ominus = -0.56\text{ V}

Four more half-equations for reactions under alkaline conditions, and their $E^\ominus$ values, are shown.

\text{Al(OH)}_4^- + 3\text{e}^- \rightleftharpoons \text{Al} + 4\text{OH}^- \quad E^\ominus = -2.35\text{ V}

\text{ClO}^- + \text{H}_2\text{O} + 2\text{e}^- \rightleftharpoons \text{Cl}^- + 2\text{OH}^- \quad E^\ominus = +0.89\text{ V}

\text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \rightleftharpoons 4\text{OH}^- \quad E^\ominus = +0.40\text{ V}

\text{Zn(OH)}_4^{2-} + 2\text{e}^- \rightleftharpoons \text{Zn} + 4\text{OH}^- \quad E^\ominus = -1.22\text{ V}

Select two oxidising agents that can oxidise $\text{Fe}^{2+}$ ions to $\text{Fe}^{3+}$ ions under alkaline conditions.

Write an equation, and give the $E^\ominus_{\text{cell}}$ value, for each of the two reactions that occur.

oxidising agent 1: ..............................

equation: ...........................................................................................................................

E^\ominus_{\text{cell}} = \text{.............................. V}

oxidising agent 2: ..............................

equation: ...........................................................................................................................

E^\ominus_{\text{cell}} = \text{.............................. V}

4M Similar

DifficultyMedium-Hard

Worked solution

Working

Under alkaline conditions, $\text{Fe}^{2+}$ is present as $\text{Fe(OH)}_2$ and $\text{Fe}^{3+}$ as $\text{Fe(OH)}_3$ . An oxidising agent must have a reduction potential more positive than $-0.56\text{ V}$ . From the list, $\text{ClO}^-\ (+0.89\text{ V})$ and $\text{O}_2\ (+0.40\text{ V})$ qualify.

Answer

Oxidising agent 1: $\text{ClO}^-$

2\text{Fe(OH)}_2 + \text{ClO}^- + \text{H}_2\text{O} \rightarrow \text{Cl}^- + 2\text{Fe(OH)}_3

E^\ominus_{\text{cell}} = +0.89 - (-0.56) = +1.45\text{ V}

Oxidising agent 2: $\text{O}_2$

4\text{Fe(OH)}_2 + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{Fe(OH)}_3

E^\ominus_{\text{cell}} = +0.40 - (-0.56) = +0.96\text{ V}

Final answer

ClO^- and O2; Ecell = +1.45 V and +0.96 V; equations as above.

Detailed explanation

Background Concept

Electrode potentials measure the tendency of a half-reaction to occur as a reduction. The more positive the $E$ value, the stronger the oxidising agent. For a spontaneous redox reaction, the cell potential $E_{\text{cell}} = E_{\text{reduction}} - E_{\text{oxidation}}$ must be positive.

Understanding the Question

Under alkaline conditions, $\text{Fe}^{2+}$ is present as $\text{Fe(OH)}_2$ and $\text{Fe}^{3+}$ as $\text{Fe(OH)}_3$ . The half-reaction $\text{Fe(OH)}_3 + \text{e}^- \rightleftharpoons \text{Fe(OH)}_2 + \text{OH}^-$ has $E^\ominus = -0.56\text{ V}$ . We must choose two oxidising agents from the list that can oxidise $\text{Fe(OH)}_2$ to $\text{Fe(OH)}_3$ , write the balanced equation for each, and give $E_{\text{cell}}$ .

Approach

An oxidising agent must have a reduction potential more positive than $-0.56\text{ V}$ . Compare the given $E^\ominus$ values. $\text{ClO}^-\ (+0.89\text{ V})$ and $\text{O}_2\ (+0.40\text{ V})$ qualify; Al and Zn do not. Then combine the reduction half-equation of the oxidising agent with the reverse of the $\text{Fe(OH)}_3/\text{Fe(OH)}_2$ half-equation, balancing electrons.

Step-by-Step Reasoning

For $\text{ClO}^-$ :
Reduction: $\text{ClO}^- + \text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{Cl}^- + 2\text{OH}^-\quad E^\ominus = +0.89\text{ V}$
Oxidation of $\text{Fe(OH)}_2$ : $\text{Fe(OH)}_2 + \text{OH}^- \rightarrow \text{Fe(OH)}_3 + \text{e}^-$ (reverse of given half-equation)
Multiply the oxidation by 2 and add:
$2\text{Fe(OH)}_2 + 2\text{OH}^- + \text{ClO}^- + \text{H}_2\text{O} + 2\text{e}^- \rightarrow 2\text{Fe(OH)}_3 + 2\text{e}^- + \text{Cl}^- + 2\text{OH}^-$
Cancel $2\text{OH}^-$ and $2\text{e}^-$ :
$2\text{Fe(OH)}_2 + \text{ClO}^- + \text{H}_2\text{O} \rightarrow 2\text{Fe(OH)}_3 + \text{Cl}^-$
$E_{\text{cell}} = +0.89 - (-0.56) = +1.45\text{ V}$ .

For $\text{O}_2$ :
Reduction: $\text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \rightarrow 4\text{OH}^-\quad E^\ominus = +0.40\text{ V}$
Oxidation: $4\text{Fe(OH)}_2 + 4\text{OH}^- \rightarrow 4\text{Fe(OH)}_3 + 4\text{e}^-$
Add and cancel $4\text{OH}^-$ and $4\text{e}^-$ :
$4\text{Fe(OH)}_2 + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{Fe(OH)}_3$
$E_{\text{cell}} = +0.40 - (-0.56) = +0.96\text{ V}$ .

Al and Zn are not chosen because their $E^\ominus$ values are much more negative than $-0.56\text{ V}$ , so they would act as reducing agents rather than oxidising agents for $\text{Fe}^{2+}$ .

Key Takeaways

A species can oxidise another only if its reduction potential is more positive than the reduction potential of the couple being oxidised. The cell potential is the difference between the two reduction potentials.

Common Mistakes

Choosing Al or Zn because they are metals.
Writing the Fe half-reaction in the wrong direction.
Not balancing electrons when combining half-equations.
Forgetting to cancel $\text{OH}^-$ ions in the final equation.

Things to Be Careful About

Use the alkaline-condition half-equations exactly as given.
$E_{\text{cell}}$ must be positive for a feasible reaction.
The mark scheme requires both the correct oxidising agents and the correct $E_{\text{cell}}$ values, as well as the balanced equations.

Techniques used

compare electrode potentials to choose oxidising agentscombine half-equationscalculate Ecellbalance redox equations under alkaline conditions

Q4MediumTransition Elements

Transition metal atoms and transition metal ions form complexes by combining with ligands.

(a)

Explain why transition elements form complex ions.

1M Similar

DifficultyEasy

Worked solution

Answer

Transition elements have empty (d) orbitals that are energetically accessible, allowing them to accept lone pairs from ligands and form dative covalent bonds.

Final answer

Empty (d) orbitals are energetically accessible and can accept lone pairs from ligands to form dative bonds.

Detailed explanation

Background Concept

Complex ion formation involves a central metal ion acting as a Lewis acid (electron pair acceptor) and ligands acting as Lewis bases (electron pair donors). The bond formed is a dative covalent (coordinate) bond. For this to occur, the metal ion must have available orbitals of suitable energy to accept the electron pairs.

Understanding the Question

The question asks for the fundamental reason why transition elements specifically are able to form complex ions. The command word is "explain", so a reason must be given, not just a statement.

Approach

Identify what makes transition metals special in terms of their electronic structure — they have partially filled d-orbitals, and importantly, they have empty d-orbitals (or d-orbitals that can accommodate additional electron pairs) that are energetically accessible for bonding.

Step-by-Step Reasoning

Transition metal atoms have electron configurations involving (n-1)d and ns orbitals. When they form ions, the d-orbitals remain partially filled, and there are empty d-orbitals (or d-orbitals of appropriate energy) available. These empty d-orbitals can accept lone pairs of electrons from ligands (such as H₂O, NH₃, CN⁻, Cl⁻), forming coordinate (dative) covalent bonds. This is the defining feature that allows transition metals to form complex ions.

Key Takeaways

The key requirement for complex formation is available orbitals of suitable energy to accept electron pairs.
Transition metals have empty (or partially filled) d-orbitals that are energetically accessible.
The bond formed is a dative covalent bond (coordinate bond).

Common Mistakes

Saying only "they have d-orbitals" without specifying they are empty or energetically accessible.
Saying they form ionic bonds with ligands (the bond is dative covalent).
Mentioning variable oxidation states as the reason for complex formation (that is a consequence, not the cause).

Things to Be Careful About

The mark scheme specifically requires mention of "empty (d) orbitals" and either "energetically accessible" or their ability to accept lone pairs / form dative bonds. Simply saying "they have d-orbitals" is insufficient.

Techniques used

identify role of d-orbitals in bondingexplain dative covalent bond formation

(b)

$\text{Co}^{2+}$ ions form complex ion G.

Each G ion contains two $\text{Co}^{2+}$ ions, both of which are octahedrally coordinated.

Each G ion contains one $\text{O}_2$ molecule, which donates one pair of electrons to each $\text{Co}^{2+}$ ion, and one $\text{NH}_2^-$ ion, which donates one pair of electrons to each $\text{Co}^{2+}$ ion.

The remaining ligands are $\text{NH}_3$ molecules.

5M

(i)

Deduce the formula of complex ion G. Include its overall charge.

formula of G: .........................................................

2M Similar

DifficultyMedium

Worked solution

Working

Each $\text{Co}^{2+}$ is octahedrally coordinated (coordination number = 6).

The $\text{O}_2$ molecule donates one pair to each $\text{Co}^{2+}$ : occupies 1 site per Co.
The $\text{NH}_2^-$ ion donates one pair to each $\text{Co}^{2+}$ : occupies 1 site per Co.

Remaining sites per $\text{Co}^{2+}$ : $6 - 1 - 1 = 4$ sites occupied by $\text{NH}_3$ .

Total $\text{NH}_3$ ligands: $4 \times 2 = 8$ .

Charge: $2(+2) + 0 + (-1) + 8(0) = +3$ .

Answer

$[\text{Co}_2\text{O}_2\text{NH}_2(\text{NH}_3)_8]^{3+}$

Final answer

[Co₂O₂NH₂(NH₃)₈]³⁺

Detailed explanation

Background Concept

In a complex ion, the coordination number is the total number of coordinate bonds from ligands to the central metal ion. An octahedral complex has a coordination number of 6. When a ligand bridges two metal centres (as $\text{O}_2$ and $\text{NH}_2^-$ do here), it donates one electron pair to each metal, occupying one coordination site on each. The overall charge of the complex is the sum of the metal ion charges and the ligand charges.

Understanding the Question

We are told the complex contains 2 $\text{Co}^{2+}$ ions (both octahedral), one $\text{O}_2$ molecule bridging both (donating one pair to each), one $\text{NH}_2^-$ ion bridging both (donating one pair to each), and the rest are $\text{NH}_3$ molecules. We must deduce the full formula including charge.

Approach

Determine how many coordination sites each ligand occupies on each Co.
Subtract these from 6 (octahedral) to find how many $\text{NH}_3$ per Co.
Multiply by 2 for the total number of $\text{NH}_3$ .
Sum all charges for the overall charge.

Step-by-Step Reasoning

Each Co²⁺ has coordination number 6 (octahedral).
$\text{O}_2$ donates 1 pair to each Co → 1 site used per Co.
$\text{NH}_2^-$ donates 1 pair to each Co → 1 site used per Co.
Sites remaining per Co for $\text{NH}_3$ : $6 - 1 - 1 = 4$ .
Total $\text{NH}_3$ in the complex: $4 \times 2 = 8$ .
Charge calculation: $2 \times (+2) + 0$ (for $\text{O}_2$ ) $+ (-1)$ (for $\text{NH}_2^-$ ) $+ 8 \times 0$ (for $\text{NH}_3$ ) $= +4 - 1 = +3$ .
Formula: $[\text{Co}_2\text{O}_2\text{NH}_2(\text{NH}_3)_8]^{3+}$ .

Key Takeaways

Bridging ligands occupy one coordination site on each metal centre they bridge.
The overall charge is found by summing all ionic charges within the complex.
Octahedral geometry always means coordination number 6.

Common Mistakes

Forgetting that $\text{O}_2$ and $\text{NH}_2^-$ each occupy a site on BOTH cobalt centres (not just one).
Miscounting $\text{NH}_3$ as 4 total instead of 8.
Getting the charge wrong by forgetting $\text{NH}_2^-$ contributes $-1$ .

Things to Be Careful About

The mark scheme requires all three components for 2 marks: $\text{Co}_2\text{O}_2\text{NH}_2$ (the bridging ligands), 8 ammonia molecules, and the correct $3+$ charge. Getting any two of these earns 1 mark.

Techniques used

deduce ligand count from coordination numbercalculate overall charge from component charges

(ii)

The d-orbitals of the $\text{Co}^{2+}$ ions present in complex ion G are split. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level in each $\text{Co}^{2+}$ ion.


number of d-orbitals at a higher energy level
number of d-orbitals at a lower energy level

1M Similar

DifficultyEasy

Worked solution

Answer


number of d-orbitals at a higher energy level	2
number of d-orbitals at a lower energy level	3

Final answer

Higher: 2, Lower: 3

Detailed explanation

Background Concept

In an octahedral complex, the five d-orbitals of the central metal ion are split into two groups by the electrostatic interaction with the six ligands. The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (the $e_g$ set) point directly at the ligands and are raised in energy. The $d_{xy}$ , $d_{xz}$ , and $d_{yz}$ orbitals (the $t_{2g}$ set) point between the ligands and are lowered in energy relative to the barycentre.

Understanding the Question

The question asks for the number of d-orbitals at each energy level in an octahedral $\text{Co}^{2+}$ complex. This is a straightforward recall of the splitting pattern.

Approach

Recall that octahedral splitting gives a 3 + 2 pattern: 3 orbitals at lower energy ( $t_{2g}$ ) and 2 orbitals at higher energy ( $e_g$ ).

Step-by-Step Reasoning

Five d-orbitals split in an octahedral field.
The $e_g$ set ( $d_{x^2-y^2}$ , $d_{z^2}$ ): 2 orbitals at higher energy.
The $t_{2g}$ set ( $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ): 3 orbitals at lower energy.
The mark scheme requires BOTH values for the single mark.

Key Takeaways

Octahedral splitting always produces a 3 + 2 pattern regardless of the metal ion.
The 3 lower orbitals are $t_{2g}$ and the 2 upper are $e_g$ .

Common Mistakes

Confusing octahedral (3 lower, 2 higher) with tetrahedral (2 lower, 3 higher) splitting.
Only giving one value (both are needed for the mark).

Things to Be Careful About

The mark scheme explicitly states "BOTH" are required — giving only one number earns zero marks.

Techniques used

recall octahedral d-orbital splitting pattern

(iii)

$\text{Co}^{2+}$ ions form a different complex ion, M.

Each M ion contains two $\text{Co}^{2+}$ ions, both of which are octahedrally coordinated, but the ligands are different from the ligands in G.

Explain why G and M have different colours.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

G and M have different ligands, so the energy gap ( $\Delta E$ ) between the split d-orbitals is different in each complex. This means a different frequency (wavelength/energy) of visible light is absorbed, resulting in different observed colours.

Final answer

Different ligands cause different ΔE (d-d energy gap), so different frequencies of visible light are absorbed.

Detailed explanation

Background Concept

The colour of a transition metal complex arises from d-d electron transitions. When visible light strikes the complex, an electron in a lower-energy d-orbital absorbs a photon and is promoted to a higher-energy d-orbital. The energy of the absorbed photon equals $\Delta E$ , the splitting between the two d-orbital sets. The colour observed is the complementary colour of the light absorbed. The magnitude of $\Delta E$ depends on the identity of the ligands (spectrochemical series), the oxidation state of the metal, and the geometry of the complex.

Understanding the Question

We are told that G and M both contain two octahedrally coordinated $\text{Co}^{2+}$ ions but have different ligands. We must explain why they have different colours. The command word is "explain", requiring cause-and-effect reasoning.

Approach

The chain of reasoning is: different ligands → different $\Delta E$ → different frequency of light absorbed → different colour observed.

Step-by-Step Reasoning

G and M have different ligands (stated in the question).
Different ligands produce different crystal field splitting energies ( $\Delta E$ ) — this is the first mark.
Since $\Delta E = hf = hc/\lambda$ , a different $\Delta E$ means a different frequency (or wavelength or energy) of visible light is absorbed — this is the second mark.
The colour observed is the complementary colour of the absorbed light, so different absorbed wavelengths give different observed colours.

Key Takeaways

Colour in transition metal complexes is due to d-d transitions.
The energy gap $\Delta E$ determines which wavelength is absorbed.
Ligand identity is a key factor in determining $\Delta E$ .
The observed colour is complementary to the absorbed colour.

Common Mistakes

Saying "different ligands give different colours" without mentioning the energy gap or absorbed light.
Not specifying that the light absorbed is in the visible region.
Confusing absorbed colour with observed colour.

Things to Be Careful About

The mark scheme requires both the $\Delta E$ point AND the different frequency/wavelength/energy of visible light absorbed point. The word "visible" is important — the mark scheme specifically mentions visible light.

Techniques used

link ligand identity to d-orbital splitting magnitudeconnect energy gap to wavelength of light absorbed

(c)

Cadmium forms complex ion X, $[\text{Cd(NH}_3)_4]^{2+}$ .

When a solution containing $\text{CN}^-$ ions is added to an aqueous solution of X, a ligand exchange reaction takes place, forming complex ion Y. Y contains no $\text{NH}_3$ ligands and no $\text{H}_2\text{O}$ ligands.

Y is in a much higher concentration in the mixture than X.

The oxidation state and coordination number of cadmium do not change in this reaction.

3M

(i)

Write an ionic equation for this reaction, using the formulae of the complex ions.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

[\text{Cd(NH}_3)_4]^{2+} + 4\text{CN}^- \rightarrow [\text{Cd(CN)}_4]^{2-} + 4\text{NH}_3

Answer

$[\text{Cd(NH}_3)_4]^{2+} + 4\text{CN}^- \rightarrow [\text{Cd(CN)}_4]^{2-} + 4\text{NH}_3$

Final answer

[Cd(NH₃)₄]²⁺ + 4CN⁻ → [Cd(CN)₄]²⁻ + 4NH₃

Detailed explanation

Background Concept

Ligand exchange reactions occur when one set of ligands in a complex is replaced by another set. The stability constant $K_{\text{stab}}$ quantifies the extent of complex formation — a larger $K_{\text{stab}}$ means the complex is more stable and forms preferentially. In this question, $\text{CN}^-$ displaces $\text{NH}_3$ from the cadmium complex because $[\text{Cd(CN)}_4]^{2-}$ has a much larger $K_{\text{stab}}$ than $[\text{Cd(NH}_3)_4]^{2+}$ .

Understanding the Question

We are told:

X is $[\text{Cd(NH}_3)_4]^{2+}$
$\text{CN}^-$ is added, causing ligand exchange
Y contains no $\text{NH}_3$ and no $\text{H}_2\text{O}$ ligands
Oxidation state and coordination number of Cd do not change

Since Cd stays in the +2 oxidation state and coordination number 4, and all $\text{NH}_3$ are replaced by $\text{CN}^-$ , Y must be $[\text{Cd(CN)}_4]^{2-}$ .

Approach

Identify the product complex Y from the constraints.
Write the balanced ionic equation showing all $\text{NH}_3$ replaced by $\text{CN}^-$ .

Step-by-Step Reasoning

Cd in X: oxidation state +2, coordination number 4, four $\text{NH}_3$ ligands.
After exchange: Cd still +2, still CN=4, all ligands are $\text{CN}^-$ (since Y has no $\text{NH}_3$ or $\text{H}_2\text{O}$ ).
Charge of Y: $(+2) + 4(-1) = -2$ , so Y is $[\text{Cd(CN)}_4]^{2-}$ .
Equation: $[\text{Cd(NH}_3)_4]^{2+} + 4\text{CN}^- \rightarrow [\text{Cd(CN)}_4]^{2-} + 4\text{NH}_3$
Check charge balance: left side = +2 + 4(-1) = -2; right side = -2 + 0 = -2. ✓
Check atom balance: 4 NH₃ on each side, 4 CN on each side, 1 Cd on each side. ✓

Key Takeaways

In ligand exchange, the metal centre's oxidation state and coordination number are often preserved.
The charge of the product complex must be calculated from the metal oxidation state and ligand charges.
Ionic equations for complex formation/exchange must be balanced in both atoms and charge.

Common Mistakes

Writing $[\text{Cd(CN)}_4]^{2+}$ (wrong charge — forgetting each $\text{CN}^-$ contributes -1).
Writing the equation with $\text{NaCN}$ instead of $\text{CN}^-$ (it asks for an ionic equation).
Forgetting to include the 4 $\text{NH}_3$ on the product side.

Things to Be Careful About

The equation must use the formulae of the complex ions as given (i.e., write $[\text{Cd(NH}_3)_4]^{2+}$ not just $\text{Cd}^{2+}$ ). The mark scheme gives 1 mark for the correct equation and allows ECF for the second mark.

Techniques used

deduce formula of product complex from ligand exchangebalance ionic equation with charges

(ii)

Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All the ligands in Z are $\text{Cl}^-$ ions.

When $\text{NaCl(aq)}$ is added to a solution of X, very little Z forms.

Write the three cadmium complexes, $\mathbf{X}$ , $\mathbf{Y}$ and $\mathbf{Z}$ , in order of increasing stability constant, $K_{\text{stab}}$ .

\text{..........................................} \quad \text{..........................................} \quad \text{..........................................}

\text{smallest value of } K_{\text{stab}} \hspace{150pt} \text{largest value of } K_{\text{stab}}

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{Z} < \text{X} < \text{Y}

(smallest $K_{\text{stab}}$ ) $\quad$ (largest $K_{\text{stab}}$ )

That is: $[\text{CdCl}_4]^{2-} < [\text{Cd(NH}_3)_4]^{2+} < [\text{Cd(CN)}_4]^{2-}$

Final answer

Z < X < Y

Detailed explanation

Background Concept

The stability constant $K_{\text{stab}}$ measures how readily a complex forms from its constituent ions. A larger $K_{\text{stab}}$ means the complex is more thermodynamically stable. In a competition between two ligands for the same metal centre, the ligand forming the more stable complex (higher $K_{\text{stab}}$ ) will displace the other. If adding a ligand causes very little of the new complex to form, the new complex has a lower $K_{\text{stab}}$ than the existing one.

Understanding the Question

We have three complexes:

X: $[\text{Cd(NH}_3)_4]^{2+}$
Y: $[\text{Cd(CN)}_4]^{2-}$
Z: $[\text{CdCl}_4]^{2-}$

From the question: Y forms readily from X (Y is in much higher concentration than X when CN⁻ is added), so $K_{\text{stab}}(Y) > K_{\text{stab}}(X)$ . Very little Z forms when NaCl is added to X, so $K_{\text{stab}}(Z) < K_{\text{stab}}(X)$ .

Approach

Use the experimental evidence to rank the stability constants:

CN⁻ displaces NH₃ → Y more stable than X.
Cl⁻ does NOT displace NH₃ → Z less stable than X.

Step-by-Step Reasoning

When CN⁻ is added to X, the reaction proceeds to form Y in much higher concentration than X remains. This means the equilibrium strongly favours Y, so $K_{\text{stab}}(Y) > K_{\text{stab}}(X)$ .
When Cl⁻ is added to X, very little Z forms. This means the equilibrium does not favour Z over X, so $K_{\text{stab}}(Z) < K_{\text{stab}}(X)$ .
Combining: $K_{\text{stab}}(Z) < K_{\text{stab}}(X) < K_{\text{stab}}(Y)$ .
Order of increasing $K_{\text{stab}}$ : Z, X, Y.

Key Takeaways

A ligand that displaces another from a complex forms the more stable complex (higher $K_{\text{stab}}$ ).
If a ligand cannot displace the existing ligands, its complex has a lower $K_{\text{stab}}$ .
The spectrochemical series predicts relative ligand strengths: CN⁻ > NH₃ > Cl⁻.

Common Mistakes

Reversing the order (putting Y as smallest).
Confusing the order of increasing with decreasing $K_{\text{stab}}$ .

Things to Be Careful About

The question asks for increasing order (smallest to largest). The mark scheme shows Z, X, Y from left to right.

Techniques used

rank complexes by stability from experimental evidence

(d)

Ethanedioate ions, $\text{C}_2\text{O}_4^{2-}$ , form complexes with transition element ions.

The concentration of $\text{C}_2\text{O}_4^{2-}$ ions can be found by reaction with acidified $\text{Cr}_2\text{O}_7^{2-}$ ions. $\text{C}_2\text{O}_4^{2-}$ ions are protonated and form $\text{HOOCCOOH}$ molecules which are oxidised by $\text{Cr}_2\text{O}_7^{2-}$ .

The half-equations are shown.

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightleftharpoons 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

2\text{CO}_2 + 2\text{H}^+ + 2\text{e}^- \rightleftharpoons \text{HOOCCOOH}

3M

(i)

Construct an equation for the reaction between acidified $\text{Cr}_2\text{O}_7^{2-}$ and $\text{HOOCCOOH}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Reduction (×1): $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

Oxidation (×3): $3\text{HOOCCOOH} \rightarrow 6\text{CO}_2 + 6\text{H}^+ + 6\text{e}^-$

Adding and cancelling $6\text{e}^-$ and $6\text{H}^+$ :

Answer

\text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{HOOCCOOH} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{CO}_2

Final answer

Cr₂O₇²⁻ + 8H⁺ + 3HOOCCOOH → 2Cr³⁺ + 7H₂O + 6CO₂

Detailed explanation

Background Concept

To combine half-equations into an overall redox equation, the number of electrons lost in oxidation must equal the number gained in reduction. One half-equation is multiplied by an appropriate factor so that electrons cancel. Species appearing on both sides (such as H⁺ or H₂O) are then simplified.

Understanding the Question

We are given two half-equations and asked to construct the overall equation for the reaction between acidified $\text{Cr}_2\text{O}_7^{2-}$ and $\text{HOOCCOOH}$ . The reduction half-equation is written as a reduction (gaining electrons), and the oxidation half-equation is written in reverse (the given equation shows $\text{CO}_2$ being reduced to $\text{HOOCCOOH}$ , so we reverse it to show $\text{HOOCCOOH}$ being oxidised to $\text{CO}_2$ ).

Approach

Write the reduction half-equation as given (×1, involves 6e⁻).
Reverse the oxidation half-equation and multiply by 3 (to give 6e⁻ lost).
Add together, cancelling electrons.
Simplify H⁺ terms (14 on left, 6 on right → net 8 on left).

Step-by-Step Reasoning

Reduction: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
Oxidation (reversed and ×3): $3\text{HOOCCOOH} \rightarrow 6\text{CO}_2 + 6\text{H}^+ + 6\text{e}^-$
Add: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 3\text{HOOCCOOH} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{CO}_2 + 6\text{H}^+$
Cancel $6\text{H}^+$ from both sides: $\text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{HOOCCOOH} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{CO}_2$
Check charge: left = -2 + 8(+1) + 0 = +6; right = 2(+3) + 0 + 0 = +6. ✓

Key Takeaways

Always ensure electrons cancel before simplifying other species.
H⁺ can appear on both sides and must be simplified to the net amount.
Check both atom balance and charge balance in the final equation.

Common Mistakes

Forgetting to multiply the oxidation half-equation by 3.
Writing 14H⁺ instead of 8H⁺ (failing to cancel the 6H⁺ produced by oxidation).
Reversing the wrong half-equation.

Things to Be Careful About

The equation must be fully balanced in atoms and charge. The mark scheme gives only 1 mark, so the equation must be correct in its entirety.

Techniques used

combine half-equations by multiplying to balance electronssimplify H⁺ terms in overall equation

(ii)

A $25.0\text{ cm}^3$ sample of a solution of $\text{Na}_2\text{C}_2\text{O}_4$ reacts with exactly $16.20\text{ cm}^3$ of an acidified solution of $0.0500\text{ mol dm}^{-3}$ $\text{K}_2\text{Cr}_2\text{O}_7$ .

Calculate the concentration of the solution of $\text{Na}_2\text{C}_2\text{O}_4$ .

[\text{Na}_2\text{C}_2\text{O}_4] = \text{.......................................... mol dm}^{-3}

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Moles of $\text{Cr}_2\text{O}_7^{2-}$ :

n = 0.0500 \times \frac{16.20}{1000} = 8.10 \times 10^{-4} \text{ mol}

From the equation, ratio $\text{Cr}_2\text{O}_7^{2-}$ : $\text{HOOCCOOH}$ = 1 : 3

Moles of $\text{C}_2\text{O}_4^{2-}$ (= moles of $\text{HOOCCOOH}$ ):

n = 8.10 \times 10^{-4} \times 3 = 2.43 \times 10^{-3} \text{ mol}

Concentration of $\text{Na}_2\text{C}_2\text{O}_4$ :

[\text{C}_2\text{O}_4^{2-}] = \frac{2.43 \times 10^{-3}}{25.0/1000} = 2.43 \times 10^{-3} \times 40 = 0.0972 \text{ mol dm}^{-3}

Answer

[\text{Na}_2\text{C}_2\text{O}_4] = 0.0972 \text{ mol dm}^{-3}

Final answer

0.0972 mol dm⁻³

Detailed explanation

Background Concept

Redox titrations use the stoichiometric ratio from the balanced equation to relate the moles of one reactant to the moles of another. The dichromate-ethanedioate titration has a 1:3 ratio (one $\text{Cr}_2\text{O}_7^{2-}$ reacts with three $\text{C}_2\text{O}_4^{2-}$ ). The concentration of the unknown is found from $c = n/V$ .

Understanding the Question

A $25.0 \text{ cm}^3$ sample of $\text{Na}_2\text{C}_2\text{O}_4$ reacts with exactly $16.20 \text{ cm}^3$ of $0.0500 \text{ mol dm}^{-3}$ $\text{K}_2\text{Cr}_2\text{O}_7$ . We must find the concentration of the $\text{Na}_2\text{C}_2\text{O}_4$ solution.

Approach

Calculate moles of $\text{Cr}_2\text{O}_7^{2-}$ from its concentration and volume.
Use the 1:3 ratio to find moles of $\text{C}_2\text{O}_4^{2-}$ .
Divide by the volume of the oxalate solution to get concentration.

Step-by-Step Reasoning

Moles of $\text{Cr}_2\text{O}_7^{2-}$ : $0.0500 \times 16.20/1000 = 8.10 \times 10^{-4}$ mol.
From the balanced equation in (d)(i): 1 mol $\text{Cr}_2\text{O}_7^{2-}$ reacts with 3 mol $\text{HOOCCOOH}$ (equivalent to 3 mol $\text{C}_2\text{O}_4^{2-}$ since each oxalate ion is protonated to give one ethanedioic acid molecule).
Moles of $\text{C}_2\text{O}_4^{2-}$ : $8.10 \times 10^{-4} \times 3 = 2.43 \times 10^{-3}$ mol.
This amount was in $25.0 \text{ cm}^3 = 0.0250 \text{ dm}^3$ .
Concentration: $2.43 \times 10^{-3} / 0.0250 = 0.0972 \text{ mol dm}^{-3}$ .
The mark scheme notes this can be calculated as $2.43 \times 10^{-3} \times 40 = 0.0972$ (multiplying by 40 is equivalent to dividing by 0.025).

Key Takeaways

The 1:3 ratio comes directly from the balanced equation constructed in (d)(i).
Each $\text{C}_2\text{O}_4^{2-}$ ion gives one $\text{HOOCCOOH}$ molecule upon protonation, so moles of oxalate = moles of ethanedioic acid.
Always convert cm³ to dm³ when using $n = cV$ .

Common Mistakes

Using a 1:1 ratio instead of 1:3.
Forgetting to convert cm³ to dm³.
Dividing by 16.20 instead of 25.0 (using the wrong volume for the unknown).

Things to Be Careful About

The answer should be given to at least 2 significant figures. The mark scheme allows ECF from part (d)(i) if the ratio used is consistent with whatever equation was written there. The final answer of 0.0972 mol dm⁻³ is correct to 3 s.f.

Techniques used

calculate moles from concentration and volumeapply stoichiometric ratio from balanced equationcalculate concentration from moles and volume

Q5MediumTransition Elements

The shapes of four different complexes, P, Q, R and S, are shown in Table 5.1.

The symbol J represents an atom or ion of a transition element.

The symbol L is used to represent a monodentate ligand.

Table 5.1

(a)

Label one bond angle on each of complexes P, Q, R and S, and identify the size of the angle in degrees.

2M Similar

DifficultyEasy

Worked solution

Answer

P: 109.5° (between any two bonds)
Q: 90° (between adjacent bonds) or 180° (between opposite bonds)
R: 90° (between adjacent bonds) or 180° (between opposite bonds)
S: 180°

(Note: The angle must be labelled from bond to bond in the diagram.)

Final answer

P: 109.5°, Q: 90° (or 180°), R: 90° (or 180°), S: 180°

Detailed explanation

Background Concept

The geometry of a coordination complex is determined primarily by its coordination number (the number of ligand donor atoms bonded to the central metal ion). For coordination number 4, the two common geometries are tetrahedral and square planar. For coordination number 6, the geometry is octahedral. For coordination number 2, the geometry is linear. Each geometry has characteristic bond angles: tetrahedral is approximately 109.5°, square planar and octahedral have 90° (and 180° for opposite ligands), and linear is 180°.

Understanding the Question

The question provides diagrams of four complexes (P, Q, R, S) with a central transition metal atom J and monodentate ligands L. Part (a) asks to identify one bond angle for each complex and state its size in degrees. The mark scheme emphasizes that the angle must be labelled from bond to bond, meaning it is the angle between two adjacent coordinate bonds (or opposite bonds where applicable), not from the central atom to a ligand.

Approach

Identify the geometry of each complex from its diagram, then recall the standard bond angle(s) for that geometry. Ensure the description clearly indicates the angle is measured between bonds.

Step-by-Step Reasoning

Complex P: The diagram shows a central atom with four ligands in a 3D arrangement with one bond in the plane, one solid wedge (forward), and one dashed wedge (back). This is a tetrahedral geometry. The bond angle in a perfect tetrahedron is 109.5° (or 109°).
Complex Q: The diagram shows a central atom with four ligands in a flat cross shape. This is a square planar geometry. The angle between adjacent bonds is 90°. (The angle between opposite bonds is 180°, which is also acceptable if labelled correctly).
Complex R: The diagram shows a central atom with six ligands: four in a square plane, one above, and one below. This is an octahedral geometry. The angle between adjacent bonds (e.g., axial and equatorial) is 90°. (The angle between opposite axial bonds is 180°).
Complex S: The diagram shows a central atom with two ligands on opposite sides in a straight line. This is a linear geometry. The bond angle is 180°.

Key Takeaways

Tetrahedral complexes (CN=4) have bond angles of 109.5°.
Square planar (CN=4) and octahedral (CN=6) complexes have 90° angles between adjacent ligands.
Linear complexes (CN=2) have a 180° bond angle.
When labelling bond angles, always measure between the bonds themselves, not from the central atom.

Common Mistakes

Labelling the angle from the central atom to a ligand rather than from bond to bond.
Confusing tetrahedral (109.5°) with square planar (90°) angles.
Forgetting that square planar and octahedral geometries also have 180° angles between opposite ligands.

Things to Be Careful About

Ensure the angle value matches the geometry correctly (e.g., do not write 90° for a tetrahedral complex).
The mark scheme accepts 109° or 109.5° for tetrahedral, and 90° or 180° for square planar/octahedral depending on which angle is labelled.

Techniques used

identify bond angles in common complex geometrieslabel bond angles from bond to bond

(b)

Identify the shapes of complexes P, Q, R and S.

P: ...............................................................................................................................................

Q: ...............................................................................................................................................

R: ...............................................................................................................................................

S: ...............................................................................................................................................

2M Similar

DifficultyEasy

Worked solution

Answer

P: tetrahedral
Q: square planar
R: octahedral
S: linear

Final answer

P: tetrahedral, Q: square planar, R: octahedral, S: linear

Detailed explanation

Background Concept

The shape of a coordination complex is defined by the spatial arrangement of the ligands around the central metal ion. Common shapes include:

Linear: Coordination number 2, bond angle 180°.
Tetrahedral: Coordination number 4, bond angles 109.5°. Ligands are arranged at the corners of a tetrahedron.
Square planar: Coordination number 4, bond angles 90°. Ligands are arranged in a flat square around the central metal.
Octahedral: Coordination number 6, bond angles 90° (and 180°). Ligands are arranged at the vertices of an octahedron.

Understanding the Question

Part (b) asks to identify the shape of each of the four complexes (P, Q, R, S) shown in the table. This requires matching the visual representation and coordination number to the standard geometric names.

Approach

Count the number of ligands (coordination number) and observe the 2D/3D representation to assign the correct geometric shape name.

Step-by-Step Reasoning

Complex P: Has 4 ligands. The diagram uses solid lines, a solid wedge (coming out of the page), and a dashed wedge (going into the page), which is the standard representation for a tetrahedral geometry.
Complex Q: Has 4 ligands arranged in a flat cross. This represents a square planar geometry.
Complex R: Has 6 ligands (4 in a plane, 1 above, 1 below). This represents an octahedral geometry.
Complex S: Has 2 ligands in a straight line. This represents a linear geometry.

Key Takeaways

Coordination number 2 = linear.
Coordination number 4 = tetrahedral or square planar (distinguished by 3D representation).
Coordination number 6 = octahedral.

Common Mistakes

Confusing tetrahedral and square planar for coordination number 4 complexes.
Miscounting ligands in 3D representations (e.g., missing the dashed or wedged ligand).

Things to Be Careful About

Use the exact terminology: 'tetrahedral', 'square planar', 'octahedral', 'linear'. Do not use '3D' or 'flat' as shape names.

Techniques used

identify complex shapes from coordination number and geometry

(c)

Two L ligands are exchanged with two different monodentate ligands X and Y in each of complexes P, Q, R and S.

Identify all the complexes which form new complexes that show geometrical isomerism.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Q and R

Final answer

Q and R

Detailed explanation

Background Concept

Geometrical isomerism occurs when ligands can be arranged in different spatial positions relative to each other, leading to isomers that are not superimposable. This is common in:

Square planar complexes (MA₂BC): Can show cis (adjacent, 90°) and trans (opposite, 180°) isomerism.
Octahedral complexes (MA₄BC): Can show cis (adjacent, 90°) and trans (opposite, 180°) isomerism for the two different ligands B and C.
Tetrahedral complexes (MA₂BC or MA₂B₂): Generally do not show geometrical isomerism because all positions are adjacent (109.5°) and any arrangement can be rotated to match another.
Linear complexes (MA₂B): Do not show geometrical isomerism because the two positions are always opposite (180°).

Understanding the Question

Two L ligands are replaced by two different monodentate ligands X and Y. We need to identify which of the original complexes (P, Q, R, S) will form new complexes that exhibit geometrical isomerism.

Approach

Substitute two L's with X and Y in each geometry and check if cis/trans (or equivalent) isomers are possible.

Step-by-Step Reasoning

Complex P (tetrahedral, now JXYL₂): In a tetrahedron, all vertices are equivalent and adjacent. Any arrangement of X, Y, and two L's is superimposable on any other by rotation. No geometrical isomerism.
Complex Q (square planar, now JXYL₂): The two different ligands X and Y can be adjacent to each other (cis isomer, 90° apart) or opposite each other (trans isomer, 180° apart). Shows geometrical isomerism.
Complex R (octahedral, now JXYL₄): The ligands X and Y can be adjacent (cis isomer, 90° apart) or opposite (trans isomer, 180° apart) in the octahedral arrangement. Shows geometrical isomerism.
Complex S (linear, now JXYL₂): The two different ligands X and Y are always opposite each other (180°). There is only one possible arrangement. No geometrical isomerism.

Key Takeaways

Square planar and octahedral complexes with the formula MA₂BC (or MA₄BC) show geometrical isomerism.
Tetrahedral and linear complexes with the same ligand sets do not show geometrical isomerism.

Common Mistakes

Assuming tetrahedral complexes can show geometrical isomerism.
Forgetting that octahedral complexes can show cis/trans isomerism when two ligands are different.

Things to Be Careful About

Ensure the substitution results in a formula that actually allows isomerism (e.g., MA₂BC, not MA₃B or MA₂B₂ where all are same). Here, JXYL₂ and JXYL₄ both allow it.

Techniques used

determine geometrical isomerism in substituted complexes

(d)

Three L ligands are exchanged with three different monodentate ligands X, Y and Z in each of complexes P, Q and R.

Identify all the complexes which form new complexes that show optical isomerism.

1M Similar

DifficultyMedium

Worked solution

Answer

P and R

Final answer

P and R

Detailed explanation

Background Concept

Optical isomerism occurs when a molecule is chiral, meaning it is non-superimposable on its mirror image. This requires the absence of a plane of symmetry (or centre of inversion).

Tetrahedral complexes (MABCD): If all four ligands are different, the complex is chiral and shows optical isomerism.
Square planar complexes: Always have a plane of symmetry (the molecular plane itself), so they never show optical isomerism.
Octahedral complexes (MA₃BCD): Can show optical isomerism. The three identical ligands (L) can be arranged in a facial (fac, all at 90° to each other) or meridional (mer, in a plane with 180° between two) arrangement. Both fac-MA₃BCD and mer-MA₃BCD lack a plane of symmetry when B, C, and D are all different, making them chiral.

Understanding the Question

Three L ligands are replaced by three different monodentate ligands X, Y, and Z. We need to identify which complexes (P, Q, R) will form new complexes that exhibit optical isomerism.

Approach

Evaluate the chirality of the substituted complexes: P (tetrahedral, JXYZL), Q (square planar, JXYZL), and R (octahedral, JXYZL₃).

Step-by-Step Reasoning

Complex P (tetrahedral, now JXYZL): The central atom J is bonded to four different ligands (X, Y, Z, L). This is the classic case of a chiral tetrahedral center (like a carbon with four different groups). The complex and its mirror image are non-superimposable. Shows optical isomerism.
Complex Q (square planar, now JXYZL): All square planar complexes lie in a single plane, which acts as a plane of symmetry. Therefore, they are always superimposable on their mirror images. Does not show optical isomerism.
Complex R (octahedral, now JXYZL₃): With three different ligands X, Y, Z and three identical ligands L, the complex can form fac and mer isomers. In both cases, because X, Y, and Z are all different, there is no plane of symmetry that can reflect the complex onto itself. For example, in the fac isomer, a plane through J and X would reflect Y onto Z, but Y ≠ Z, so no symmetry plane exists. Thus, the isomers are chiral. Shows optical isomerism.

Key Takeaways

Tetrahedral complexes with 4 different ligands (MABCD) are chiral.
Square planar complexes are never chiral.
Octahedral complexes with 3 different monodentate ligands (MA₃BCD) are chiral and show optical isomerism.

Common Mistakes

Assuming square planar complexes can be chiral.
Thinking octahedral MA₃BCD is always achiral; forgetting that different ligands break the planes of symmetry present in MA₃B₃.

Things to Be Careful About

Optical isomerism requires non-superimposability on the mirror image, which means no plane of symmetry.
In octahedral complexes, ensure the ligand set actually breaks all symmetry planes (here, X, Y, Z being different is key).

Techniques used

determine optical isomerism in substituted complexes

Q6MediumHydrocarbonsAnalytical TechniquesCarboxylic Acids and Derivatives

Benzene, $\text{C}_6\text{H}_6$ , reacts with chloroethane, $\text{C}_2\text{H}_5\text{Cl}$ , in the presence of a suitable catalyst to form ethylbenzene, $\text{C}_6\text{H}_5\text{C}_2\text{H}_5$ . In the presence of the catalyst, the ion $\text{C}_2\text{H}_5^+$ is formed. This ion reacts with benzene.

(a)

Complete the equation for the reaction of $\text{C}_2\text{H}_5\text{Cl}$ with this catalyst to form $\text{C}_2\text{H}_5^+$ as one product.

\text{C}_2\text{H}_5\text{Cl} + \text{.......................................} \rightarrow \text{C}_2\text{H}_5^+ + \text{.......................................}

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{C}_2\text{H}_5\text{Cl} + \text{AlCl}_3 \rightarrow \text{C}_2\text{H}_5^+ + \text{AlCl}_4^-

Final answer

C2H5Cl + AlCl3 -> C2H5+ + AlCl4-

Detailed explanation

Background Concept

In Friedel–Crafts alkylation of benzene, an alkyl halide reacts with a haloarene or benzene in the presence of a Lewis acid catalyst (a halogen carrier) such as aluminium chloride ( $\text{AlCl}_3$ ) or iron(III) chloride ( $\text{FeCl}_3$ ). The Lewis acid has an incomplete octet or accessible empty orbitals capable of accepting a lone pair of electrons from the halogen atom of the haloalkane. This polarises and weakens the carbon–halogen bond, generating a carbocation electrophile (e.g., $\text{C}_2\text{H}_5^+$ ) along with a tetrachloroaluminate or tetrachloroferrate(III) complex anion.

Understanding the Question

The question asks to complete the equation where chloroethane reacts with a suitable catalyst to generate the ethyl cation ( $\text{C}_2\text{H}_5^+$ ) as one product, which subsequently acts as the electrophile in electrophilic aromatic substitution.

Approach

Select an appropriate Friedel–Crafts catalyst, typically $\text{AlCl}_3$ or $\text{FeCl}_3$ .
Identify the chloride ion abstraction: the catalyst accepts $\text{Cl}^-$ from $\text{C}_2\text{H}_5\text{Cl}$ .
Write the resulting complex anion, which is $\text{AlCl}_4^-$ (or $\text{FeCl}_4^-$ if $\text{FeCl}_3$ is chosen).

Step-by-Step Reasoning

Reactant: $\text{C}_2\text{H}_5\text{Cl}$
Catalyst: $\text{AlCl}_3$
The Lewis acid $\text{AlCl}_3$ forms a dative bond with the chlorine atom of chloroethane and pulls the chloride ion away:

\text{C}_2\text{H}_5\text{Cl} + \text{AlCl}_3 \rightarrow \text{C}_2\text{H}_5^+ + \text{AlCl}_4^-

Alternatively, using $\text{FeCl}_3$ :

\text{C}_2\text{H}_5\text{Cl} + \text{FeCl}_3 \rightarrow \text{C}_2\text{H}_5^+ + \text{FeCl}_4^-

Both mass and charge are conserved.

Key Takeaways

Halogen carriers (Lewis acids) abstract halogen atoms from alkyl halides to generate powerful carbocation electrophiles for Friedel–Crafts alkylation.

Common Mistakes

Omitting the negative charge on the complex ion, writing $\text{AlCl}_4$ instead of $\text{AlCl}_4^-$ .
Using an incorrect formula for the catalyst, such as $\text{AlCl}_2$ .

Things to Be Careful About

Ensure both electrical charge and atoms are balanced across the reaction arrow.

Techniques used

identify the halogen carrier catalystwrite an equation for carbocation electrophile generation

(b)

Ethylbenzene reacts with more $\text{C}_2\text{H}_5\text{Cl}$ , forming a mixture containing 1,2-diethylbenzene and 1,4-diethylbenzene.

2M

(i)

Draw the structures of 1,2-diethylbenzene and 1,4-diethylbenzene.

Fig. 6.0

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Final answer

1,2-diethylbenzene has ethyl groups on adjacent carbons (1,2-positions); 1,4-diethylbenzene has ethyl groups on opposite carbons (1,4-positions).

Detailed explanation

Background Concept

Disubstituted benzene derivatives exhibit positional isomerism. When naming disubstituted benzenes, ring carbons are numbered to give substituents the lowest possible locants:

1,2-disubstitution corresponds to adjacent ring carbons (ortho).
1,3-disubstitution corresponds to carbons separated by one unsubstituted carbon (meta).
1,4-disubstitution corresponds to carbons opposite each other (para).

Understanding the Question

The candidate must draw the skeletal or structural formulas of 1,2-diethylbenzene and 1,4-diethylbenzene in the corresponding boxes of Fig. 6.0.

Approach

Draw a benzene ring (a regular hexagon with an alternating double-bond system or a circle inside).
For 1,2-diethylbenzene, attach two ethyl groups (e.g., $-\text{CH}_2\text{CH}_3$ or $-\text{C}_2\text{H}_5$ ) to two adjacent ring carbons.
For 1,4-diethylbenzene, attach two ethyl groups to positions 1 and 4 (opposite positions) of the ring.

Step-by-Step Reasoning

1,2-diethylbenzene has ethyl groups attached to carbon-1 and carbon-2.
1,4-diethylbenzene has ethyl groups attached to carbon-1 and carbon-4.
Ensure bonds connect from the aromatic ring carbons directly to the carbon atoms of the ethyl groups (e.g., ring bonded to $\text{C}_2\text{H}_5$ or $-\text{CH}_2\text{CH}_3$ ).

Key Takeaways

Numbering of the benzene ring determines positional nomenclature (1,2- = ortho, 1,4- = para).

Common Mistakes

Drawing 1,3-diethylbenzene instead of 1,4-diethylbenzene.
Inverting connection atoms when writing condensed formulas, e.g., writing the bond to the hydrogen atom rather than the carbon atom.

Things to Be Careful About

Clear representation of the benzene ring (do not forget the delocalised ring circle or the three alternating double bonds).

Techniques used

draw aromatic positional isomersrepresent organic structures correctly

(ii)

Explain why there is very little 1,3-diethylbenzene in the product mixture.

1M Similar

(c)

1,2-diethylbenzene can be oxidised to benzene-1,2-dioic acid, $\text{C}_6\text{H}_4\text{(COOH)}_2$ .

Fig. 6.3

4M

(i)

State the reagent and conditions used for this reaction.

1M Similar

(ii)

Complete the overall equation for this reaction.

An atom of oxygen from the oxidising agent is represented as [O].

All of the atoms in the two ethyl groups are fully oxidised in this reaction.

\text{................................. (1,2-diethylbenzene)} + \text{................. [O]} \rightarrow \text{C}_6\text{H}_4\text{(COOH)}_2 + \text{................................} + \text{................................}

2M Similar

DifficultyMedium

Worked solution

Working

1,2-diethylbenzene has molecular formula $\text{C}_{10}\text{H}_{14}$ or $\text{C}_6\text{H}_4(\text{C}_2\text{H}_5)_2$ .
Benzene-1,2-dioic acid is $\text{C}_6\text{H}_4(\text{COOH})_2$ , which has formula $\text{C}_8\text{H}_6\text{O}_4$ .

The two terminal methyl carbons of the ethyl groups are fully oxidised to $2\text{CO}_2$ .
Hydrogen balance:

Reactants have 14 H atoms.
Products have 6 H atoms in $\text{C}_6\text{H}_4(\text{COOH})_2$ .
The remaining $14 - 6 = 8$ H atoms form $4\text{H}_2\text{O}$ .

Oxygen balance:

Products contain $4$ (in dioic acid) $+ 2 \times 2$ (in $2\text{CO}_2$ ) $+ 4$ (in $4\text{H}_2\text{O}$ ) $= 4 + 4 + 4 = 12$ oxygen atoms.
Therefore, $12\text{ [O]}$ are needed.

Answer

\text{C}_{10}\text{H}_{14} + 12\text{ [O]} \rightarrow \text{C}_6\text{H}_4(\text{COOH})_2 + 2\text{CO}_2 + 4\text{H}_2\text{O}

Final answer

C10H14 + 12 [O] -> C6H4(COOH)2 + 2CO2 + 4H2O

Detailed explanation

Background Concept

When an alkyl group longer than a methyl group (such as ethyl, $-\text{CH}_2\text{CH}_3$ ) attached to an aromatic ring undergoes complete oxidation, the benzylic carbon becomes the carboxyl carbon ( $-\text{COOH}$ ). The extra carbon atoms in the chain are completely oxidised to carbon dioxide ( $\text{CO}_2$ ), and surplus hydrogen atoms are oxidised to water ( $\text{H}_2\text{O}$ ).

Understanding the Question

The question specifies that "all of the atoms in the two ethyl groups are fully oxidised in this reaction." We need to deduce the missing formulae and balance the equation using $[\text{O}]$ .

Approach

Write the formula of 1,2-diethylbenzene: $\text{C}_6\text{H}_4(\text{C}_2\text{H}_5)_2$ or $\text{C}_{10}\text{H}_{14}$ .
Identify the oxidation products of the extra carbons: each ethyl group loses one carbon as $\text{CO}_2$ , giving $2\text{CO}_2$ .
Balance hydrogen atoms using $\text{H}_2\text{O}$ .
Balance oxygen atoms to find the coefficient of $[\text{O}]$ .

Step-by-Step Reasoning

Reactant: $\text{C}_{10}\text{H}_{14}$ (10 carbons, 14 hydrogens).
Benzene-1,2-dioic acid: $\text{C}_6\text{H}_4(\text{COOH})_2 = \text{C}_8\text{H}_6\text{O}_4$ (8 carbons, 6 hydrogens, 4 oxygens).
Carbons unaccounted for: $10 - 8 = 2$ carbons $\rightarrow 2\text{CO}_2$ .
Hydrogens unaccounted for: $14 - 6 = 8$ hydrogens $\rightarrow 4\text{H}_2\text{O}$ .
Total oxygen atoms in products:
- In $\text{C}_6\text{H}_4(\text{COOH})_2$ : 4 O atoms
- In $2\text{CO}_2$ : $2 \times 2 = 4$ O atoms
- In $4\text{H}_2\text{O}$ : $4 \times 1 = 4$ O atoms
- Total $= 4 + 4 + 4 = 12$ O atoms.
Hence, $12\text{ [O]}$ are required on the left-hand side.

Key Takeaways

Any carbon atom in an alkyl side chain beyond the benzylic carbon is oxidised to $\text{CO}_2$ during exhaustive oxidation.

Common Mistakes

Forgetting that the remaining carbons from the ethyl groups form $\text{CO}_2$ .
Incorrect balancing of $[\text{O}]$ due to miscounting the oxygens in the carboxylic acid groups.

Things to Be Careful About

Ensure the formula of 1,2-diethylbenzene is written accurately ( $\text{C}_{10}\text{H}_{14}$ or $\text{C}_6\text{H}_4(\text{C}_2\text{H}_5)_2$ ).

Techniques used

construct and balance redox equations with [O]

(iii)

Predict the number of peaks in the carbon-13 NMR spectrum of benzene-1,2-dioic acid.

1M Similar

(d)

The proton ( $^1\text{H}$ ) NMR spectra of ethylbenzene, $\text{C}_6\text{H}_5\text{C}_2\text{H}_5$ , in $\text{CDCl}_3$ and of benzene-1,2-dioic acid, $\text{C}_6\text{H}_4\text{(COOH)}_2$ , in $\text{CDCl}_3$ are shown. They have not been identified.

Fig. 6.1

Fig. 6.2

6M

(i)

Explain the use of $\text{CDCl}_3$ , instead of $\text{CHCl}_3$ , as the solvent when obtaining these spectra.

1M Similar

(ii)

Identify the substance shown by the spectrum in Fig. 6.1, and complete Table 6.1.

substance: ..........................................................................................................................

Table 6.1

	peak at $\delta = 1.2$	peak at $\delta = 2.6$
name of splitting pattern
group responsible for peak
explanation of splitting pattern

3M Similar

DifficultyMedium

Worked solution

Answer

substance: ethylbenzene (or $\text{C}_6\text{H}_5\text{C}_2\text{H}_5$ )

Table 6.1

	peak at $\delta = 1.2$	peak at $\delta = 2.6$
name of splitting pattern	triplet	quartet
group responsible for peak	$-\text{CH}_3$	$-\text{CH}_2-$
explanation of splitting pattern	2 protons on neighbouring carbon (split by $-\text{CH}_2-$ )	3 protons on neighbouring carbon (split by $-\text{CH}_3$ )

Final answer

Substance: ethylbenzene. Table: delta 1.2 is triplet, CH3, 2 neighbouring H; delta 2.6 is quartet, CH2, 3 neighbouring H.

Detailed explanation

Background Concept

In $^1\text{H}$ NMR spectroscopy:

Chemical shift ( $\delta$ ) reflects the chemical environment of the protons.
Spin-spin splitting follows the $(n+1)$ $(n + 1)$ rule, where $n$ $n$ is the number of non-equivalent protons on adjacent carbon atoms:
- An adjacent $-\text{CH}_2-$ group ( $n = 2$ ) splits a signal into $2 + 1 = 3$ peaks (a triplet).
- An adjacent $-\text{CH}_3$ group ( $n = 3$ ) splits a signal into $3 + 1 = 4$ peaks (a quartet).
Typical chemical shifts:
- Alkyl $-\text{CH}_3$ attached to an alkyl group: $\delta \approx 0.9 - 1.4\text{ ppm}$ .
- Benzylic $-\text{CH}_2-$ attached directly to a benzene ring: $\delta \approx 2.3 - 3.0\text{ ppm}$ .
- Aromatic ring protons: $\delta \approx 6.5 - 8.5\text{ ppm}$ .

Understanding the Question

Fig. 6.1 shows a spectrum with a multiplet around $\delta = 7.2$ (aromatic protons), a quartet at $\delta = 2.6$ , and a triplet at $\delta = 1.2$ . We must identify the substance (ethylbenzene vs benzene-1,2-dioic acid) and complete Table 6.1.

Approach

Benzene-1,2-dioic acid has only two types of protons: aromatic protons and carboxylic acid protons ( $-\text{COOH}$ at $\delta > 10\text{ ppm}$ ). It has no aliphatic protons at $\delta < 3\text{ ppm}$ .
Ethylbenzene has an ethyl group with a $-\text{CH}_2-$ and a $-\text{CH}_3$ , perfectly matching the triplet and quartet.
Complete the table using the $(n+1)$ rule.

Step-by-Step Reasoning

Substance: ethylbenzene ( $\text{C}_6\text{H}_5\text{C}_2\text{H}_5$ ).
Peak at $\delta = 1.2$ :
- Splitting pattern: triplet.
- Group responsible: $-\text{CH}_3$ .
- Explanation: The $-\text{CH}_3$ protons are adjacent to the $-\text{CH}_2-$ carbon bearing 2 protons ( $n = 2$ ), so $n + 1 = 3$ .
Peak at $\delta = 2.6$ :
- Splitting pattern: quartet.
- Group responsible: $-\text{CH}_2-$ .
- Explanation: The $-\text{CH}_2-$ protons are adjacent to the $-\text{CH}_3$ carbon bearing 3 protons ( $n = 3$ ), so $n + 1 = 4$ .

Key Takeaways

An ethyl group ( $-\text{CH}_2\text{CH}_3$ ) attached to an electron-withdrawing group or aromatic ring produces a characteristic triplet-quartet pair.

Common Mistakes

Inverting the assignment (assigning $\delta = 1.2$ to $-\text{CH}_2-$ and $\delta = 2.6$ to $-\text{CH}_3$ ). The $-\text{CH}_2-$ group is directly bonded to the benzene ring and is therefore more deshielded (higher $\delta$ ).

Things to Be Careful About

Make sure the explanation clearly refers to the number of protons on the neighbouring carbon atom, not the number of protons in the group itself.

Techniques used

interpret splitting patterns using the (n+1) ruleassign proton NMR chemical shifts to functional groupsidentify compound from 1H NMR spectrum

(iii)

Identify the substance shown by the spectrum in Fig. 6.2, and complete Table 6.2.

substance: ..........................................................................................................................

Table 6.2

	peak at $\delta = 7.8$	peak at $\delta = 13.1$
group responsible for peak

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

substance: benzene-1,2-dioic acid (or $\text{C}_6\text{H}_4(\text{COOH})_2$ )

Table 6.2

	peak at $\delta = 7.8$	peak at $\delta = 13.1$
group responsible for peak	aromatic protons / $\text{C}_6\text{H}_4$ (benzene ring protons)	$-\text{COOH}$ / carboxylic acid protons

Final answer

Substance: benzene-1,2-dioic acid. Table: delta 7.8 is benzene ring protons (C6H4); delta 13.1 is carboxylic acid (-COOH) protons.

Detailed explanation

Background Concept

Protons attached to different functional groups have characteristic chemical shifts:

Protons attached directly to an aromatic benzene ring typically appear in the region $\delta = 6.0 - 9.0\text{ ppm}$ .
Carboxylic acid protons ( $-\text{COOH}$ ) are strongly deshielded due to the highly electronegative oxygen atoms and hydrogen bonding, appearing as a broad peak at very high chemical shift, typically $\delta = 10.0 - 13.5\text{ ppm}$ .

Understanding the Question

Fig. 6.2 shows two peaks: a sharp multiplet at $\delta \approx 7.8\text{ ppm}$ and a broad peak at $\delta \approx 13.1\text{ ppm}$ . We need to identify the compound and the groups responsible for each peak.

Approach

Identify the substance: benzene-1,2-dioic acid (the other substance, ethylbenzene, was identified in (d)(ii)).
Match $\delta = 7.8\text{ ppm}$ to the aromatic protons on the benzene ring ( $\text{C}_6\text{H}_4$ ).
Match $\delta = 13.1\text{ ppm}$ to the carboxylic acid protons ( $-\text{COOH}$ ).

Step-by-Step Reasoning

Substance: benzene-1,2-dioic acid.
The peak at $\delta = 7.8$ corresponds to the four protons attached to the benzene ring ( $\,-\text{C}_6\text{H}_4-$ ).
The very downfield peak at $\delta = 13.1$ is characteristic of the two acidic protons in the $-\text{COOH}$ groups.

Key Takeaways

Carboxylic acid protons are observed at extremely low field (high $\delta$ , typically 10–13.5 ppm).

Common Mistakes

Misidentifying the peak at $\delta = 13.1$ as an aldehyde proton (which appears at $\delta \approx 9.0 - 10.0\text{ ppm}$ ).

Techniques used

assign proton NMR chemical shifts to aromatic and carboxylic acid environmentsidentify compound from 1H NMR spectrum

(iv)

When $\text{D}_2\text{O}$ is used as a solvent, the spectrum obtained is different from the spectrum in Fig. 6.2.

Describe this difference and explain your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The peak at $\delta = 13.1$ ( $-\text{COOH}$ peak) disappears because the acidic proton exchanges with deuterium from $\text{D}_2\text{O}$ .

Final answer

The peak at delta = 13.1 (-COOH peak) disappears because the proton exchanges with deuterium.

Detailed explanation

Background Concept

Protons bonded to electronegative atoms such as oxygen ( $-\text{OH}$ , $-\text{COOH}$ ) or nitrogen ( $-\text{NH}_2$ ) are labile (exchangeable). When deuterium oxide ( $\text{D}_2\text{O}$ ) is added to the sample, these labile protons undergo rapid exchange with deuterium atoms from $\text{D}_2\text{O}$ :

\text{R}-\text{COOH} + \text{D}_2\text{O} \rightleftharpoons \text{R}-\text{COOD} + \text{HOD}

Because deuterium nuclei ( $^2\text{H}$ ) do not resonate at the frequency of $^1\text{H}$ , the signal corresponding to the labile proton disappears from the $^1\text{H}$ NMR spectrum (known as the $\text{D}_2\text{O}$ shake test).

Understanding the Question

Describe how the spectrum of benzene-1,2-dioic acid (Fig. 6.2) changes when $\text{D}_2\text{O}$ is used and explain why.

Approach

State the visual difference: the peak at $\delta = 13.1$ ( $-\text{COOH}$ ) disappears.
State the reason: chemical exchange of the proton with deuterium ( $\,^1\text{H}/\text{D}$ exchange).

Step-by-Step Reasoning

The carboxylic acid proton is labile.
In the presence of $\text{D}_2\text{O}$ , the reaction $\text{RCOOH} + \text{D}_2\text{O} \rightarrow \text{RCOOD} + \text{HOD}$ takes place.
Deuterium is not detected in $^1\text{H}$ NMR, so the $-\text{COOH}$ peak at $\delta = 13.1$ is eliminated.

Key Takeaways

$\text{D}_2\text{O}$ exchange is a standard analytical test used in $^1\text{H}$ NMR to confirm the presence of $-\text{OH}$ or $-\text{COOH}$ protons by their disappearance.

Common Mistakes

Stating that the peak moves to another chemical shift rather than disappearing completely.
Claiming aromatic protons exchange with $\text{D}_2\text{O}$ (aromatic C-H bonds are not labile under neutral conditions).

Things to Be Careful About

Clearly name both parts: the specific peak that disappears AND the exchange mechanism with deuterium.

Techniques used

explain D2O exchange of labile protons in 1H NMR

(e)

Benzene-1,2-dioic acid can be used to produce K.

Fig. 6.4

Suggest the name of this type of reaction.

1M Similar

Q7Medium-HardNitrogen CompoundsCarboxylic Acids and DerivativesOrganic Synthesis

A reaction scheme is shown in Fig. 7.1.

The reagents needed for reaction 2 and reaction 3 are stated.

Reaction 5 takes place when $\text{C}_2\text{H}_5\text{NH}_2$ is mixed with compound V. No special conditions are required.

Fig. 7.1

(a)

Identify compound U which contains only three elements.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

CH₃CN (ethanenitrile)

Final answer

CH3CN

Detailed explanation

Background Concept

Nitriles (R-C≡N) contain carbon, hydrogen, and nitrogen (three elements). They can be prepared from halogenoalkanes and reduced to primary amines using LiAlH₄ or catalytic hydrogenation. They also undergo hydrolysis in the presence of dilute acid or alkali to form carboxylic acids (or their salts).

Understanding the Question

Compound U is converted to ethanoic acid (CH₃COOH) in reaction 1 and to ethylamine (C₂H₅NH₂) in reaction 2 using LiAlH₄. U contains only three elements. We must identify U.

Approach

LiAlH₄ is a strong reducing agent that reduces nitriles to primary amines. The reduction of a nitrile with two carbons yields ethylamine (C₂H₅NH₂), so U must have two carbons and a nitrile group: CH₃CN. Hydrolysis of CH₃CN yields ethanoic acid (CH₃COOH), which matches reaction 1. The elements in CH₃CN are C, H, and N (three elements), satisfying the condition.

Step-by-Step Reasoning

Reaction 2 uses LiAlH₄ to convert U to C₂H₅NH₂. LiAlH₄ reduces nitriles to primary amines: R-CN + 4[H] → R-CH₂NH₂. For the product to be C₂H₅NH₂, the nitrile must be CH₃CN.
Reaction 1 converts U to CH₃COOH. Hydrolysis of CH₃CN with dilute acid gives CH₃COOH + NH₄⁺. This is consistent.
CH₃CN contains C, H, N — exactly three elements.

Key Takeaways

Nitriles are versatile intermediates: reduction gives primary amines, hydrolysis gives carboxylic acids. Recognising LiAlH₄ as a nitrile-reducing agent is key.

Common Mistakes

Writing CH₃CONH₂ (ethanamide) — reduction of an amide gives an amine, but hydrolysis of an amide gives a carboxylic acid and ammonia, not just the acid. Also, amides are less commonly the direct precursor in this specific two-step pattern compared to nitriles.
Forgetting that nitrile hydrolysis requires acidic or basic conditions.

Things to Be Careful About

Ensure the carbon chain length matches the products. CH₃CN has 2 carbons, matching both CH₃COOH (2 carbons) and C₂H₅NH₂ (2 carbons).

Techniques used

deduce structure from reaction schemerecognise nitrile reduction to primary amine

(b)

Describe the reagents and conditions for reaction 1.

1M Similar

(c)

Identify compound V.

1M Similar

(d)

Complete the equation for reaction 3.

\text{CH}_3\text{COOH} + \text{SOCl}_2 \rightarrow \text{......................................................................................}

1M Similar

(e)

Identify compound W.

1M Similar

(f)

Describe the conditions for reaction 4.

1M Similar

(g)

Suggest the reagent needed for reaction 6.

1M Similar

(h)

Complete Table 7.1 by adding the reaction numbers, 1, 2, 3, 4, 5 and 6, to the right-hand column. Use the reaction numbers given in Fig. 7.1.

Each of the numbers 1, 2, 3, 4, 5 and 6 should be used once only.

Table 7.1

type of reaction	reaction number(s)
hydrolysis
addition
reduction
substitution

4M Similar

DifficultyMedium

Worked solution

Answer

type of reaction	reaction number(s)
hydrolysis	1
addition	2
reduction	2, 6
substitution	3, 4, 5

Final answer

hydrolysis: 1; addition: 2; reduction: 2, 6; substitution: 3, 4, 5

Detailed explanation

Background Concept

Organic reactions can be classified into several types:

Hydrolysis: cleavage of a bond by water (often acid- or base-catalysed).
Addition: two molecules combine to form one, typically across a multiple bond (C=C, C≡C, C=O, C≡N).
Reduction: gain of hydrogen or loss of oxygen.
Substitution: an atom or group is replaced by another.

Some reactions can be classified in more than one way. For example, the reduction of a nitrile with H₂ is both an addition (of H₂) and a reduction.

Understanding the Question

Classify reactions 1–6 in the scheme into hydrolysis, addition, reduction, and substitution. Each number used once only, but a number can appear in multiple rows if it fits multiple categories.

Approach

Analyze each reaction:

Reaction 1: CH₃CN → CH₃COOH. Hydrolysis of nitrile.
Reaction 2: CH₃CN → C₂H₅NH₂. Addition of H₂ (or reduction by LiAlH₄). Fits both addition and reduction.
Reaction 3: CH₃COOH → CH₃COCl. Substitution of -OH by -Cl.
Reaction 4: C₂H₅Br + C₂H₅NH₂ → C₂H₅NHC₂H₅. Nucleophilic substitution.
Reaction 5: CH₃COCl + C₂H₅NH₂ → C₂H₅NHCOCH₃. Nucleophilic addition-elimination (overall substitution of -Cl by -NHC₂H₅).
Reaction 6: C₂H₅NHCOCH₃ → C₂H₅NHC₂H₅. Reduction of amide.

Step-by-Step Reasoning

Hydrolysis: Reaction 1 (nitrile to carboxylic acid). Answer: 1.
Addition: Reaction 2 (nitrile to amine involves addition of H across the C≡N bond). Answer: 2.
Reduction: Reaction 2 (gain of H, loss of N triple bond character) and Reaction 6 (amide to amine, gain of H). Answer: 2, 6.
Substitution: Reaction 3 (OH replaced by Cl), Reaction 4 (Br replaced by amine), Reaction 5 (Cl replaced by amine). Answer: 3, 4, 5.

Key Takeaways

Reactions can belong to multiple categories. Always check if a reaction fits more than one classification. Nitrile reduction is both addition and reduction. Acyl chloride formation and amine reactions are substitutions.

Common Mistakes

Missing that reaction 2 is both addition and reduction.
Classifying reaction 5 as addition only — it is an addition-elimination, which is overall a substitution.
Missing reaction 6 as a reduction.

Things to Be Careful About

The mark scheme allows a reaction number to appear in multiple rows. Ensure all valid classifications are included. Reaction 5 is technically addition-elimination, but at A-Level it is often classified as substitution (nucleophilic acyl substitution).

Techniques used

classify reaction types in a synthesis scheme

(i)

Compare the basicities of $\text{C}_2\text{H}_5\text{NHCOCH}_3$ , $\text{C}_2\text{H}_5\text{NHC}_2\text{H}_5$ and $\text{NH}_3$ .

Explain your answer.

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

most basic [space] least basic

4M Similar

DifficultyMedium-Hard

Worked solution

Answer

Most basic: C₂H₅NHC₂H₅ Least basic: C₂H₅NHCOCH₃

Explanation:

Basicity depends on the availability of the lone pair on nitrogen to accept a proton (H⁺).
C₂H₅NHC₂H₅ is more basic than NH₃ because the two ethyl groups have a positive inductive effect (+I), donating electron density to the nitrogen atom, making the lone pair more available to accept a proton.
C₂H₅NHCOCH₃ is less basic than NH₃ because the lone pair on nitrogen is delocalised into the carbonyl (C=O) group by resonance, making it less available to accept a proton.

Final answer

C2H5NHC2H5 > NH3 > C2H5NHCOCH3

Detailed explanation

Background Concept

Basicity of nitrogen compounds depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton (H⁺). Two main factors affect this:

Inductive effect: Electron-donating groups (like alkyl groups) push electron density towards nitrogen, increasing basicity. Electron-withdrawing groups decrease basicity.
Delocalisation (resonance): If the lone pair is delocalised into an adjacent π-system (like a C=O group in amides), it is less available to accept a proton, decreasing basicity.

The general order of basicity is: aliphatic amines > ammonia > aromatic amines > amides.

Understanding the Question

Compare the basicities of N-ethylethanamine (C₂H₅NHC₂H₅, a secondary amine), ammonia (NH₃), and N-ethyl ethanamide (C₂H₅NHCOCH₃, an amide). Explain the order.

Approach

Rank the three compounds: secondary amine > ammonia > amide.
Explain why the amine is more basic than ammonia: +I effect of ethyl groups.
Explain why the amide is less basic than ammonia: delocalisation of the lone pair into the C=O group.
Link basicity to lone pair availability for proton acceptance.

Step-by-Step Reasoning

Ranking: C₂H₅NHC₂H₅ (most basic) > NH₃ > C₂H₅NHCOCH₃ (least basic).
Amine vs ammonia: In C₂H₅NHC₂H₅, the two ethyl groups are electron-donating (+I effect). They push electron density towards the nitrogen atom, increasing the electron density on the lone pair. This makes the lone pair more attractive to a proton (H⁺), so the amine is more basic than ammonia.
Amide vs ammonia: In C₂H₅NHCOCH₃, the nitrogen lone pair is adjacent to a carbonyl group (C=O). The lone pair is delocalised into the π-system of the C=O bond (resonance). This delocalisation reduces the electron density on nitrogen, making the lone pair less available to accept a proton. Thus, the amide is less basic than ammonia.
Key phrase: Basicity is linked to the ability of the lone pair (or p orbital) on nitrogen to accept/donate to a proton (H⁺).

Key Takeaways

Alkyl groups increase basicity via the +I inductive effect.
Adjacent carbonyl groups decrease basicity via lone pair delocalisation (resonance).
Always link basicity to lone pair availability for proton acceptance.

Common Mistakes

Writing that amides are basic because nitrogen has a lone pair — failing to mention delocalisation.
Saying alkyl groups "attract" electrons — they donate electrons (+I effect), they do not attract.
Forgetting to link the explanation to proton acceptance (H⁺).
Ranking ammonia incorrectly (e.g., thinking amides are more basic than ammonia).

Things to Be Careful About

Use precise terminology: "positive inductive effect" or "electron-donating", not just "push electrons".
Mention "delocalised into C=O group" for amides, not just "resonance".
The mark scheme requires four points: ranking (M1), lone pair/proton acceptance (M2), +I effect for amine (M3), delocalisation for amide (M4). Ensure all are covered.

Techniques used

compare basicities using inductive and delocalisation effectsexplain lone pair availability for proton acceptance

Q8MediumHydroxy CompoundsNitrogen Compounds

(a)

An aqueous solution of phenol, $\text{C}_6\text{H}_5\text{OH}$ , is acidic at $298\text{ K}$ .

Explain why phenol is more acidic than water.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The lone pair on the oxygen atom (in a p orbital) is delocalised into the delocalised $\pi$ -system of the benzene ring. This delocalisation weakens the O–H bond in phenol, making it easier to lose a proton. Alternatively, the resulting phenoxide ion ( $\text{C}_6\text{H}_5\text{O}^-$ ) is stabilised by delocalisation of the negative charge into the ring, making it more stable than the hydroxide ion ( $\text{OH}^-$ ).

Final answer

Lone pair on O delocalised into ring; phenoxide ion stabilised by delocalisation.

Detailed explanation

Background Concept

Phenol ( $\text{C}_6\text{H}_5\text{OH}$ ) is a weak acid, meaning it partially dissociates in water to release $\text{H}^+$ ions. The acidity depends on the ease of breaking the O–H bond and the stability of the resulting anion (conjugate base). In phenol, the oxygen atom is attached to an $sp^2$ -hybridised carbon of the benzene ring. The oxygen atom has a lone pair of electrons in a p orbital that is parallel to the p orbitals forming the delocalised $\pi$ -system of the ring. This allows the lone pair to overlap and delocalise into the ring system.

Understanding the Question

The question asks to explain why phenol is more acidic than water. Acidity is a measure of the tendency to donate a proton ( $\text{H}^+$ ). We need to compare the O–H bond strength and the stability of the conjugate bases ( $\text{C}_6\text{H}_5\text{O}^-$ vs $\text{OH}^-$ ).

Approach

Describe the electronic interaction between the oxygen lone pair and the benzene ring (delocalisation).
Explain how this affects the O–H bond (weakens it).
Explain how this affects the stability of the phenoxide ion compared to the hydroxide ion.

Step-by-Step Reasoning

M1: In phenol, the oxygen atom has a lone pair of electrons. Because the oxygen is attached to the benzene ring, this lone pair is in a p orbital that overlaps with the delocalised $\pi$ -system of the ring. This means the lone pair is delocalised into the ring.
M2: This delocalisation has two consequences that increase acidity:
- It withdraws electron density from the O–H bond, weakening it and making it easier to break heterolytically to release $\text{H}^+$ .
- When the proton is lost, the resulting phenoxide ion ( $\text{C}_6\text{H}_5\text{O}^-$ ) has a negative charge on the oxygen. This negative charge can also delocalise into the benzene ring (across the ortho and para positions). This delocalisation stabilises the phenoxide ion more than the hydroxide ion ( $\text{OH}^-$ ), which has no such stabilisation. A more stable conjugate base means the forward reaction (dissociation) is more favourable.

Key Takeaways

Phenol is more acidic than water/alcohols due to delocalisation of the oxygen lone pair into the aromatic ring.
This delocalisation weakens the O–H bond and stabilises the phenoxide anion.

Common Mistakes

Saying "the ring is electron-withdrawing" without explaining the mechanism (delocalisation of lone pair).
Saying "phenol has a weaker bond" without explaining why (delocalisation).
Forgetting to mention the stability of the anion/conjugate base.

Things to Be Careful About

Use precise terminology: "lone pair on oxygen", "delocalised into the ring", "stabilised anion".
Do not say the ring "pulls" electrons; say the lone pair "delocalises" or "overlaps".

Techniques used

explain delocalisation of lone paircompare stability of conjugate bases

(b)

3M

(i)

Name the two products formed when phenol reacts with an excess of $\text{Br}_2\text{(aq)}$ .

............................................................... and ...............................................................

1M Similar

(ii)

Draw the structures of the two isomeric organic products, with $M_r = 139$ , that are formed when phenol reacts with $\text{HNO}_3\text{(aq)}$ at room temperature.

Fig. 8.1

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The two isomeric organic products are 2-nitrophenol (ortho-nitrophenol) and 4-nitrophenol (para-nitrophenol).

Final answer

2-nitrophenol and 4-nitrophenol (structures drawn)

Detailed explanation

Background Concept

Nitration of phenol with dilute nitric acid ( $\text{HNO}_3\text{(aq)}$ ) at room temperature is a mild reaction. Unlike benzene, which requires a mixture of concentrated $\text{HNO}_3$ and $\text{H}_2\text{SO}_4$ (nitrating mixture) and heat, phenol is reactive enough to be nitrated by dilute acid. The –OH group directs substitution to the ortho (2, 6) and para (4) positions. At room temperature, mononitration occurs, producing a mixture of 2-nitrophenol and 4-nitrophenol. The molecular formula for nitrophenol is $\text{C}_6\text{H}_5\text{NO}_3$ ( $M_r = 12\times6 + 1\times5 + 14 + 16\times3 = 72+5+14+48 = 139$ ), which matches the question.

Understanding the Question

Draw the structures of the two isomeric organic products with $M_r = 139$ formed from phenol and dilute $\text{HNO}_3$ . These are the mono-nitration products.

Approach

Determine the position of the nitro group: ortho (2-position) and para (4-position).
Draw the benzene ring with an –OH group and a – $\text{NO}_2$ group at the respective positions.

Step-by-Step Reasoning

The reaction is electrophilic substitution. The –OH group is an ortho/para director.
At room temperature with dilute acid, mononitration occurs.
The products are 2-nitrophenol (ortho) and 4-nitrophenol (para).
Structure 1: Benzene ring with –OH at position 1 and – $\text{NO}_2$ at position 2 (ortho).
Structure 2: Benzene ring with –OH at position 1 and – $\text{NO}_2$ at position 4 (para).
Note: 3-nitrophenol (meta) is a minor product and usually not considered a major isomer in this context unless specified, but the question asks for two isomers, implying the major ortho and para products. Also $M_r$ calculation confirms mononitration ( $139 - 94 = 45$ , mass of $\text{NO}_2$ minus H is $46-1=45$ ). Phenol $M_r = 94$ . $94+45=139$ .

Key Takeaways

Dilute nitric acid nitrates phenol at room temperature to give ortho and para isomers.
The molecular mass helps confirm the degree of substitution (mononitration).

Common Mistakes

Drawing the nitro group incorrectly (e.g., N bonded to O instead of C, or wrong connectivity $\text{O}_2\text{N}-$ vs $-\text{NO}_2$ ).
Drawing 3-nitrophenol (meta) as a major product.
Drawing dinitro products (which would have higher $M_r$ ).

Things to Be Careful About

Ensure the benzene ring is drawn correctly (circle or alternating double bonds).
Show all atoms and bonds in the functional groups if required, but skeletal/semi-skeletal is usually fine for rings. The mark scheme shows semi-skeletal with explicit –OH and – $\text{NO}_2$ / $\text{O}_2\text{N}-$ .
4-nitrophenol: –OH at top, – $\text{NO}_2$ (or $\text{O}_2\text{N}-$ ) at bottom (para position).
2-nitrophenol: –OH at top right, – $\text{NO}_2$ at bottom right (ortho position).

Techniques used

draw structural formulas of nitrophenolsidentify isomers from molecular formula

(iii)

Write the equation for the reaction between phenol, $\text{C}_6\text{H}_5\text{OH}$ , and sodium metal.

1M Similar

DifficultyEasy

Worked solution

Answer

$2\text{C}_6\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_6\text{H}_5\text{ONa} + \text{H}_2$

Final answer

2C6H5OH + 2Na -> 2C6H5ONa + H2

Detailed explanation

Background Concept

Phenol is acidic enough to react with reactive metals like sodium to produce hydrogen gas and a salt (sodium phenoxide), similar to how alcohols react with sodium but faster. The reaction is a redox reaction where sodium is oxidised to $\text{Na}^+$ and protons from the phenol are reduced to $\text{H}_2$ .

Understanding the Question

Write the balanced chemical equation for the reaction between phenol and sodium metal.

Approach

Reactant: $\text{C}_6\text{H}_5\text{OH}$ and $\text{Na}$ .
Products: Sodium phenoxide ( $\text{C}_6\text{H}_5\text{ONa}$ ) and hydrogen gas ( $\text{H}_2$ ).
Balance the equation.

Step-by-Step Reasoning

Phenol reacts with sodium: $\text{C}_6\text{H}_5\text{OH} + \text{Na} \rightarrow \text{C}_6\text{H}_5\text{ONa} + \frac{1}{2}\text{H}_2$ .
To avoid fractions, multiply by 2: $2\text{C}_6\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_6\text{H}_5\text{ONa} + \text{H}_2$ .

Key Takeaways

Acids (including phenol) react with metals to form salts and hydrogen gas.
Balance equations carefully, especially with diatomic hydrogen.

Common Mistakes

Writing $\text{C}_6\text{H}_5\text{OH} + \text{Na} \rightarrow \text{C}_6\text{H}_5\text{ONa} + \text{H}$ (wrong hydrogen formula).
Unbalanced equation.
Writing water as a product (that's acid-base with alkalis, not metal).

Things to Be Careful About

State symbols are not explicitly required by the mark scheme but good practice: $(\text{s})$ for Na, $(\text{l})$ or $(\text{aq})$ for phenol (usually liquid or solution), $(\text{s})$ for salt, $(\text{g})$ for $\text{H}_2$ . The mark scheme doesn't show them, so plain formulae are fine.
Ensure the formula for sodium phenoxide is correct: $\text{C}_6\text{H}_5\text{ONa}$ .

Techniques used

write balanced equation for acid-metal reaction

(c)

Phenol can be produced from phenylamine in a two-step synthesis.

\text{phenylamine} \xrightarrow{\text{step one}} \text{intermediate compound} \xrightarrow{\text{step two}} \text{phenol}

Describe the reagents and conditions needed in each step.

step one:

reagents: ....................................................................................................................................

conditions: .................................................................................................................................

step two:

reagents: ....................................................................................................................................

conditions: .................................................................................................................................

2M Similar

DifficultyMedium

Worked solution

Answer

Step one:

Reagents: $\text{NaNO}_2$ and $\text{HCl}$ (or $\text{HNO}_2$ and $\text{HCl}$ )
Conditions: Temperature $\leqslant 10\ ^{\circ}\text{C}$

Step two:

Reagents: $\text{H}_2\text{O}$ (water)
Conditions: Warm / Temperature $> 10\ ^{\circ}\text{C}$

Final answer

Step 1: NaNO2/HCl, <10 C. Step 2: H2O, warm.

Detailed explanation

Background Concept

Phenylamine (aniline, $\text{C}_6\text{H}_5\text{NH}_2$ ) can be converted to phenol via a diazonium salt intermediate. This is a two-step process:

Diazotisation: Phenylamine reacts with nitrous acid ( $\text{HNO}_2$ ) at low temperatures to form a benzenediazonium salt ( $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$ ). Nitrous acid is unstable, so it is generated in situ by reacting sodium nitrite ( $\text{NaNO}_2$ ) with hydrochloric acid ( $\text{HCl}$ ).
Hydrolysis: The diazonium salt is unstable and decomposes in water when warmed to release nitrogen gas and form phenol.

Understanding the Question

Describe the reagents and conditions for the two steps to convert phenylamine to phenol.

Approach

Identify Step 1 as diazotisation. List reagents ( $\text{NaNO}_2 + \text{HCl}$ ) and low temperature condition ( $< 10\ ^{\circ}\text{C}$ ).
Identify Step 2 as hydrolysis. List reagent (water) and warm temperature condition.

Step-by-Step Reasoning

Step one: Phenylamine reacts with nitrous acid to form benzenediazonium chloride. Since $\text{HNO}_2$ $HNO_{2}$ is not stable, $\text{NaNO}_2$ $NaNO_{2}$ and $\text{HCl}$ $HCl$ are used. The reaction must be kept cold ( $\leqslant 10\ ^{\circ}\text{C}$ $⩽ 10^{\circ} C$ ) because diazonium salts decompose exothermically at higher temperatures.
- Reagents: $\text{NaNO}_2$ and $\text{HCl}$ (or $\text{HNO}_2$ and $\text{HCl}$ ).
- Conditions: $T \leqslant 10\ ^{\circ}\text{C}$ (ice bath).
Step two: The benzenediazonium salt is treated with water and warmed. The diazonium group ( $-\text{N}_2^+$ $- N_{2}^{+}$ ) is replaced by $-\text{OH}$ $- OH$ , releasing $\text{N}_2$ $N_{2}$ gas.
- Reagents: $\text{H}_2\text{O}$ .
- Conditions: Warm / $T > 10\ ^{\circ}\text{C}$ .

Key Takeaways

Synthesis of phenol from phenylamine involves a diazonium salt intermediate.
Diazotisation requires cold temperatures ( $< 10\ ^{\circ}\text{C}$ ) and $\text{NaNO}_2/\text{HCl}$ .
Hydrolysis requires water and warming.

Common Mistakes

Forgetting the temperature condition for step one (crucial for safety and yield).
Using concentrated $\text{HNO}_3$ / $\text{H}_2\text{SO}_4$ for step one (that's nitration, not diazotisation).
Writing $\text{NaOH}$ for step two (that would just deprotonate phenol, not form it from diazonium salt; hydrolysis uses water).

Things to Be Careful About

Reagents for step one: $\text{NaNO}_2 + \text{HCl}$ is the standard way to write it. $\text{HNO}_2 + \text{HCl}$ is also accepted.
Conditions: Must specify temperature. $\leqslant 10\ ^{\circ}\text{C}$ for step one, warm/ $>10\ ^{\circ}\text{C}$ for step two.
The question asks for reagents and conditions separately.

Techniques used

propose multi-step synthesisidentify diazotisation conditionsidentify hydrolysis conditions for diazonium salts

Chemistry 9701/43 — October/November 2024

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