Chemistry 9701/42 — May/June 2023

Background Concept

The thermal decomposition of a carbonate ion involves the redistribution of electrons within the ion. The carbonate ion, $\text{CO}_3^{2-}$, has a central carbon bonded to three oxygen atoms: one $\text{C=O}$ double bond and two $\text{C–O}^-$ single bonds. On heating, the ion breaks down into $\text{CO}_2$ and $\text{O}^{2-}$.

Curly arrows in mechanisms always show the movement of an electron pair (not a single electron). A curly arrow starts from a source of electrons (a lone pair or a bond) and points to where those electrons end up.

Understanding the Question

The question asks you to add two curly arrows to the given structure of the carbonate ion to show how it decomposes into $\text{CO}_2$ and $\text{O}^{2-}$. You need to show the electron redistribution that converts two $\text{C–O}^-$ single bonds into one $\text{C=O}$ double bond (in $\text{CO}_2$) and releases an oxide ion.

Approach

Work backwards from the products: $\text{CO}_2$ has two $\text{C=O}$ double bonds, and $\text{O}^{2-}$ has three lone pairs. Starting from the carbonate ion (one $\text{C=O}$, two $\text{C–O}^-$), you need to:

1. Convert one $\text{C–O}^-$ into a $\text{C=O}$ — a lone pair on that oxygen moves to form the pi bond.
2. Break the other $\text{C–O}$ bond heterolytically — the bonding pair moves onto the oxygen, giving it an extra lone pair and a 2− charge.

Step-by-Step Reasoning

Arrow 1: Start from a lone pair on one of the $\text{O}^-$ atoms and point it toward the $\text{C–O}$ bond. This shows the lone pair being donated to form a new $\pi$ bond between C and that O, converting the single bond into a double bond.

Arrow 2: Start from the $\text{C–O}$ bond on the other oxygen and point it toward that oxygen atom. This shows the bonding pair of electrons moving entirely onto the oxygen, breaking the bond heterolytically and giving that oxygen a full negative charge of 2− (it already had one negative charge plus the bonding pair = $\text{O}^{2-}$).

The net result: the carbon now has two double bonds to oxygen (= $\text{CO}_2$) and one oxygen leaves as $\text{O}^{2-}$.

Key Takeaways

• Curly arrows always start from electrons (lone pair or bond) and end at the destination.
• In decomposition of carbonate, one oxygen donates a lone pair to form a double bond while the other oxygen takes the bonding pair to leave as $\text{O}^{2-}$.
• This is a heterolytic cleavage — both electrons from the broken bond go to one atom.

Common Mistakes

• Drawing the arrow from the bond to the carbon (wrong direction — the electrons go to oxygen, not carbon).
• Drawing only one arrow instead of two.
• Starting an arrow from a positive charge or from an atom rather than from electrons.
• Pointing an arrow to an atom rather than to a bond or lone pair.

Things to Be Careful About

• Ensure both arrows are clearly drawn with correct direction.
• The arrow must start from a lone pair (on $\text{O}^-$) or from a bond, never from an atom symbol.
• State symbols are not required here as the question shows the mechanism.

Background Concept

Group 2 metal carbonates decompose on heating: $\text{MCO}_3(\text{s}) \rightarrow \text{MO}(\text{s}) + \text{CO}_2(\text{g})$. The ease of this decomposition depends on how strongly the metal cation distorts (polarises) the carbonate ion.

A cation with high charge density (small radius, high charge) strongly attracts the electron cloud of the large $\text{CO}_3^{2-}$ anion, polarising it. This weakens the $\text{C–O}$ bond within the carbonate ion, making it easier to break and release $\text{CO}_2$. The result is lower thermal stability.

Understanding the Question

The command word is "Describe" and "Explain". You must state the trend (what happens to thermal stability down the group) and then give the reason using the concepts of ionic size and polarising power.

Approach

1. State the trend: thermal stability increases from Mg to Ba.
2. Explain why: the cation gets larger → charge density decreases → less polarisation of $\text{CO}_3^{2-}$ → the C–O bond is less weakened → more energy needed to decompose.

Step-by-Step Reasoning

M1 — The trend: Thermal stability increases down Group 2. $\text{MgCO}_3$ decomposes at the lowest temperature; $\text{BaCO}_3$ at the highest.

M2 — The reason (size/charge density): Down the group, the number of electron shells in the $\text{M}^{2+}$ ion increases, so the ionic radius increases. Since the charge remains +2, the charge density (charge/volume ratio) decreases.

M3 — The consequence (polarisation): A cation with lower charge density exerts a weaker electrostatic pull on the electron cloud of the large $\text{CO}_3^{2-}$ anion. Less polarisation means the $\text{C–O}$ bonds within the carbonate ion are less distorted and less weakened, so a higher temperature (more energy) is required to break them and release $\text{CO}_2$.

Key Takeaways

• Thermal stability of Group 2 carbonates increases down the group.
• The explanation always follows: larger cation → lower charge density → less polarisation → harder to decompose.
• This is a classic anion-polarisation argument applicable to nitrates and carbonates alike.

Common Mistakes

• Saying "the charge increases down the group" — the charge stays +2; it is the radius that changes.
• Confusing polarisation of the anion with lattice energy arguments (both are valid but this question asks for the polarisation explanation).
• Stating the trend backwards (stability decreases).

Things to Be Careful About

• You must give all three points (trend, size/charge density, polarisation) for full marks.
• Use the word "polarisation" or "distortion" — not just "attraction".
• The mark scheme accepts either "ionic radius increases" or "charge density decreases" for M2.

Background Concept

Lattice energy ($\Delta H^\ominus_{\text{latt}}$) is a thermodynamic quantity that measures the strength of the ionic bonds in a crystal lattice. It is defined as the enthalpy change when one mole of an ionic compound is formed from its constituent ions in the gas phase. Since gaseous ions are infinitely separated (no interaction), bringing them together to form a solid lattice releases energy — hence lattice energy is always exothermic (negative).

Understanding the Question

The command word is "Define". A definition must include all the qualifying conditions precisely. For lattice energy, the mark scheme requires two elements: (1) energy released when one mole of ionic solid forms, and (2) from gaseous ions under standard conditions.

Approach

Recall the standard definition verbatim, ensuring both the "one mole of ionic solid formed" and "from gaseous ions" components are present.

Step-by-Step Reasoning

M1: "Energy released when one mole of an ionic solid (or compound) is formed" — this specifies what happens (formation) and the quantity (one mole).

M2: "From gaseous ions (under standard conditions)" — this specifies the starting state. The ions must be in the gas phase because lattice energy measures only the ionic bonding, not any other interactions.

Both elements are needed for full marks. Omitting "gaseous" or "one mole" loses a mark.

Key Takeaways

• Lattice energy is always negative (exothermic) by this definition.
• The key qualifiers are: one mole, ionic solid, gaseous ions, standard conditions.
• Do not confuse with lattice dissociation energy (the reverse process, which is endothermic).

Common Mistakes

• Saying "energy required to break apart" — that is lattice dissociation energy, not lattice formation energy.
• Omitting "gaseous" — if you say "from its elements" you are describing enthalpy of formation, not lattice energy.
• Saying "one mole of ions" instead of "one mole of ionic solid/compound".

Things to Be Careful About

• The definition must specify the state of the ions (gaseous) — this is the most commonly missed element.
• "Standard conditions" is part of the full definition but the mark scheme may accept without it depending on the exact wording.

Background Concept

The decomposition reaction is: $\text{MCO}_3(\text{s}) \rightarrow \text{MO}(\text{s}) + \text{CO}_2(\text{g})$

Using a Born-Haber-type cycle, $\Delta H_{\text{decomp}}$ can be expressed in terms of lattice energies:
$\Delta H_{\text{decomp}} = \Delta H^\ominus_{\text{latt}}(\text{MCO}_3) - \Delta H^\ominus_{\text{latt}}(\text{MO}) + \text{other terms}$

More precisely, $\Delta H_{\text{decomp}}$ is approximately equal to the lattice energy of the carbonate (which must be overcome) minus the lattice energy of the oxide (which is released). Since both are negative (exothermic), $\Delta H_{\text{decomp}} \approx \Delta H^\ominus_{\text{latt}}(\text{MCO}_3) - \Delta H^\ominus_{\text{latt}}(\text{MO})$.

The key insight is that lattice energy depends on the size of the ions: smaller ions give more exothermic (more negative) lattice energies. As the cation gets larger down the group, both lattice energies become less exothermic, but they change by different amounts because the anions have different sizes.

Understanding the Question

The question states that both $\Delta H^\ominus_{\text{latt}}(\text{MCO}_3)$ and $\Delta H^\ominus_{\text{latt}}(\text{MO})$ become less exothermic down the group, but they change by different amounts. You must deduce how $\Delta H_{\text{decomp}}$ changes and explain using anion size arguments.

Approach

1. Recognise that $\Delta H_{\text{decomp}}$ depends on the difference between the two lattice energies.
2. Compare the sizes of the anions: $\text{O}^{2-}$ is much smaller than $\text{CO}_3^{2-}$.
3. The smaller anion means the oxide lattice energy is more sensitive to changes in cation size (the cation radius is a larger fraction of the total interionic distance).
4. Therefore $\Delta H^\ominus_{\text{latt}}(\text{MO})$ decreases (becomes less exothermic) faster than $\Delta H^\ominus_{\text{latt}}(\text{MCO}_3)$.
5. Since $\Delta H_{\text{decomp}} \approx \Delta H^\ominus_{\text{latt}}(\text{MCO}_3) - \Delta H^\ominus_{\text{latt}}(\text{MO})$, and the subtracted term becomes less negative faster, the result becomes more positive.

Step-by-Step Reasoning

Point 1 — Direction of change: $\Delta H_{\text{decomp}}$ becomes more positive (or less negative) down the group. This is consistent with increasing thermal stability.

Point 2 — Relative anion sizes: The oxide ion ($\text{O}^{2-}$) is much smaller than the carbonate ion ($\text{CO}_3^{2-}$). The interionic distance in MO is dominated by the cation radius change, whereas in $\text{MCO}_3$ the large carbonate ion means the cation radius change is a smaller proportion of the total distance.

Point 3 — Consequence for lattice energy change: Because $\text{O}^{2-}$ is smaller, $\Delta H^\ominus_{\text{latt}}(\text{MO})$ is more sensitive to the increasing cation size and becomes less exothermic faster than $\Delta H^\ominus_{\text{latt}}(\text{MCO}_3)$ down the group.

Since $\Delta H_{\text{decomp}}$ involves subtracting $\Delta H^\ominus_{\text{latt}}(\text{MO})$ (which is negative), a faster decrease in its magnitude means the overall $\Delta H_{\text{decomp}}$ increases (becomes more positive).

Key Takeaways

• Lattice energy is more sensitive to cation size when the anion is small.
• $\Delta H_{\text{decomp}}$ depends on the difference between two lattice energies, so you must compare their rates of change.
• This connects the polarisation argument (part a) with the thermodynamic (lattice energy) argument.

Common Mistakes

• Saying $\Delta H_{\text{decomp}}$ becomes more negative (wrong direction).
• Saying the carbonate ion is smaller than the oxide ion (backwards).
• Not linking the anion size difference to the rate of change of lattice energy.
• Confusing this with the polarisation explanation — both give the same trend but through different reasoning.

Things to Be Careful About

• The mark scheme gives 2 marks for all three points, 1 mark for any two. Ensure you state the direction, the size comparison, and the consequence.
• "ORA" (or reverse argument) is accepted for the size comparison — you could say the carbonate is larger instead of the oxide is smaller.
• Use precise language: "less exothermic faster" or "changes more" rather than vague "changes differently".

Background Concept

To combine two half-equations into an overall redox equation, the number of electrons lost in the oxidation half-equation must equal the number gained in the reduction half-equation. You multiply each half-equation by an appropriate factor so that the electrons cancel when the equations are added together.

Understanding the Question

Two half-equations are given:

• Oxidation: $\text{H}_2\text{O} + \text{SO}_3^{2-} \rightarrow \text{SO}_4^{2-} + 2\text{H}^+ + 2\text{e}^-$
• Reduction: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

You must give the overall balanced ionic equation.

Approach

1. Find the LCM of the electrons: LCM(2, 5) = 10.
2. Multiply the oxidation by 5 and the reduction by 2.
3. Add the two equations and cancel electrons, then cancel any species appearing on both sides.

Step-by-Step Reasoning

Step 1: Multiply oxidation by 5:
$5\text{H}_2\text{O} + 5\text{SO}_3^{2-} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10\text{e}^-$

Step 2: Multiply reduction by 2:
$2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$

Step 3: Add left sides and right sides:
Left: $5\text{H}_2\text{O} + 5\text{SO}_3^{2-} + 2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^-$
Right: $5\text{SO}_4^{2-} + 10\text{H}^+ + 10\text{e}^- + 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$

Step 4: Cancel $10\text{e}^-$ from both sides.

Step 5: Cancel $\text{H}^+$: 16 on left − 10 on right = 6 $\text{H}^+$ remaining on left.

Step 6: Cancel $\text{H}_2\text{O}$: 8 on right − 5 on left = 3 $\text{H}_2\text{O}$ remaining on right.

Final equation: $2\text{MnO}_4^- + 6\text{H}^+ + 5\text{SO}_3^{2-} \rightarrow 2\text{Mn}^{2+} + 3\text{H}_2\text{O} + 5\text{SO}_4^{2-}$

Check: atoms balance (Mn: 2=2, S: 5=5, O: 8+15=23 left, 12+3+20=35... let me recount. Left: 2×4 + 6×0 + 5×3 = 8+15 = 23 O. Right: 2×0 + 3×1 + 5×4 = 3+20 = 23 O ✓. H: 6 left, 6 right ✓. Charge: 2(−1)+6(+1)+5(−2) = −2+6−10 = −6 left. Right: 2(+2)+5(−2) = +4−10 = −6 ✓.

Key Takeaways

• Always equalise electrons first, then add and cancel.
• After cancelling electrons, check for common species on both sides (here $\text{H}^+$ and $\text{H}_2\text{O}$).
• Verify the final equation by checking atom and charge balance.

Common Mistakes

• Forgetting to cancel $\text{H}^+$ or $\text{H}_2\text{O}$ that appear on both sides.
• Writing the equation with electrons still present.
• Incorrect stoichiometric coefficients from arithmetic errors in the cancellation step.

Things to Be Careful About

• The equation must be balanced for both atoms and charge.
• State symbols are not required by the mark scheme for this equation.
• This is the ionic equation, not the full molecular equation.

Background Concept

Percentage purity is calculated as:
$\%\text{ purity} = \frac{\text{mass of pure substance}}{\text{mass of impure sample}} \times 100$

In a redox titration, the moles of titrant give the moles of analyte via the stoichiometric ratio from the balanced equation. If only an aliquot of the original solution is titrated, the result must be scaled up to the full volume.

Understanding the Question

A $250\text{ cm}^3$ solution contains $3.40\text{ g}$ of impure $\text{K}_2\text{SO}_3$. A $25.0\text{ cm}^3$ aliquot requires $22.40\text{ cm}^3$ of $0.0250\text{ mol dm}^{-3}$ $\text{MnO}_4^-$. Find the percentage purity.

Approach

1. Calculate moles of $\text{MnO}_4^-$ from concentration × volume.
2. Use the 2:5 ratio from the ionic equation to find moles of $\text{SO}_3^{2-}$ in the aliquot.
3. Scale up by ×10 (250/25) to get moles in the full solution.
4. Convert to mass using $M_r$ of $\text{K}_2\text{SO}_3$.
5. Divide by the mass of the impure sample and multiply by 100.

Step-by-Step Reasoning

Step 1 — Moles of $\text{MnO}_4^-$:
$n = c \times V = 0.0250 \times \frac{22.40}{1000} = 5.60 \times 10^{-4} \text{ mol}$

Step 2 — Moles of $\text{SO}_3^{2-}$ in $25.0\text{ cm}^3$:
From the balanced equation: $2\text{MnO}_4^- : 5\text{SO}_3^{2-}$
$n(\text{SO}_3^{2-}) = 5.60 \times 10^{-4} \times \frac{5}{2} = 1.40 \times 10^{-3} \text{ mol}$

Step 3 — Moles in $250\text{ cm}^3$:
The $25.0\text{ cm}^3$ aliquot is $\frac{1}{10}$ of the total, so multiply by 10:
$n(\text{SO}_3^{2-})_{\text{total}} = 1.40 \times 10^{-3} \times 10 = 1.40 \times 10^{-2} \text{ mol}$

Step 4 — Mass of $\text{K}_2\text{SO}_3$:
$M_r = 2(39.1) + 32.1 + 3(16.0) = 78.2 + 32.1 + 48.0 = 158.3$
$m = 1.40 \times 10^{-2} \times 158.3 = 2.216 \text{ g}$

Step 5 — Percentage purity:
$\% = \frac{2.216}{3.40} \times 100 = 65.2\%$

The mark scheme accepts 65.2% or 65.3% (depending on rounding of intermediate steps) to a minimum of 2 significant figures.

Key Takeaways

• Always identify the stoichiometric ratio from the balanced equation.
• Remember to scale from the aliquot to the full solution volume.
• The moles of $\text{SO}_3^{2-}$ equal the moles of $\text{K}_2\text{SO}_3$ (1:1 ratio within the compound).

Common Mistakes

• Forgetting to scale from $25\text{ cm}^3$ to $250\text{ cm}^3$ (gives 6.52% instead of 65.2%).
• Using the wrong mole ratio (e.g. 2:1 instead of 2:5).
• Using the wrong $M_r$ (e.g. for $\text{K}_2\text{SO}_4$ instead of $\text{K}_2\text{SO}_3$).
• Not converting $\text{cm}^3$ to $\text{dm}^3$ when calculating moles.

Things to Be Careful About

• The mark scheme allows ecf (error carried forward) at each step, so a correct method with an arithmetic error still earns method marks.
• Give the final answer to at least 2 significant figures (3 s.f. preferred).
• The ratio is 5 $\text{SO}_3^{2-}$ per 2 $\text{MnO}_4^-$, not the other way around.