Chemistry 9701/42 — February/March 2018

Background Concept

Group 2 elements form hydroxides with the general formula $\text{M(OH)}_2$. Their solubility in water follows a clear trend down the group: it increases from magnesium hydroxide (sparingly soluble) to barium hydroxide (moderately soluble). This is the opposite of the trend for Group 2 sulfates, which become less soluble down the group.

Understanding the Question

The question asks you to state the trend — a simple recall of the syllabus fact. No explanation is required for part (i); the explanation is asked for separately in part (ii).

Approach

Recall the trend from the syllabus: solubility of Group 2 hydroxides increases down the group.

Step-by-Step Reasoning

The trend is:
$\text{Mg(OH)}_2 < \text{Ca(OH)}_2 < \text{Sr(OH)}_2 < \text{Ba(OH)}_2$
in terms of solubility. Magnesium hydroxide is only slightly soluble; barium hydroxide dissolves readily. The single mark is awarded for stating that solubility increases down the group.

Key Takeaways

• Group 2 hydroxides: solubility increases down the group.
• Group 2 sulfates: solubility decreases down the group.
  This contrast is a classic exam point.

Common Mistakes

• Stating that solubility decreases down the group — this is the sulfate trend, not the hydroxide trend.
• Confusing the trend with that of thermal stability (which also increases down the group for carbonates and nitrates).

Things to Be Careful About

Give the direction clearly: "increases down the group". Do not just say "varies" — the mark requires the specific direction.

Background Concept

The solubility of a salt is governed by the enthalpy change of solution, $\Delta H_\text{sol}$, which is the sum of two opposing energy terms:
$\Delta H_\text{sol} = \text{lattice energy} + \text{hydration energy}$

• Lattice energy is the energy released when gaseous ions come together to form a solid lattice (always negative, exothermic).
• Hydration energy is the energy released when gaseous ions are surrounded by water molecules (also negative, exothermic).

A more negative (more exothermic) $\Delta H_\text{sol}$ favours dissolution, so the salt is more soluble.

Understanding the Question

Part (i) established the trend (solubility increases down the group). Now you must explain why using the energy terms. The mark scheme rewards three distinct points: both energies decrease, lattice decreases more, and the consequence for $\Delta H_\text{sol}$.

Approach

Consider how the ionic radius changes down the group, then how that affects lattice energy and hydration energy separately, and finally the balance between them.

Step-by-Step Reasoning

1. Both energies decrease down the group.
   As we go from $\text{Mg}^{2+}$ to $\text{Ba}^{2+}$, the cation radius increases and charge density decreases. Both lattice energy and hydration energy become less negative (decrease in magnitude).

2. Lattice energy decreases more than hydration energy.
   Lattice energy depends on the sum of the cation and anion radii ($r_+ + r_-$) — the anion ($\text{OH}^-$) is the same throughout, but the cation grows. Hydration energy depends mainly on the cation radius alone. Since the anion is constant, the fractional change in the cation radius affects lattice energy more strongly because it appears in the denominator alongside the anion radius. Hence lattice energy falls more steeply.

3. Consequence for $\Delta H_\text{sol}$.
   Since lattice energy decreases more than hydration energy, the sum $\Delta H_\text{sol} = \text{lattice} + \text{hydration}$ becomes more negative (more exothermic) down the group. A more exothermic dissolution favours solubility, so solubility increases.

Key Takeaways

• Solubility is determined by the balance between lattice and hydration energies.
• When comparing down a group, identify which term changes more.
• A more negative $\Delta H_\text{sol}$ means greater solubility.

Common Mistakes

• Saying hydration energy decreases more than lattice energy — this is backwards.
• Only stating that both energies decrease, without the comparison (this loses the second mark).
• Not linking the energy change to solubility (loses the third mark).

Things to Be Careful About

Use precise wording: "lattice energy decreases more than hydration energy". The mark scheme explicitly requires this comparison. Also make the link to $\Delta H_\text{sol}$ becoming more negative/exothermic explicit.

Background Concept

The solubility product, $K_{\text{sp}}$, is the equilibrium constant for the dissolution of a sparingly soluble ionic solid. For a salt $\text{A}_m\text{B}_n$ that dissociates as $\text{A}_m\text{B}_n(\text{s}) \rightleftharpoons m\text{A}^{n+}(\text{aq}) + n\text{B}^{m-}(\text{aq})$, the expression is:
$K_{\text{sp}} = [\text{A}^{n+}]^m[\text{B}^{m-}]^n$
The solid is not included in the expression.

Understanding the Question

Write the $K_{\text{sp}}$ expression for $\text{Sr(OH)}_2$. This is a direct recall/application question worth 1 mark.

Approach

Write the dissociation equation, then construct the equilibrium expression with each ion concentration raised to its stoichiometric coefficient.

Step-by-Step Reasoning

The dissociation is:
$\text{Sr(OH)}_2(\text{s}) \rightleftharpoons \text{Sr}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})$
The coefficient of $\text{Sr}^{2+}$ is 1, so its concentration appears to the power 1. The coefficient of $\text{OH}^-$ is 2, so its concentration is squared:
$K_{\text{sp}} = [\text{Sr}^{2+}][\text{OH}^-]^2$

Key Takeaways

• The $K_{\text{sp}}$ expression uses stoichiometric coefficients as powers.
• The solid phase is excluded from the expression.

Common Mistakes

• Forgetting to square $[\text{OH}^-]$ — this is the most common error.
• Including $[\text{Sr(OH)}_2]$ in the expression — solids are never included.

Things to Be Careful About

The square on $[\text{OH}^-]$ is essential. The mark scheme gives exactly $K_{\text{sp}} = [\text{Sr}^{2+}][\text{OH}^-]^2$.

Background Concept

The solubility of a salt is the number of moles that dissolve per cubic decimetre of solution. For $\text{Sr(OH)}_2$, each formula unit that dissolves produces one $\text{Sr}^{2+}$ ion and two $\text{OH}^-$ ions. Therefore, if the solubility is $s$, then $[\text{Sr}^{2+}] = s$ and $[\text{OH}^-] = 2s$.

Understanding the Question

You are given the solubility $s = 3.37 \times 10^{-2} \text{ mol dm}^{-3}$ at $0\,^\circ\text{C}$ and asked to calculate $K_{\text{sp}}$ with units. The mark scheme gives 1 mark for the numerical value and 1 mark for the units.

Approach

1. Write $[\text{Sr}^{2+}] = s$ and $[\text{OH}^-] = 2s$.
2. Substitute into $K_{\text{sp}} = [\text{Sr}^{2+}][\text{OH}^-]^2$.
3. Calculate and determine the units.

Step-by-Step Reasoning

1. $[\text{Sr}^{2+}] = 3.37 \times 10^{-2} \text{ mol dm}^{-3}$.
2. $[\text{OH}^-] = 2 \times 3.37 \times 10^{-2} = 6.74 \times 10^{-2} \text{ mol dm}^{-3}$.
3. Substitute:
   $K_{\text{sp}} = (3.37 \times 10^{-2})(6.74 \times 10^{-2})^2$
   $= (3.37 \times 10^{-2})(4.543 \times 10^{-3})$
   $= 1.531 \times 10^{-4} \approx 1.5 \times 10^{-4}$
4. Units: $\text{mol dm}^{-3} \times (\text{mol dm}^{-3})^2 = \text{mol}^3 \text{ dm}^{-9}$.

An alternative compact form: for an $\text{M(OH)}_2$ salt, $K_{\text{sp}} = 4s^3 = 4(3.37 \times 10^{-2})^3 = 1.53 \times 10^{-4}$.

Key Takeaways

• For a 1:2 salt, $K_{\text{sp}} = 4s^3$ where $s$ is the solubility.
• Always determine the units of $K_{\text{sp}}$ from the expression, not from memory.

Common Mistakes

• Forgetting to double the $\text{OH}^-$ concentration (using $s$ instead of $2s$).
• Giving the wrong units, e.g. $\text{mol dm}^{-3}$ instead of $\text{mol}^3 \text{ dm}^{-9}$.
• Rounding too early — keep intermediate values to enough significant figures.

Things to Be Careful About

The mark scheme accepts $1.5 \times 10^{-4}$ (2 significant figures, matching the data). The units are a separate mark — do not forget them. The final answer must include both value and units.

Background Concept

Metal peroxides contain the peroxide ion $\text{O}_2^{2-}$. On heating, the O–O bond breaks and the peroxide decomposes to the metal oxide (containing $\text{O}^{2-}$) and oxygen gas. The general equation is:
$2\text{MO}_2 \rightarrow 2\text{MO} + \text{O}_2$

Understanding the Question

The stem tells you that peroxides $\text{MO}_2$ decompose to give a single gas and the solid oxide $\text{MO}$. You must write the balanced equation for strontium peroxide.

Approach

Write the skeleton equation $\text{SrO}_2 \rightarrow \text{SrO} + \text{O}_2$, then balance it.

Step-by-Step Reasoning

1. Skeleton: $\text{SrO}_2 \rightarrow \text{SrO} + \text{O}_2$.
2. Oxygen atoms: left has 2, right has 1 (in SrO) + 2 (in O2) = 3. Not balanced.
3. Multiply $\text{SrO}_2$ and $\text{SrO}$ by 2:
   $2\text{SrO}_2 \rightarrow 2\text{SrO} + \text{O}_2$
4. Check: Sr 2 = 2; O 4 = 2 + 2 = 4. Balanced.

Key Takeaways

• Peroxide decomposition: $2\text{MO}_2 \rightarrow 2\text{MO} + \text{O}_2$.
• The oxygen product is $\text{O}_2$ gas, hence the factor of 2 on the peroxide.

Common Mistakes

• Writing $\text{SrO}_2 \rightarrow \text{SrO} + \text{O}_2$ without balancing (oxygen doesn't balance).
• Forgetting state symbols — the mark scheme expects (s) for solids and (g) for oxygen.

Things to Be Careful About

The equation must be balanced. The mark scheme gives exactly $2\text{SrO}_2 \rightarrow 2\text{SrO} + \text{O}_2$.

Background Concept

Thermal stability of ionic compounds depends on the polarising power of the cation. A small, highly charged cation (high charge density) strongly distorts the electron cloud of a nearby anion — this is called polarisation. When a cation polarises the peroxide ion $\text{O}_2^{2-}$, it pulls electron density away from the O–O bond, weakening it. A weaker O–O bond means the peroxide decomposes at a lower temperature.

Understanding the Question

Part (i) gave the decomposition equation for $\text{SrO}_2$. Now you must predict how the decomposition temperature of $\text{MO}_2$ changes down Group 2 and explain using polarisation. The mark scheme rewards three points: the trend, the charge-density change, and the link to polarisation/bond weakening.

Approach

1. Identify how cation size/charge density changes down the group.
2. Relate this to polarising power.
3. Link polarisation to O–O bond strength and hence decomposition temperature.

Step-by-Step Reasoning

1. Trend: down the group, the cation radius increases, so charge density decreases.
2. Polarising power: a cation with lower charge density polarises the $\text{O}_2^{2-}$ ion less.
3. Effect on the O–O bond: less polarisation means the O–O bond is weakened less — it remains relatively strong.
4. Consequence: a higher temperature is needed to break the O–O bond and cause decomposition. Therefore the decomposition temperature increases down the group.

This mirrors the familiar trend for Group 2 carbonates and nitrates, where thermal stability increases down the group for the same reason.

Key Takeaways

• Thermal stability (decomposition temperature) increases down Group 2.
• The underlying cause is decreasing cation charge density → less polarisation of the anion.
• Polarisation weakens the anion's internal bonds, lowering the decomposition temperature.

Common Mistakes

• Stating the temperature decreases down the group (reversing the trend).
• Saying charge density increases down the group (it decreases).
• Not making the link from polarisation to the O–O bond being weakened less.

Things to Be Careful About

Use the precise phrase "charge density of the cation decreases" and connect it explicitly to "less polarisation of the $\text{O}_2^{2-}$ ion" and "weakens the O–O bond less". The mark scheme requires all three links.

Background Concept

Ethanedioates (oxalates) contain the $\text{C}_2\text{O}_4^{2-}$ ion. On heating, the metal ethanedioate decomposes to the metal oxide and a mixture of carbon monoxide and carbon dioxide. The two carbon atoms in the ethanedioate ion are distributed between CO and CO2.

Understanding the Question

The stem states that ethanedioates decompose to give a mixture of two different gases and the solid oxide $\text{MO}$. You must complete the equation for barium ethanedioate by filling in the three products.

Approach

1. The solid product is $\text{BaO}$ (given in the stem).
2. The two gases must account for the two carbon atoms and the remaining oxygen atoms.
3. Balance the equation.

Step-by-Step Reasoning

1. Write the skeleton with the known product $\text{BaO}$:
   $\text{BaC}_2\text{O}_4 \rightarrow \text{BaO} + \text{?} + \text{?}$
2. The two carbon atoms must form two carbon-containing gases. The natural products are CO and CO2.
3. Check the oxygen balance: left has 4 O atoms. Right: BaO contributes 1, CO contributes 1, CO2 contributes 2 — total 4. Balanced.
4. Check atoms: Ba 1 = 1; C 2 = 1 + 1 = 2; O 4 = 1 + 1 + 2 = 4. Balanced.

So the complete equation is:
$\text{BaC}_2\text{O}_4(\text{s}) \rightarrow \text{BaO}(\text{s}) + \text{CO}(\text{g}) + \text{CO}_2(\text{g})$

Key Takeaways

• Ethanedioates decompose to oxide + CO + CO2.
• The carbon atoms split between CO and CO2.

Common Mistakes

• Writing $\text{CO}_2$ twice instead of CO and CO2 — this would not balance (too many O atoms).
• Forgetting that two different gases are produced (the stem specifies this).
• Not including state symbols.

Things to Be Careful About

The mark scheme gives exactly $\text{BaC}_2\text{O}_4 \rightarrow \text{BaO} + \text{CO} + \text{CO}_2$. Include state symbols for full clarity.

Background Concept

Ethanedioic acid (oxalic acid, $\text{H}_2\text{C}_2\text{O}_4$) is a reducing agent. Acidified manganate(VII) ions, $\text{MnO}_4^-$, are a strong oxidising agent. In acid, $\text{MnO}_4^-$ is reduced to $\text{Mn}^{2+}$:
$\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$
Meanwhile, ethanedioate is oxidised to carbon dioxide:
$\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2\text{e}^-$
The manganate(VII) ion is intensely purple; the $\text{Mn}^{2+}$ product is almost colourless (pale pink). The carbon dioxide is a gas, so bubbles are seen.

Understanding the Question

The question asks for two observations during the reaction. It does not ask for the equation or the mechanism — just what you would see. The mark scheme gives one mark for the colour change and one for the gas evolution.

Approach

Identify what happens to each reactant: the purple $\text{MnO}_4^-$ is consumed (colour change) and $\text{CO}_2$ gas is produced (bubbles).

Step-by-Step Reasoning

1. Colour change: The purple/pink colour of $\text{MnO}_4^-$ disappears as it is reduced to colourless $\text{Mn}^{2+}$. The solution decolourises.
2. Gas evolution: The ethanedioate is oxidised to $\text{CO}_2$, which escapes as bubbles.

Both observations are visible without any special apparatus.

Key Takeaways

• $\text{MnO}_4^-$ (purple) → $\text{Mn}^{2+}$ (colourless) is the classic redox indicator colour change.
• Ethanedioic acid is oxidised to $\text{CO}_2$ — a reducing agent.

Common Mistakes

• Saying "turns colourless" without specifying the initial purple colour — be precise about the colour change.
• Missing the gas evolution — both observations are needed for full marks.
• Confusing this with the reaction of manganate(VII) with an alkene (which also decolourises but does not evolve gas).

Things to Be Careful About

The mark scheme wording is "the KMnO4 would decolourise" and "bubbles / gas evolution would be seen". Use "decolourise" (not "turn clear" alone) and specify the gas as carbon dioxide where possible.