Chemistry 9701/13 — May/June 2025

Background Concept

Orbitals are regions of space in which there is a high probability of finding an electron. Different types of orbital have different shapes: s orbitals are spherically shaped, p orbitals are dumbbell-shaped (two lobes), and d orbitals have more complex shapes. Each s orbital can hold a maximum of two electrons, and an atom can have more than one occupied s orbital (e.g. 1s, 2s, 3s, 4s).

In a ground-state atom, electrons fill subshells in a specific order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc. The question asks for the atom that has a total of five electrons in spherically shaped orbitals — that is, five electrons distributed among all its occupied s orbitals.

Understanding the Question

The question gives four neutral atoms — sodium, fluorine, boron, and potassium — and asks which one has exactly five electrons in s orbitals. The phrase “isolated gaseous atom” tells us to consider a single neutral atom in its ground state, not an ion and not an atom in a compound. The key idea is recognising that “spherically shaped orbitals” means s orbitals, then counting all s electrons, not just the outer-shell ones.

Approach

1. Identify that spherically shaped orbitals are s orbitals.
2. Write the full ground-state electron configuration for each option.
3. Count the electrons in the s subshells (1s, 2s, 3s, 4s, …).
4. Compare the totals and select the atom with five s electrons.

Step-by-Step Reasoning

• Sodium, Na (Z = 11): configuration $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^1$. The s electrons are in 1s, 2s, and 3s: $2 + 2 + 1 = 5$. This matches the requirement.
• Fluorine, F (Z = 9): configuration $1\text{s}^2\,2\text{s}^2\,2\text{p}^5$. The s electrons are $2 + 2 = 4$. Fluorine has five electrons in total in its outer shell, but only four are in s orbitals.
• Boron, B (Z = 5): configuration $1\text{s}^2\,2\text{s}^2\,2\text{p}^1$. The s electrons are $2 + 2 = 4$. Boron has five electrons in total, but one is in a p orbital.
• Potassium, K (Z = 19): configuration $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6\,4\text{s}^1$. The s electrons are in 1s, 2s, 3s, and 4s: $2 + 2 + 2 + 1 = 7$.

Only sodium gives the required total of five s electrons, so the correct option is A.

Key Takeaways

• s orbitals are the spherically shaped orbitals; p and d orbitals have different shapes.
• “Total number of electrons in s orbitals” means adding electrons from every occupied s subshell, not just the valence shell.
• Writing full electron configurations and counting carefully is the reliable method for this type of question.

Common Mistakes

• Counting only the outer-shell electron: sodium has one outer s electron, but the question asks for the total across all occupied s orbitals.
• Confusing total electrons with s electrons: boron and fluorine each have five total electrons, but neither has five s electrons.
• Thinking p orbitals are spherical: p orbitals are dumbbell-shaped, so their electrons must not be counted.
• Forgetting that potassium has both 3s and 4s electrons; its s-electron total is 7, not 1.

Things to Be Careful About

• Use the correct filling order: 4s fills before 3d for potassium, so potassium’s configuration ends in $4\text{s}^1$, not $3\text{d}^1$.
• The question specifies an isolated gaseous atom, so use the neutral ground-state configuration, not an ion or an excited-state configuration.
• Read “five electrons occupying spherically shaped orbitals” as five electrons in s orbitals, not five s orbitals (each s orbital holds two electrons).

Background Concept

The activation energy $E_A$ is the minimum energy that colliding particles must possess for a reaction to occur. It is determined by the reaction pathway (the specific bonds being broken and formed) and is an intrinsic property of the reaction mechanism. Temperature does not alter this energy barrier.

What temperature does affect is the distribution of kinetic energies among the particles. At a higher temperature, the Boltzmann distribution shifts to the right and flattens, meaning a greater proportion of particles have energy $\geq E_A$. This results in more successful collisions per unit time and therefore a faster reaction rate.

On an enthalpy profile diagram, $E_A$ is the vertical distance from the reactants level to the peak (transition state). On a concentration–time graph, a faster reaction is shown by a steeper initial gradient (concentration drops more rapidly).

Understanding the Question

The question presents two diagrams for a reaction: an enthalpy profile (Diagram 1) showing an exothermic reaction with a marked $E_A$, and a concentration–time graph (Diagram 2) showing exponential decay of reactant concentration. Four options (A–D) each show a pair of diagrams drawn to the same scale. The task is to identify which pair correctly represents the effect of increasing temperature on this reaction.

The command word is implicit: "which diagrams... show the effect" — we must select the option where both diagrams are consistent with a temperature increase.

Approach

Apply two rules:

1. $E_A$ must remain unchanged (same peak height relative to reactants) — this eliminates any option showing a different barrier.
2. The concentration–time curve must be steeper (faster rate) while maintaining the same general shape (same reaction order) — this eliminates options showing no change or a wrong shape.

Check each option against both criteria.

Step-by-Step Reasoning

Option A: The enthalpy diagram shows the same $E_A$ as the original. The concentration–time graph also appears unchanged (same gradient as the original). This shows no effect of temperature on rate — incorrect.

Option B: The enthalpy diagram shows the same $E_A$. The concentration–time graph is steeper, suggesting a faster rate. However, examining the curve shape more carefully, it drops extremely steeply and then levels off almost immediately in a way inconsistent with the same reaction order (it looks like a different kinetic profile rather than simply a faster version of the same exponential decay). This is not a correct representation of the same reaction at higher temperature.

Option C: The enthalpy diagram shows the same $E_A$ (peak at the same height above reactants, same $\Delta H$). The concentration–time graph shows a steeper initial gradient than the original, indicating a faster reaction, while maintaining the same exponential decay shape (same order). This correctly represents increased temperature: same energy barrier, faster rate. ✓

Option D: The enthalpy diagram shows a larger $E_A$ (higher peak). This would represent a different reaction pathway or the effect of removing a catalyst — not the effect of temperature. Temperature does not change $E_A$. The concentration–time graph also appears unchanged. Incorrect on both counts.

Therefore, C is the correct answer.

Key Takeaways

• Activation energy is a fixed property of the reaction pathway; it does not change with temperature.
• Increasing temperature increases the rate (steeper concentration–time gradient) without altering the energy barrier.
• On a Boltzmann distribution, temperature shifts the curve rightward, increasing the fraction of particles exceeding $E_A$, but $E_A$ itself is unchanged.
• When interpreting paired diagrams, both must be consistent with the single change being tested.

Common Mistakes

• Confusing temperature with catalyst effects: A catalyst lowers $E_A$ (shown by a lower peak); temperature does not. Students who select D are confusing these two factors.
• Thinking $E_A$ changes with temperature: The barrier height is determined by the mechanism, not by how hot the system is.
• Selecting A: Students may think that because the concentration–time graph doesn't change shape (still exponential), nothing has changed. But the gradient must be steeper for a faster reaction.
• Misreading the scale: The question states diagrams are drawn to the same scale, so visual comparison of peak heights and gradients is valid.

Things to Be Careful About

• Ensure you compare the same features across options: peak height relative to reactants line (for $E_A$) and initial gradient of the curve (for rate).
• The products level and reactants level should remain unchanged in the enthalpy diagram (same $\Delta H$), as temperature does not alter the thermodynamics of the reaction in this context.
• The concentration–time graph should start at the same initial concentration (same $y$-intercept) since we are comparing the same reaction, just at a different temperature.

Background Concept

In organic reaction mechanisms, different types of elementary steps can occur. A Brønsted–Lowry acid–base reaction is specifically defined as a proton (H$^+$) transfer: the acid is the proton donor, and the base is the proton acceptor. In mechanism diagrams, this is represented by a curly arrow originating from a lone pair on the base and pointing to the hydrogen atom, with a second curly arrow showing the heterolytic cleavage of the H–X bond (electrons moving to X).

Other common steps in nucleophilic acyl substitution (the reaction of an acid chloride with an amine) include nucleophilic addition (a nucleophile attacks an electrophilic carbon, often breaking a $\pi$-bond) and elimination (a lone pair reforms a $\pi$-bond, expelling a leaving group). These steps involve the transfer of electron pairs to form or break bonds to atoms other than hydrogen, so they are not Brønsted–Lowry acid–base reactions.

Understanding the Question

The question provides the overall equation for the reaction between ethylamine (a primary amine) and propanoyl chloride (an acid chloride) to form an amide and an ammonium salt. It then shows the four-step mechanism with curly arrows and asks which steps are Brønsted–Lowry acid–base reactions. The task is to analyze each step and identify where a proton (H$^+$) is transferred from one species to another.

Approach

To solve this, examine each step in the mechanism and look for a proton transfer. Specifically, check if a curly arrow starts at a lone pair on an atom (the base) and ends at a hydrogen atom, and if another curly arrow shows the bond between that hydrogen and another atom breaking (the acid donating the proton). If no hydrogen is involved in the bond-making/bond-breaking process, it is not a Brønsted–Lowry acid–base step.

Step-by-Step Reasoning

• Step 1: The lone pair on the nitrogen atom of ethylamine attacks the electrophilic carbonyl carbon of propanoyl chloride. Simultaneously, the $\pi$-electrons of the C=O double bond move onto the oxygen atom. This is a nucleophilic addition step. No proton is transferred; electrons move to carbon and oxygen. Not a Brønsted–Lowry acid–base reaction.
• Step 2: The lone pair on the negatively charged oxygen moves down to reform the C=O $\pi$-bond, and the C–Cl bond breaks heterolytically, releasing a chloride ion (Cl$^-$). This is an elimination step. No proton is transferred. Not a Brønsted–Lowry acid–base reaction.
• Step 3: The intermediate has a positively charged nitrogen with an N–H bond. A chloride ion (Cl$^-$), acting as a base, uses its lone pair to attack and accept the proton (H$^+$) from the nitrogen. The N–H bond breaks, and the electrons remain on nitrogen to neutralize its positive charge, forming the neutral amide and HCl. This is a clear proton transfer (Cl$^-$ accepts H$^+$, the protonated amide donates H$^+$). This is a Brønsted–Lowry acid–base reaction.
• Step 4: A second molecule of ethylamine has a lone pair on its nitrogen. This lone pair attacks the hydrogen atom of the HCl molecule produced in Step 3. The H–Cl bond breaks, with the electrons going to chlorine to form Cl$^-$. The ethylamine accepts the proton to form the ethylammonium ion (CH$_3$CH$_2$NH$_3^+$), which pairs with Cl$^-$ to form ethylammonium chloride. This is another proton transfer (ethylamine accepts H$^+$, HCl donates H$^+$). This is a Brønsted–Lowry acid–base reaction.

Therefore, steps 3 and 4 are the Brønsted–Lowry acid–base reactions. The correct option is B.

Key Takeaways

• A Brønsted–Lowry acid–base reaction in a mechanism is always a proton transfer (H$^+$ moves from one species to another).
• Nucleophilic addition and elimination steps involve electron pair movement to form/break bonds to atoms like carbon, oxygen, or chlorine, but do not involve proton transfer.
• Always look for the curly arrow pointing to a hydrogen atom and the adjacent bond breaking to identify acid–base steps.

Common Mistakes

• Confusing nucleophilic attack with acid–base reactions: Students may see a lone pair attacking something and assume it is a base reacting with an acid. However, if the lone pair attacks a carbon atom (nucleophilic attack) rather than a hydrogen atom (protonation), it is not a Brønsted–Lowry acid–base reaction.
• Missing the second acid–base step: Step 4 is often overlooked. The overall equation shows two moles of amine reacting, but only one is incorporated into the amide product. The second mole acts as a base to neutralize the HCl produced, which is exactly what Step 4 depicts. Forgetting to check Step 4 leads to choosing option C (3 only).

Things to Be Careful About

• Ensure you distinguish between the Lewis definition of an acid/base (electron pair acceptor/donor) and the Brønsted–Lowry definition (proton donor/acceptor). Steps 1 and 2 involve Lewis acid-base interactions (formation of dative bonds), but only steps 3 and 4 are Brønsted–Lowry.
• Read the curly arrows carefully: an arrow pointing to H means proton transfer; an arrow pointing to C or another atom means a different type of step.

Background Concept

Redox reactions are identified by changes in oxidation number. Oxidation number is a bookkeeping charge assigned by rules: uncombined element 0; oxygen usually $-2$; hydrogen usually $+1$ (but $-1$ in metal hydrides); the sum of oxidation numbers in a neutral species is 0. In any redox reaction, the total number of electrons lost by the reducing agent equals the total number gained by the oxidising agent. Therefore, a balanced redox equation must balance both atoms and electron transfer. The question also uses stoichiometric coefficients: $2$ formula units of X produce $4$ Cr atoms, so each X contains $2$ Cr atoms.

Understanding the Question

We are told exactly how much of each species reacts: $2\text{X} + 3\text{Si} \rightarrow 4\text{Cr} + 3\text{(silicon compound)}$. The silicon compound has Si in oxidation state $+4$. Since only Cr and Si change oxidation state, the other element in X (O or H) must be unchanged in the silicon product. We need choose among four chromium compounds. The options differ only in the non-chromium element and its amount: $\text{O}_3$, $\text{H}_6$, $\text{H}_4$, $\text{O}_4$. The correct formula must give a balanced atom count and a balanced electron transfer.

Approach

Use the redox bookkeeping first. Three Si atoms are oxidised from 0 to $+4$, so they release $3 \times 4 = 12$ electrons. Those $12$ electrons must reduce the $4$ Cr atoms produced. If each Cr in X has oxidation state $+x$, then $4x = 12$, so $x = +3$. Thus X contains Cr in the $+3$ state. A neutral compound with two $\text{Cr}^{3+}$ ions needs the other atoms to supply $-6$ charge. With oxygen ($-2$), that is three O atoms, giving $\text{Cr}_2\text{O}_3$. Then confirm the full equation balances. Check the other options against the same electron and atom balance.

Step-by-Step Reasoning

1. From the coefficient $2\text{X} \rightarrow 4\text{Cr}$, each formula unit of X contains $2$ Cr atoms.
2. The $3$ Si atoms go from oxidation state 0 to $+4$: total electrons lost = $3 \times 4 = 12$.
3. These $12$ electrons reduce $4$ Cr atoms from $+x$ to 0: $4x = 12$, so $x = +3$. So Cr in X is $+3$.
4. X is neutral. $2$ Cr at $+3$ contribute $+6$. The remaining atoms must contribute $-6$.
   • Oxygen has oxidation state $-2$, so $3$ O atoms give $-6$: $\text{Cr}_2\text{O}_3$.
   • Hydrogen would need to be $-1$ (as in a metal hydride) to give $-6$, giving $\text{Cr}_2\text{H}_6$. But in the silicon product $\text{SiH}_4$, H is $+1$ and Si would be $-4$, not $+4$. Also H would change oxidation state, contradicting the statement.
   • $\text{Cr}_2\text{H}_4$ would require Cr $+2$ if H is $-1$; then $4$ Cr atoms would gain only $8$ electrons, not the $12$ supplied by Si.
5. Check option A by balancing:

Atoms: $\text{Cr}: 4 = 4$; $\text{O}: 6 = 6$; $\text{Si}: 3 = 3$. Oxidation states: Cr $+3 \rightarrow 0$ (gain $3$ e each, $12$ e total); Si $0 \rightarrow +4$ (lose $4$ e each, $12$ e total). Oxygen remains $-2$ throughout.
6. Option D, $\text{Cr}_2\text{O}_4$, would have Cr $+4$ ($2\text{Cr} + 4(-2) = 0$), so $4$ Cr atoms would need $16$ electrons, but Si supplies only $12$; it also cannot form $3$ identical silicon compounds with Si in the $+4$ state from $8$ O atoms. So D is impossible.

Therefore $\text{X} = \text{Cr}_2\text{O}_3$, option A.

Key Takeaways

• Stoichiometric coefficients give atom ratios that can identify an unknown formula.
• Redox reactions must balance electron transfer as well as atoms.
• Oxidation numbers are bookkeeping charges; the "only elements that change oxidation state" constraint is a powerful filter.
• A neutral compound's oxidation numbers must sum to zero.

Common Mistakes

• Assuming hydrogen is always $+1$. In metal hydrides it is $-1$, but in silanes it is $+1$; either way, H would change oxidation state, which is forbidden here.
• Forgetting to check electron balance: a formula can balance atoms but not redox (as with $\text{Cr}_2\text{O}_4$).
• Thinking the silicon compound must be $\text{SiO}_2$ without checking; here the $+4$ state and oxygen as the unchanged element make $\text{SiO}_2$ the only sensible product.
• Ignoring the phrase "only Cr and Si change oxidation state"; it eliminates any compound containing H, because H would have to change between X and the silicon product.

Things to Be Careful About

• Use oxidation-number rules consistently: O is $-2$ in both $\text{Cr}_2\text{O}_3$ and $\text{SiO}_2$; Si is 0 as an element and $+4$ in $\text{SiO}_2$; Cr is $+3$ in $\text{Cr}_2\text{O}_3$ and 0 as an element.
• The electron transfer must be equal: $4 \text{ Cr} \times 3 \text{ e} = 3 \text{ Si} \times 4 \text{ e} = 12 \text{ e}$.
• The question gives mole ratios, so the coefficients in the balanced equation must match exactly: $2, 3, 4, 3$.
• For an MCQ, after finding a candidate that satisfies all conditions, confirm the other options fail at least one condition.

Background Concept

This question uses Le Chatelier's principle: when a system at equilibrium is disturbed, the position of equilibrium shifts to minimise the disturbance. For a temperature change, the equilibrium shifts in the direction that absorbs or releases heat to oppose the change. Here the forward reaction is exothermic ($\Delta H = -9.5\text{ kJ mol}^{-1}$), so the backward reaction is endothermic. Raising the temperature adds heat to the system, so the equilibrium shifts in the endothermic direction, i.e. backwards: $\text{H}_2 + \text{I}_2 \leftarrow 2\text{HI}$. This consumes $\text{HI}$ and produces $\text{H}_2$ and $\text{I}_2$.

The stoichiometry of the reaction is also essential: the equation shows that two moles of $\text{HI}$ decompose to give one mole of $\text{H}_2$ and one mole of $\text{I}_2$. Therefore any change in $[\text{HI}]$ is twice the corresponding change in $[\text{H}_2]$ or $[\text{I}_2]$.

Understanding the Question

We are given the equilibrium concentrations at 450°C: $[\text{H}_2] = 1.02$, $[\text{I}_2] = 0.02$, $[\text{HI}] = 0.98$ (all in mol dm$^{-3}$). The temperature is then increased at constant volume and a new equilibrium is reached. We are told that one component has a concentration of $0.80\text{ mol dm}^{-3}$. The task is to find the new $[\text{H}_2]$.

The command is essentially a calculation disguised as a multiple-choice question. The key first step is to decide which component falls to 0.80. Since increasing the temperature shifts the equilibrium backwards, $\text{HI}$ must decrease and both $\text{H}_2$ and $\text{I}_2$ must increase. Only $\text{HI}$ can therefore be the component that falls to 0.80.

Approach

1. Use the sign of $\Delta H$ to decide the direction of the shift when temperature increases.
2. Identify which species falls to 0.80.
3. Calculate the decrease in $[\text{HI}]$.
4. Use the 2:1 stoichiometric ratio to find the increase in $[\text{H}_2]$.
5. Add this increase to the original $[\text{H}_2]$.

Step-by-Step Reasoning

• The forward reaction is exothermic because $\Delta H$ is negative. Increasing the temperature favours the endothermic backward reaction.
• In the backward reaction, $\text{HI}$ is consumed and $\text{H}_2$ and $\text{I}_2$ are formed. Hence $[\text{HI}]$ falls; $[\text{H}_2]$ and $[\text{I}_2]$ rise.
• The component that falls to 0.80 must be $\text{HI}$: $0.98 \rightarrow 0.80$.
• Decrease in $[\text{HI}] = 0.98 - 0.80 = 0.18\text{ mol dm}^{-3}$.
• From $2\text{HI} \rightarrow \text{H}_2 + \text{I}_2$, the amount of $\text{H}_2$ formed is half the amount of $\text{HI}$ decomposed: $0.18/2 = 0.09\text{ mol dm}^{-3}$.
• New $[\text{H}_2] = 1.02 + 0.09 = 1.11\text{ mol dm}^{-3}$.

This corresponds to option C.

Why the other options are wrong:

• A (0.80) is simply the new concentration of HI, not H2.
• B (0.93) would result from subtracting 0.09 from 1.02, which is the wrong direction.
• D (1.20) would result from adding 0.18 to 1.02, ignoring the 2:1 stoichiometric ratio.

Key Takeaways

• For an exothermic forward reaction, increasing temperature shifts equilibrium to the left.
• Identify which species can actually decrease before doing any arithmetic.
• Use the stoichiometric coefficients, not a 1:1 assumption, when converting a change in one concentration into a change in another.

Common Mistakes

• Assuming that H2 falls because the question mentions a fall; in fact H2 is produced by the backward shift.
• Using a 1:1 ratio between HI and H2 instead of 2:1.
• Confusing the sign of $\Delta H$: a negative $\Delta H$ means the forward reaction is exothermic, so increasing temperature shifts equilibrium to the endothermic (backward) direction.
• Forgetting that the temperature change is at constant volume; here that simply means no concentration change from volume change, but the equilibrium shift still occurs.

Things to Be Careful About

• Keep units as mol dm$^{-3}$ throughout.
• The stoichiometric ratio is $2\text{HI} : 1\text{H}_2 : 1\text{I}_2$, so divide the HI change by 2.
• The new concentration is found by adding the increase to the original concentration, not by using the new HI concentration directly.
• In an MCQ, always check that your answer matches one of the options; if not, revisit the direction of shift or the stoichiometry.

Background Concept

The standard enthalpy change of formation, $\Delta H_f^\ominus$, is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (298 K and 1 atm). By definition, an element in its standard state — here $\text{O}_2\text{(g)}$ — has $\Delta H_f^\ominus = 0$.

Hess's law states that the enthalpy change of a reaction depends only on the initial and final states, not on the route taken. When formation enthalpies are known, the reaction enthalpy is found by:

$\Delta H^\ominus_{\text{reaction}} = \Sigma \Delta H_f^\ominus(\text{products}) - \Sigma \Delta H_f^\ominus(\text{reactants})$

Each formation enthalpy is multiplied by the stoichiometric coefficient of that substance in the balanced equation.

Understanding the Question

The question gives the balanced equation for the oxidation of ammonia and a table of standard enthalpies of formation for $\text{NH}_3\text{(g)}$, $\text{NO(g)}$ and $\text{H}_2\text{O(g)}$. Oxygen gas is an element in its standard state, so its $\Delta H_f^\ominus$ is zero and it does not appear in the table. We must calculate the standard enthalpy change for the whole reaction.

Approach

Apply the Hess's law formula directly:

1. Sum the formation enthalpies of the products, each multiplied by its coefficient.
2. Sum the formation enthalpies of the reactants, each multiplied by its coefficient.
3. Subtract the reactant sum from the product sum.

Step-by-Step Reasoning

1. Products — $4\text{NO(g)} + 6\text{H}_2\text{O(g)}$:
   $4(+90.3) + 6(-241.8) = 361.2 - 1450.8 = -1089.6\text{ kJ mol}^{-1}$

2. Reactants — $4\text{NH}_3\text{(g)} + 5\text{O}_2\text{(g)}$:
   $4(-46.1) + 5(0) = -184.4\text{ kJ mol}^{-1}$

3. Reaction enthalpy:
   $\Delta H^\ominus = -1089.6 - (-184.4) = -905.2\text{ kJ mol}^{-1}$

The negative sign shows the reaction is exothermic — ammonia oxidation releases heat.

Key Takeaways

• Always multiply each $\Delta H_f^\ominus$ by its stoichiometric coefficient.
• Elements in their standard states contribute zero.
• The formula $\Delta H^\ominus = \Sigma \Delta H_f^\ominus(\text{products}) - \Sigma \Delta H_f^\ominus(\text{reactants})$ is the standard route from formation data.

Common Mistakes

• Forgetting coefficients: using $+90.3$ instead of $4(+90.3)$.
• Sign error in subtraction: forgetting that subtracting $-184.4$ means adding $184.4$.
• Ignoring O2: some mistakenly assign a value to $\text{O}_2$ instead of zero.
• Reversing the subtraction: using reactants minus products gives the wrong sign.

Things to Be Careful About

• Watch the sign of each formation enthalpy carefully — $\text{NH}_3$ and $\text{H}_2\text{O}$ are negative, $\text{NO}$ is positive.
• Units are kJ mol⁻¹ for the reaction as written.
• Option A (+905.2) is the trap for getting the sign wrong; option D (−1274.0) is the trap for miscounting coefficients.

Background Concept

Boltzmann distribution curves show the distribution of molecular energies in a sample of gas at a given temperature. The area under the curve represents the total number of molecules (which remains constant). The peak of the curve corresponds to the most probable energy. The curve starts at the origin (no molecules have zero energy) and tails off to the right.
The activation energy ($E_a$) is the minimum energy required for a successful collision (a reaction to occur). It is represented by a vertical line on the distribution graph. The area under the curve to the right of the $E_a$ line represents the fraction of molecules with energy greater than or equal to $E_a$.

Understanding the Question

The question asks to identify the curve and line corresponding to two separate conditions:

1. Decomposition at a higher temperature.
2. Decomposition with a catalyst present.
   We are given two curves (P and Q) representing different temperatures, and two vertical lines (X and Y) representing different activation energies.

Approach

1. Recall how temperature affects the Boltzmann distribution curve: higher temperature means higher average kinetic energy, shifting the curve to the right and lowering the peak.
2. Recall how a catalyst affects the activation energy: a catalyst lowers the activation energy by providing an alternative pathway.
3. Match these effects to the given curves (P, Q) and lines (X, Y) to find the correct option.

Step-by-Step Reasoning

1. Temperature effect: When the temperature of a gas increases, the molecules have more kinetic energy on average. The Boltzmann distribution curve changes: the peak becomes lower (fewer molecules have the most probable energy), the curve becomes broader, and it shifts to the right towards higher energies. The area under the curve remains the same (total number of molecules is constant). Looking at the diagram, curve Q is broader, flatter, and shifted to the right compared to curve P. Therefore, curve Q represents the higher temperature, and curve P represents the lower temperature.
2. Catalyst effect: A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy ($E_a$). On the Boltzmann distribution graph, a lower activation energy is represented by a vertical line further to the left (lower energy value). Looking at the diagram, line X is at a lower energy than line Y. Therefore, line X represents the lower activation energy (with a catalyst), and line Y represents the higher activation energy (without a catalyst).
3. Conclusion: The higher temperature is represented by curve Q, and the catalyst present is represented by line X. This matches option C.

Key Takeaways

• Higher temperature shifts the Boltzmann distribution curve to the right, lowers the peak, and broadens the curve. The area under the curve remains constant.
• A catalyst lowers the activation energy, which is shown as a vertical line at a lower energy value on the Boltzmann distribution graph.
• The area under the curve to the right of the activation energy line represents the proportion of molecules with sufficient energy to react. Both a higher temperature and a lower activation energy increase this area, thus increasing the rate of reaction.

Common Mistakes

• Confusing the effect of temperature on the curve: thinking a higher temperature makes the peak higher (it actually makes it lower and broader because the area must remain constant).
• Confusing the effect of a catalyst on activation energy: thinking a catalyst increases activation energy (it decreases it).
• Misreading the graph: confusing which line is at a lower energy (X is to the left of Y on the energy axis, so X is lower energy).

Things to Be Careful About

• Remember that the total area under the Boltzmann distribution curve is constant regardless of temperature or catalyst. Only the shape and position change.
• A catalyst does NOT change the position of the Boltzmann distribution curve itself; it only changes the position of the activation energy line. The question asks for the curve at a higher temperature AND the line with a catalyst. These are two independent changes being asked about.
• Ensure you read the axes correctly: the x-axis is molecular energy (increasing to the right), and the y-axis is the number of molecules.

Background Concept

In covalent bonding, a single bond is always a $\sigma$ (sigma) bond, formed by the head-on overlap of atomic orbitals. A double bond consists of one $\sigma$ bond and one $\pi$ (pi) bond, formed by the side-on overlap of p-orbitals. A triple bond consists of one $\sigma$ bond and two $\pi$ bonds. Therefore, the presence of any double or triple bond in a molecule indicates the presence of $\pi$ bonds; if only single bonds are present, the molecule contains only $\sigma$ bonds.

Regarding molecular shape, VSEPR (Valence Shell Electron Pair Repulsion) theory predicts that electron pairs around a central atom arrange themselves to minimise repulsion. Atoms with four single bonds (or a combination of bonds and lone pairs summing to four regions of electron density) are sp$^3$ hybridised and adopt a tetrahedral geometry with bond angles of approximately 109.5°. When atoms with tetrahedral geometry are joined in a ring, a planar regular hexagon (with 120° internal angles) is geometrically incompatible with the ideal 109.5° bond angles. Additionally, a planar ring would force all adjacent substituents into eclipsed conformations, creating high torsional strain. To relieve both angle strain and torsional strain, six-membered rings containing sp$^3$ atoms pucker into non-planar conformations, most commonly the chair or boat forms.

Understanding the Question

The question provides a skeletal structure of a six-membered cyclic sugar molecule (a pyranose ring, such as glucose in its cyclic form). The structure contains five carbon atoms and one oxygen atom in the ring, with hydroxyl (-OH) and hydroxymethyl (-CH$_2$OH) substituents. Four statements are given about the bonding and 3D geometry of this molecule, and the task is to identify which two are correct. Statements 1 and 2 concern the types of covalent bonds present ($\sigma$ vs $\pi$), while Statements 3 and 4 concern the planarity of the ring.

Approach

1. Examine the skeletal structure for any double or triple bonds to determine whether $\pi$ bonds are present.
2. Identify the hybridisation of the atoms in the ring by counting the number of bonded atoms and lone pairs.
3. Apply VSEPR theory and knowledge of ring strain to deduce whether a ring of sp$^3$ hybridised atoms can be planar.

Step-by-Step Reasoning

• Bonding (Statements 1 and 2): Look closely at Fig. 8.1. The skeletal structure shows only single lines connecting atoms. There are no double lines (=) or triple lines ($\equiv$). Every bond shown (C–C, C–O, C–H, O–H) is a single covalent bond. Since all bonds are single, they are all $\sigma$ bonds. There are no $\pi$ bonds in this molecule. Therefore, Statement 1 ('This molecule contains only $\sigma$ bonds') is correct, and Statement 2 is incorrect.
• Ring Planarity (Statements 3 and 4): The carbon atoms in the ring are each bonded to four groups (e.g., two ring atoms, one hydrogen, and one -OH or -CH$_2$OH group). This means every ring carbon is sp$^3$ hybridised with tetrahedral geometry. The oxygen atom in the ring is bonded to two carbons and has two lone pairs, which also makes it sp$^3$ hybridised. Because all ring atoms are sp$^3$ hybridised, the ideal bond angles are ~109.5°. If the ring were planar, it would be a regular hexagon with 120° angles, causing significant angle strain. Furthermore, a planar conformation would cause severe torsional strain as the substituents on adjacent carbons would be forced into eclipsed positions. To minimise these strains, the ring puckers into a non-planar shape (such as a chair conformation). Therefore, Statement 4 ('The six-membered ring is non-planar') is correct, and Statement 3 is incorrect.
• Conclusion: Statements 1 and 4 are correct. This matches option B.

Key Takeaways

• Single bonds are always $\sigma$ bonds; $\pi$ bonds only exist in double or triple bonds.
• Atoms with four regions of electron density (sp$^3$ hybridised) have tetrahedral geometry (~109.5° bond angles).
• Six-membered rings containing only sp$^3$ atoms cannot be planar; they must pucker to achieve ideal bond angles and avoid torsional strain.

Common Mistakes

• Assuming that a hexagonal ring drawn on paper is planar: skeletal structures are 2D representations and do not accurately depict 3D geometry. A hexagon drawn flat does not mean the molecule is flat.
• Misidentifying the oxygen in the ring: the ring oxygen has two bonds and two lone pairs, making it sp$^3$ hybridised, not sp$^2$.
• Confusing $\sigma$ and $\pi$ bonds: remembering that every single bond is a $\sigma$ bond, and $\pi$ bonds are only added on top of a $\sigma$ bond in multiple bonds.

Things to Be Careful About

• Always check the skeletal structure carefully for double bonds (like C=O or C=C) which would introduce $\pi$ bonds. In this sugar molecule, it is in the cyclic hemiacetal form, so there are no carbonyl groups or alkenes.
• Do not assume planarity from the 2D drawing. Apply hybridisation and VSEPR theory to determine the true 3D shape.

Background Concept

Substances can be classified by the type of structure they form. In a giant covalent structure (also called a network covalent structure), all the atoms are joined to each other by strong covalent bonds in a continuous 3D lattice. Examples include diamond, graphite and silicon(IV) oxide. In a simple molecular structure, atoms are joined within each molecule by strong covalent bonds, but the molecules themselves are held together only by weak intermolecular forces such as van der Waals' forces, permanent dipole–dipole forces, or hydrogen bonds.

When a substance melts, the forces between its particles are overcome. For a molecular substance, this means breaking intermolecular forces, not the covalent bonds inside each molecule. For a giant covalent substance, there are no separate molecules, so melting requires breaking covalent bonds throughout the network. This is why giant covalent substances have very high melting points.

Understanding the Question

The question asks which substance has covalent bonds broken as it melts. The key distinction is between:

• bonds within a molecule (intramolecular covalent bonds), and
• forces between molecules (intermolecular forces).

You are being tested on whether you can identify which of the four options is a giant covalent solid rather than a molecular solid. The options are silicon(IV) oxide, ice, iodine and ethanol.

Approach

For each option, ask: "What kind of structure does this substance have?"

• If it is molecular, melting breaks intermolecular forces, not covalent bonds.
• If it is giant covalent, melting breaks covalent bonds.

Only one of the four substances is giant covalent, so that substance is the answer.

Step-by-Step Reasoning

1. Silicon(IV) oxide, SiO₂ — Each silicon atom is bonded to four oxygen atoms and each oxygen atom is bonded to two silicon atoms, forming a giant 3D covalent network. There are no discrete molecules. When SiO₂ melts, the Si–O covalent bonds must be broken. This matches the question.

2. Ice, H₂O — Ice is a molecular solid. Each water molecule contains O–H covalent bonds, but the molecules are held together by hydrogen bonds between the oxygen of one molecule and a hydrogen of another. When ice melts, only these hydrogen bonds are broken; the O–H covalent bonds stay intact.

3. Iodine, I₂ — Iodine is a simple molecular substance. Each I₂ molecule has a strong I–I covalent bond, but the molecules are held together in the solid by weak van der Waals' forces. Melting iodine breaks only those weak intermolecular forces, not the I–I covalent bond.

4. Ethanol, C₂H₅OH — Ethanol is a molecular liquid/solid with covalent bonds within each molecule (C–C, C–H, C–O and O–H). The molecules are held together by hydrogen bonding and van der Waals' forces. Melting ethanol breaks these intermolecular forces, not its covalent bonds.

Therefore, the only substance in which covalent bonds are broken on melting is silicon(IV) oxide.

Key Takeaways

• Giant covalent structures have high melting points because melting requires breaking covalent bonds.
• Molecular substances have lower melting points because only weak intermolecular forces are broken on melting.
• "Hydrogen bonds" are intermolecular forces, not covalent bonds, even though the word "bond" is used.
• To answer this type of question, first classify the substance as giant covalent, ionic, metallic, or molecular.

Common Mistakes

• Saying ice breaks covalent bonds: Ice is molecular; melting breaks hydrogen bonds between water molecules, not O–H covalent bonds.
• Confusing iodine's I–I covalent bond with the forces between I₂ molecules: The I–I bond is intramolecular and is not broken on melting.
• Thinking ethanol's O–H bond breaks on melting: Ethanol molecules stay intact; only intermolecular forces are overcome.
• Not recognising SiO₂ as giant covalent: SiO₂ is often written as a formula that looks molecular, but it is a network solid.

Things to Be Careful About

• Read the phrase "covalent bonds broken as it melts" carefully: it specifically refers to intramolecular covalent bonds, not intermolecular forces.
• Remember that substances like diamond, graphite and SiO₂ are giant covalent, whereas most small covalent compounds (water, iodine, ethanol, carbon dioxide) are molecular.
• Do not confuse melting with chemical reaction: melting is a physical change, so the molecules of a molecular substance remain unchanged.

Background Concept

Ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms or ions. For a given species, successive ionisation energies rise because each electron removed leaves a more positive ion, so the remaining electrons are held more strongly by the nucleus.

The energy needed to remove an electron depends on three factors: the nuclear charge (the number of protons), the distance of the electron from the nucleus, and the shielding caused by inner electrons. Shielding reduces the effective nuclear charge felt by an outer electron. An electron in a lower principal shell ($n = 2$) is closer to the nucleus and is shielded only by the $1\text{s}$ electrons, whereas an electron in a higher principal shell ($n = 3$) is further away and is shielded by all the electrons in the inner shells.

When comparing the second ionisation energies of two elements, it is essential to identify exactly which electron is removed after the first electron has already been lost.

Understanding the Question

This question compares the second ionisation energy of sodium with that of magnesium. It tests whether you can work out the electron configuration of each +1 ion and identify the shell from which the second electron is removed. The correct explanation must refer to the attraction between the nucleus and the electron being removed.

Approach

Start by writing the full electron configurations of Na and Mg. Remove one electron to form $\text{Na}^+$ and $\text{Mg}^+$. Then remove a second electron and note which shell it comes from. Compare distance from the nucleus and shielding. Use this comparison to judge each option.

Step-by-Step Reasoning

1. Sodium has atomic number 11: $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^1$. Its first ionisation removes the $3\text{s}$ electron, leaving $\text{Na}^+$ with configuration $1\text{s}^2\,2\text{s}^2\,2\text{p}^6$.

2. Magnesium has atomic number 12: $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2$. Its first ionisation removes one $3\text{s}$ electron, leaving $\text{Mg}^+$ with configuration $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^1$.

3. The second ionisation of sodium removes an electron from the $2\text{p}$ subshell. This electron is in the second principal shell, close to the nucleus, and is shielded only by the two $1\text{s}$ electrons.

4. The second ionisation of magnesium removes the remaining $3\text{s}$ electron. This electron is in the third principal shell, further from the nucleus, and is shielded by the full $1\text{s}^2\,2\text{s}^2\,2\text{p}^6$ core.

5. Therefore the electron removed from $\text{Na}^+$ experiences a much stronger attraction to the nucleus than the electron removed from $\text{Mg}^+$. This is why the second ionisation energy of sodium is larger. The actual values reflect this: the second ionisation energy of Na is about $4560\ \text{kJ mol}^{-1}$, while that of Mg is about $1450\ \text{kJ mol}^{-1}$.

Now check the options:

• A is correct because it states this greater attraction.
• B is incorrect because $\text{Mg}^+$ has the greater nuclear charge (12 protons compared with 11 in $\text{Na}^+$).
• C is incorrect because the outer electron of $\text{Na}^+$ is less shielded, not more shielded, than the outer electron of $\text{Mg}^+$.
• D is not a valid explanation: the outer electrons of the neutral atoms are both in the $3\text{s}$ orbital, so this provides no distinction; after first ionisation the electrons removed are from different shells ($2\text{p}$ for $\text{Na}^+$ and $3\text{s}$ for $\text{Mg}^+$).

Key Takeaways

To compare successive ionisation energies, always write the configurations of the ions and identify the shell from which the electron is removed. A large jump occurs when removing an electron from a completed inner shell. Shielding and distance from the nucleus can outweigh a greater nuclear charge.

Common Mistakes

Choosing B: the nuclear charge of $\text{Na}^+$ is not greater; $\text{Mg}^+$ has 12 protons. Choosing C: the outer electron of $\text{Na}^+$ is less shielded, not more shielded. Choosing D: the orbitals involved are different ($2\text{p}$ for $\text{Na}^+$ and $3\text{s}$ for $\text{Mg}^+$). Also, forgetting to consider the ion after the first ionisation.

Things to Be Careful About

Use the correct electron configurations for the ions, not the atoms. Remember that the second electron removed from Na comes from the second shell, not the third. In ionisation energy questions, compare the electron being removed, not the overall atom.

Background Concept

BF$_3$ is electron deficient: boron has only six valence electrons in three B–F bonds, so it can accept a lone pair. NH$_3$ has a lone pair on nitrogen. The reaction forms a coordinate (dative) bond from N to B, giving F$_3$B–NH$_3$.

VSEPR theory predicts shapes from electron-pair repulsion:

• BF$_3$: 3 bonding pairs, no lone pairs → trigonal planar, 120°.
• NH$_3$: 3 bonding pairs + 1 lone pair → trigonal pyramidal, ~107°.
• F$_3$B–NH$_3$: B has 4 bonding pairs; N has 4 bonding pairs. Both are tetrahedral, ~109.5°.

Understanding the Question

The question asks how the F–B–F angle and H–N–H angle change when BF$_3$ and NH$_3$ form the adduct. Need compare each angle before and after.

Approach

1. Determine shape and bond angle of BF$_3$.
2. Determine shape and bond angle of NH$_3$.
3. Determine shape around B and N in F$_3$B–NH$_3$.
4. Compare each angle.

Step-by-Step Reasoning

• In BF$_3$, B has three bond pairs and no lone pair. VSEPR: trigonal planar, F–B–F = 120°.
• In NH$_3$, N has three bond pairs and one lone pair. Lone pair repulsion compresses H–N–H to about 107°.
• In F$_3$B–NH$_3$, the N lone pair becomes a B–N coordinate bond. B now has four bond pairs, so it is tetrahedral: F–B–F ≈ 109.5° (decrease from 120°).
• N now has four bond pairs and no lone pair, so it is tetrahedral: H–N–H ≈ 109.5° (increase from ~107°).
• Therefore F–B–F decreases, H–N–H increases → B.

Key Takeaways

• A molecule with 3 bond pairs and no lone pair is trigonal planar.
• A molecule with 3 bond pairs and 1 lone pair is trigonal pyramidal.
• A molecule/ion with 4 bond pairs and no lone pair is tetrahedral.
• Formation of a coordinate bond can change the electron-pair geometry of both atoms.

Common Mistakes

• Assuming both angles decrease because both atoms become tetrahedral. N was already approximately tetrahedral in electron-pair arrangement; the lone pair compressed the bond angle below 109.5°. Removing the lone pair (by making it a bond pair) lets the H–N–H angle open up slightly.
• Forgetting that BF$_3$ is trigonal planar, not tetrahedral.

Things to Be Careful About

• The H–N–H angle in NH$_3$ is not exactly 109.5°; it is about 107° due to lone-pair repulsion.
• In F$_3$B–NH$_3$, both B and N have four bonding pairs, so both are tetrahedral with angles near 109.5°.
• A coordinate bond counts as a normal bond pair for VSEPR purposes.

Background Concept

The standard enthalpy change of reaction, ΔHr°, is the enthalpy change when the molar quantities shown in the balanced equation react under standard conditions. Standard conditions are usually taken as 298 K (25 °C) and 100 kPa (1 bar; older sources often use 1 atm = 101.325 kPa).

The standard enthalpy change of formation, ΔHf°, is a special case: it is the enthalpy change when 1 mol of a compound is formed from its elements in their standard states.

Understanding the Question

The question gives the Haber process equation:

N2(g) + 3H2(g) ⇌ 2NH3(g) ΔHr° = −92.2 kJ mol⁻¹

It asks which statement about ΔHr° is correct. This is a definition-and-interpretation question, not a calculation.

Approach

Check each option against the definition of standard enthalpy change and the stoichiometry of the balanced equation.

Step-by-Step Reasoning

• Option A: Standard enthalpy changes are measured at 298 K, not 298 °C. 298 °C is about 571 K, so this statement is incorrect.
• Option B: Standard pressure is 100 kPa, and 101 kPa is commonly used as 1 atm. This statement is correct.
• Option C: The standard enthalpy change of formation of ammonia refers to forming 1 mol of NH3 from its elements. The given equation forms 2 mol of NH3, so ΔHr° is the enthalpy change of reaction, not the standard enthalpy change of formation of ammonia. Incorrect.
• Option D: The balanced equation shows 1 mol N2 reacting with 3 mol H2, not 1 mol H2. Incorrect.

Therefore, the correct answer is B.

Key Takeaways

• Standard conditions: 298 K and 100 kPa (or 101 kPa as 1 atm).
• ΔHr° always refers to the reaction exactly as written, including its stoichiometric coefficients.
• ΔHf° is always per mole of product formed.

Common Mistakes

• Confusing 298 K with 298 °C.
• Assuming that any enthalpy change involving a product is automatically a formation enthalpy change.
• Misreading the stoichiometry as 1 mol N2 : 1 mol H2 instead of 1 mol N2 : 3 mol H2.

Things to Be Careful About

• Pay attention to the units: kJ mol⁻¹ refers to the reaction as written, not per mole of a particular reactant unless stated.
• Remember that standard pressure may be given as 100 kPa or 101 kPa; both are acceptable in this context.

Background Concept

Hydrogen peroxide decomposes according to the balanced equation

This tells us the mole ratio: every 2 mol of hydrogen peroxide produces 1 mol of oxygen gas. Under room conditions, one mole of any gas occupies $24\text{ dm}^3$ ($24\ 000\text{ cm}^3$). The concentration of a solution is the amount of solute in moles divided by the volume of solution in dm$^3$.

Understanding the Question

The question gives the volume of hydrogen peroxide solution ($20.0\text{ cm}^3$) and the volume of oxygen gas produced at room conditions ($600\text{ cm}^3$). It asks for the concentration of the original hydrogen peroxide solution. To find concentration, two conversions are needed: gas volume $\rightarrow$ moles of O$_2$, and moles of O$_2$ $\rightarrow$ moles of H$_2$O$_2$ using the reaction ratio.

Approach

Use the molar volume to convert $600\text{ cm}^3$ of oxygen into moles. Then use the balanced equation to find the moles of hydrogen peroxide decomposed. Finally, divide those moles by the volume of hydrogen peroxide solution converted into dm$^3$.

Step-by-Sep Reasoning

1. Convert oxygen volume to dm$^3$: $600\text{ cm}^3 = 0.600\text{ dm}^3$.
2. Use molar volume:

1. Apply the mole ratio. Since $2\text{ H}_2\text{O}_2 \rightarrow 1\text{ O}_2$, the moles of hydrogen peroxide decomposed are twice the moles of oxygen:

1. Convert solution volume: $20.0\text{ cm}^3 = 0.0200\text{ dm}^3$.
2. Calculate concentration:

This corresponds to option B.

Among the distractors, option C ($1.25\text{ mol dm}^{-3}$) is the value obtained if the $2:1$ stoichiometric ratio is ignored and the moles of hydrogen peroxide are taken as equal to the moles of oxygen. Options A and D arise from using a wrong solution volume or a wrong conversion factor.

Key Takeways

The essential chain for gas stoichiometry in solution is: volume of gas $\rightarrow$ moles of gas (using molar volume), moles of gas $\rightarrow$ moles of solute (using the balanced equation), then moles of solute $\rightarrow$ concentration. Always associate room conditions with $24\text{ dm}^3\text{ mol}^{-1}$.

Common Mistakes

• Forgetting the $2:1$ mole ratio and using $0.0250$ mol of H$_2$O$_2$ instead of $0.0500$ mol; this gives option C.
• Using $22.4\text{ dm}^3$ as the molar volume; that value refers to standard temperature and pressure (STP), not room conditions.
• Failing to convert $20.0\text{ cm}^3$ into $0.0200\text{ dm}^3$ before calculating concentration.
• Mixing cm$^3$ and dm$^3$; remember $1\text{ dm}^3 = 1000\text{ cm}^3$.

Things to Be Careful About

The balanced equation must show that oxygen is produced from two molecules of hydrogen peroxide, so the ratio is $2:1$. Write the volume of oxygen as a fraction of the molar volume in the same units. Keep consistent units throughout, and report the concentration with its unit, $\text{mol dm}^{-3}$.

Background Concept

The ideal gas equation $pV = nRT$ relates the pressure $p$ (in Pa), volume $V$ (in m³), amount of substance $n$ (in mol), and temperature $T$ (in K) of an ideal gas, where $R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$ is the gas constant. Because the compound is vaporised, it behaves as a gas and obeys this equation. The amount of substance is linked to the mass $m$ and the relative molecular mass $M_r$ by $n = \frac{m}{M_r}$. Combining the two relationships lets us determine $M_r$ from a single measurement of pressure, volume and temperature of a known mass of vapour.

Understanding the Question

The question gives the mass of a vaporised organic compound ($0.15\text{ g}$), the volume it occupies ($65.0\text{ cm}^3$), the temperature ($405\text{ K}$) and the pressure ($1.00 \times 10^5\text{ Pa}$), and asks which of four expressions correctly calculates the relative molecular mass $M_r$. The two traps are the algebraic rearrangement of $pV = nRT$ and the conversion of the volume from cm³ to m³ — the pressure is in Pa and $R$ is in J K⁻¹ mol⁻¹, so the volume must be in m³ for the units to work.

Approach

1. Rearrange $pV = nRT$ to make $n$ the subject: $n = \frac{pV}{RT}$.
2. Substitute $n = \frac{m}{M_r}$ and solve for $M_r$: $M_r = \frac{mRT}{pV}$.
3. Convert the volume to SI units: $65.0\text{ cm}^3 = 65.0 \times 10^{-6}\text{ m}^3$.
4. Substitute all values and match the result to the options.

Step-by-Step Reasoning

Start from the ideal gas equation:

Make $n$ the subject:

Now use the relationship between mass, moles and molar mass, $n = \frac{m}{M_r}$:

Rearrange to find $M_r$:

Substitute the given values. The volume must be converted: $65.0\text{ cm}^3 = 65.0 \times 10^{-6}\text{ m}^3$ (since $1\text{ cm}^3 = 10^{-6}\text{ m}^3$).

This is exactly option D.

Why the other options are wrong:

• A and C have the form $\frac{m \times V \times p}{RT}$. This is $m \times n$, not $M_r$ — the rearrangement is inverted, putting $pV$ in the numerator instead of the denominator.
• B has the correct arrangement but uses $65 \times 10^{-3}$ instead of $65 \times 10^{-6}$ for the volume. $10^{-3}$ would be the conversion for dm³, not cm³; $1\text{ cm}^3 = 10^{-6}\text{ m}^3$.