Chemistry 9701/14 — October/November 2025

Background Concept

In nuclide notation, a species is written as $^{A}_{Z}\text{X}$, where:

• $Z$ (the proton number, or atomic number) is the number of protons in the nucleus. It defines the element.
• $A$ (the nucleon number, or mass number) is the total number of protons plus neutrons in the nucleus.
• The number of neutrons is therefore $A - Z$.

For a neutral atom, the number of electrons equals the number of protons. For an ion, the electron count changes: a positive charge means electrons have been lost, and a negative charge means electrons have been gained.

Understanding the Question

This question gives the helium ion $^{3}_{2}\text{He}^+$ and asks which of three statements about its subatomic particles are correct. The key point is that the superscript of the ion is the mass number, the subscript is the proton number, and the charge tells us how many electrons have been lost or gained.

The three statements to judge are:

1. It contains two protons.
2. It contains three neutrons.
3. It contains one electron.

We only need to decide which statements are true and then choose the matching option.

Approach

Use the standard relationships for a nuclide $^{A}_{Z}\text{X}$:

1. Protons = $Z$.
2. Neutrons = $A - Z$.
3. Electrons = $Z$ minus the positive charge (if the ion has a positive charge).

Apply each relationship to helium and compare with the three statements.

Step-by-Step Reasoning

1. Protons: The nuclide is written with the proton number 2 as the lower-left subscript: $^{3}_{2}\text{He}^+$. Therefore there are exactly 2 protons. Statement 1 is correct.

2. Neutrons: The upper-left superscript 3 is the nucleon number: total nucleons (protons + neutrons) = 3. Since there are 2 protons, the number of neutrons is:

   $$\text{neutrons} = A - Z = 3 - 2 = 1$$

   So the ion contains one neutron, not three. Statement 2 is incorrect.

3. Electrons: A neutral helium atom has 2 electrons, matching its proton number. The ion has a charge of $+1$, meaning one electron has been removed:

   $$\text{electrons} = 2 - 1 = 1$$

   So the ion contains one electron. Statement 3 is correct.

The only true statements are 1 and 3, which corresponds to option B.

Key Takeaways

• Always read nuclide notation carefully: the lower number is the proton number and the upper number is the nucleon (mass) number.
• Neutrons are found by subtracting the proton number from the nucleon number.
• For ions, the charge must be applied to the electron count: positive charge means fewer electrons, negative charge means more electrons.
• A quick check on an ion: a $+$ charge is generated by removing the same number of electrons as the charge magnitude.

Common Mistakes

• Confusing mass number with neutron number. The mass number is protons + neutrons, not just neutrons. For $^3_2\text{He}$, the neutron number is $3 - 2 = 1$, not 3.
• Ignoring the ion charge when counting electrons. Forgetting that $\text{He}^+$ has one fewer electron than neutral He would make statement 3 appear false.
• Assuming the subscript is the neutron number. The subscript is the proton number, not the neutron number.

Things to Be Careful About

• The notation places the mass number above and the proton number below, before the element symbol. Check both numbers before doing any arithmetic.
• For a positive ion, always subtract the magnitude of the charge from the neutral electron count; for a negative ion, add it.
• Keep the units simple here: the numbers are counts of particles, and there are no state symbols or equations needed in this question.

Background Concept

Successive ionisation energies are the energies needed to remove each successive electron from a gaseous atom. The first ionisation energy removes one electron from the neutral atom; the second removes one from the singly charged ion, and so on. For a given element, these values generally increase because each electron is removed from an increasingly positive ion, but the key feature is a very large jump when the next electron has to be removed from a completed inner shell. Inner-shell electrons are much closer to the nucleus and are less shielded, so they are held far more strongly. The number of electrons removed before the large jump equals the number of outer-shell (valence) electrons.

For Period 3 elements, the outer shell is the third shell ($n = 3$). Magnesium has the configuration $\text{Mg}: 1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2$, so it has two valence electrons in the 3s subshell.

Understanding the Question

The question provides the first five ionisation energies of an unknown Period 3 element X and asks you to identify it from four options. The values are 736, 1450, 7740, 10500 and 13600 kJ mol$^{-1}$. The key is not the exact numbers but where the large jump occurs. Since X is in Period 3, its group can be deduced from the number of electrons removed before the jump, and this identifies the element.

Approach

List the successive ionisation energies and look for the largest relative increase. A large jump after the nth ionisation energy means the (n+1)th electron is removed from a new, inner shell, so the atom has n valence electrons. Then match n valence electrons to the correct Period 3 element. Check the options against this deduction.

Step-by-Step Reasoning

1. The successive ionisation energies are:
   $I_1 = 736$, $I_2 = 1450$, $I_3 = 7740$, $I_4 = 10500$, $I_5 = 13600$ kJ mol$^{-1}$.
2. Compare consecutive values. The increase from $I_2$ to $I_3$ is from 1450 to 7740, more than a fivefold rise. This is much larger than the increase from $I_1$ to $I_2$ (736 to 1450).
3. The large jump between the second and third ionisation energies shows that the third electron is removed from an inner shell. Therefore X has two electrons in its outer shell.
4. A Period 3 element with two outer electrons is in Group 2. Its electron configuration is $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2$, which is magnesium.
5. Check the other options:
   • Sodium has one outer electron, so the large jump would be between $I_1$ and $I_2$, not between $I_2$ and $I_3$.
   • Silicon has four outer electrons, so the large jump would be between $I_4$ and $I_5$.
   • Argon has eight outer electrons, so no large jump would appear among the first five ionisation energies; its first ionisation energy is also much higher than 736 kJ mol$^{-1}$.

Therefore the correct option is B, magnesium.

Key Takeaways

Successive ionisation energy data reveal the number of valence electrons: the big jump occurs immediately after all valence electrons have been removed. This lets you identify the group of an element. For Period 3, the group number equals the number of outer-shell electrons, so two valence electrons means Group 2, magnesium.

Common Mistakes

• Choosing sodium because the first ionisation energy is relatively low. The pattern of jumps, not just the first value, is what matters; sodium would show a large jump between the first and second ionisation energies.
• Misreading the jump. The jump between the 2nd and 3rd ionisation energies means two electrons are removed before reaching the core, not one.
• Assuming a large absolute value always indicates a jump. Compare successive values relative to each other.
• Forgetting that X is in Period 3. Without this, the same two-valence-electron pattern could also suggest calcium, but Period 3 restricts the answer to magnesium.

Things to Be Careful About

• Ionisation energies are quoted in kJ mol$^{-1}$; no conversion is needed here.
• Use the relative size of the jump, not the absolute values alone.
• For argon, the first five ionisation energies all remove outer-shell (3s and 3p) electrons, so no large jump appears in the data provided; also its first ionisation energy is much higher than 736 kJ mol$^{-1}$.
• In the exam, the answer is the option letter B; write it clearly.

Background Concept

At the same temperature and pressure, equal volumes of gases contain equal numbers of molecules. This means the volume ratios of gases in a reaction are the same as the mole ratios shown by the balanced equation.

At room temperature and pressure, water is a liquid. Therefore, when calculating the final volume of a gas mixture after combustion, the volume of water produced should not be counted unless the question specifically says all products are measured as gases.

Understanding the Question

We are given 10 cm3 of methanethiol gas and 60 cm3 of oxygen gas, both measured at room conditions. The methanethiol burns completely according to:

CH3SH + 3O2 → CO2 + SO2 + 2H2O

We need the final total volume of the gaseous mixture after the reaction, still at room temperature.

Approach

Use the stoichiometric coefficients as volume ratios. First determine how much oxygen is needed to react with all 10 cm3 of methanethiol. Compare this with the 60 cm3 of oxygen supplied to find the limiting reagent and the excess oxygen. Then calculate the volumes of gaseous products and add the leftover oxygen.

Step-by-Step Reasoning

1. Write the balanced equation:
   CH3SH + 3O2 → CO2 + SO2 + 2H2O

2. Convert the coefficients into volume ratios:
   1 volume CH3SH : 3 volumes O2 : 1 volume CO2 : 1 volume SO2
   Water is not counted because it is liquid at room temperature.

3. Oxygen needed for 10 cm3 CH3SH:
   10 × 3 = 30 cm3 O2

4. Oxygen available is 60 cm3, so oxygen is in excess.
   Excess oxygen = 60 − 30 = 30 cm3

5. Gaseous products formed from 10 cm3 CH3SH:
   CO2 = 10 cm3
   SO2 = 10 cm3

6. Total final gas volume:
   leftover O2 + CO2 + SO2 = 30 + 10 + 10 = 50 cm3

Key Takeaways

• Gas volume ratios equal mole ratios when all gases are at the same temperature and pressure.
• Always identify the limiting reagent before calculating the amounts of products.
• At room temperature, water is a liquid and its volume must be excluded from a gas volume calculation.
• Excess reactant remains after the reaction and contributes to the final volume.

Common Mistakes

• Counting the 20 cm3 of water vapour as a gas: this would give 70 cm3, which is option D.
• Forgetting to include the leftover oxygen: this would give only 20 cm3, which is option A.
• Assuming oxygen is the limiting reagent: this is incorrect because 60 cm3 of oxygen is more than the 30 cm3 required.
• Confusing the volume of oxygen used with the volume of oxygen remaining.

Things to Be Careful About

• Make sure all gas volumes are measured under the same conditions before using volume ratios.
• Check the coefficients in the balanced equation carefully.
• Read whether the question asks for the volume of the gas mixture or the total volume including liquids.
• At room conditions, water is liquid; at high temperatures it could be steam and would then be counted as a gas.

Background Concept

A carboxylic acid contains one or more $\text{COOH}$ groups. Each $\text{COOH}$ group releases one $\text{H}^+$ ion, so a monocarboxylic acid is monoprotic: one mole of acid reacts with exactly one mole of $\text{NaOH}$.

The empirical formula is the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual numbers of atoms. They are related by a whole-number multiplier.

Understanding the Question

We are given the mass of an organic acid (4.4 g), the volume and concentration of NaOH used to neutralise it exactly (25.0 cm³ of 2.0 mol dm⁻³), and told the acid has exactly one COOH group per molecule. We must find the empirical formula.

The key is to work backwards: from the NaOH data find the moles of acid, then the molar mass, then the molecular formula, and finally simplify to the empirical formula.

Approach

1. Calculate moles of NaOH from concentration and volume.
2. Use the 1:1 stoichiometry (one COOH per molecule) to get moles of acid.
3. Divide the given mass by the moles to get the molar mass.
4. Use the general formula of a monocarboxylic acid ($\text{C}_n\text{H}_{2n}\text{O}_2$) to find the molecular formula.
5. Divide the molecular formula by the highest common factor to get the empirical formula.

Step-by-Step Reasoning

Step 1 — Moles of NaOH
$\text{moles} = \text{concentration} \times \text{volume in dm}^3$
$= 2.0 \times 0.025 = 0.05 \text{ mol}$.

Step 2 — Moles of acid
Since the acid has one COOH group per molecule, each molecule provides one $\text{H}^+$:
$\text{acid} + \text{NaOH} \rightarrow \text{salt} + \text{H}_2\text{O}$
So moles of acid = 0.05 mol.

Step 3 — Molar mass
$M_r = \frac{\text{mass}}{\text{moles}} = \frac{4.4}{0.05} = 88 \text{ g mol}^{-1}$.

Step 4 — Molecular formula
For a monocarboxylic acid, $\text{C}_n\text{H}_{2n}\text{O}_2$:
$12n + 2n + 32 = 88$
$14n = 56$
$n = 4$
So the molecular formula is $\text{C}_4\text{H}_8\text{O}_2$ (butanoic acid).

Step 5 — Empirical formula
Divide by 2: $\text{C}_4\text{H}_8\text{O}_2 \rightarrow \text{C}_2\text{H}_4\text{O}$.
This matches option B.

Key Takeaways

• The number of COOH groups tells you the acid:base stoichiometry (1 COOH = 1 H⁺ = 1 NaOH).
• Always check whether the question asks for the molecular or empirical formula.
• The general formula for a saturated monocarboxylic acid is $\text{C}_n\text{H}_{2n}\text{O}_2$.

Common Mistakes

• Forgetting to convert cm³ to dm³: 25.0 cm³ = 0.025 dm³, not 25.0 dm³.
• Stopping at the molecular formula: The question asks for the empirical formula, so $\text{C}_4\text{H}_8\text{O}_2$ must be simplified to $\text{C}_2\text{H}_4\text{O}$.
• Assuming the acid is diprotic: The question explicitly states one COOH group per molecule.

Things to Be Careful About

• Use the correct units: volume must be in dm³ for the concentration formula.
• The empirical formula is the simplest ratio — check for a common factor before finalising.
• A molar mass of 88 g mol⁻¹ corresponds to $\text{C}_4\text{H}_8\text{O}_2$, but its empirical formula is $\text{C}_2\text{H}_4\text{O}$.

Background Concept

The shape of a covalent molecule is determined by VSEPR theory: electron pairs (bonding and lone pairs) around the central atom repel and take up positions that minimise repulsion. A molecule is planar if all of its atoms lie in the same plane. Linear molecules (two electron domains) and trigonal planar molecules (three electron domains, no lone pairs) are planar. Bent triatomic molecules such as water are also planar because any three points define a plane. Pyramidal molecules such as ammonia and phosphine are not planar because the lone pair forces the three bonded atoms out of one plane.

Hybridisation gives the same result: $\text{sp}$ centres are linear, $\text{sp}^2$ centres are trigonal planar, and $\text{sp}^3$ centres are tetrahedral or pyramidal. In ethene, each carbon is $\text{sp}^2$, so the whole molecule is planar.

Understanding the Question

The question asks which set contains only molecules whose atoms are all in one plane. It is not enough for the central atom and its immediate neighbours to be planar; every atom in the molecule must be coplanar. This is why a molecule such as propene, which contains a tetrahedral methyl group, is not acceptable even though its double-bonded carbons are planar.

Approach

For each molecule, determine the shape around every central atom using VSEPR or hybridisation. Mark each molecule as planar or non-planar. Then inspect the four options and choose the one in which every molecule is planar.

Step-by-Step Reasoning

• $\text{AlCl}_3$: aluminium has three bonding pairs and no lone pairs, so it is trigonal planar. All four atoms lie in one plane.
• $\text{BF}_3$: boron also has three bonding pairs and no lone pairs, so it is trigonal planar.
• $\text{PH}_3$: phosphorus has three bonding pairs and one lone pair, so the shape is trigonal pyramidal. The hydrogen atoms are not all in one plane, so option A fails.
• $\text{CO}_2$: carbon has two double bonds and no lone pairs, so the molecule is linear. A linear molecule is planar.
• $\text{NH}_3$: nitrogen has three bonding pairs and one lone pair, so ammonia is trigonal pyramidal, not planar. Option B fails.
• $\text{C}_2\text{H}_4$: each carbon has three sigma bonds and no lone pairs, so each carbon is $\text{sp}^2$ and trigonal planar. The whole ethene molecule is planar.
• $\text{C}_3\text{H}_6$: the molecular formula can represent propene or cyclopropane. In propene, the $\text{CH}_3$ group is tetrahedral, so not all atoms are coplanar. In cyclopropane, the ring is planar but the hydrogen atoms lie above and below the ring, so again not all atoms are coplanar. Option C fails.
• $\text{H}_2\text{O}$: oxygen has two bonding pairs and two lone pairs, so the molecule is bent. With only three atoms, all three atoms necessarily lie in one plane.

Thus only option D contains molecules that are all planar.

Key Takeaways

• VSEPR predicts shape from electron-pair repulsion; lone pairs matter.
• Linear, trigonal planar, and bent triatomic molecules are planar; pyramidal and tetrahedral molecules are not.
• A molecule with more than three atoms is planar only if every atom lies in the same plane, so tetrahedral substituents disqualify it.

Common Mistakes

• Thinking that a bent molecule such as water is not planar. A bent triatomic molecule is planar because any three atoms define a plane.
• Thinking that linear molecules are not planar. Linear is a special case of planar.
• Forgetting the lone pair in $\text{NH}_3$ and $\text{PH}_3$, which makes them pyramidal.
• Assuming $\text{C}_3\text{H}_6$ is planar because the formula can be written with a double bond. Propene contains a tetrahedral $\text{CH}_3$ group, and cyclopropane has hydrogens out of the ring plane.

Things to Be Careful About

• Check every atom, not just the central atom, when deciding whether a molecule is planar.
• Remember that lone pairs occupy space and affect the shape.
• For triatomic molecules, planarity is automatic; the shape is either linear or bent.
• $\text{AlCl}_3$ is planar as a simple molecule; do not confuse it with its dimer $\text{Al}_2\text{Cl}_6$, in which aluminium is tetrahedral.

Background Concept

Hybridisation is the mixing of atomic orbitals to form new hybrid orbitals suitable for the geometry of the molecule. The number of electron domains (regions of electron density, i.e., bonding pairs plus lone pairs) around a central atom determines its hybridisation:

• 4 electron domains $\rightarrow$ $\text{sp}^3$ (tetrahedral, ~109.5°)
• 3 electron domains $\rightarrow$ $\text{sp}^2$ (trigonal planar, ~120°)
• 2 electron domains $\rightarrow$ $\text{sp}$ (linear, 180°)

A critical rule to remember is that a single, double, or triple bond each counts as exactly ONE electron domain when determining hybridisation and molecular geometry.

Understanding the Question

The question asks to identify the types of hybridisation present among the three carbon atoms in propyne ($\text{CH}_3\text{C} \equiv \text{CH}$), based on its displayed structure. The options list combinations of $\text{sp}$, $\text{sp}^2$, and $\text{sp}^3$.

Approach

Determine the number of electron domains around each carbon atom individually. Then match the number of domains to the corresponding hybridisation type. Do not average or assume all carbons in a molecule share the same hybridisation.

Step-by-Step Reasoning

1. Identify the structure of propyne: $\text{CH}_3-\text{C} \equiv \text{C}-\text{H}$. There are three carbon atoms to analyse.
2. Carbon 1 (terminal alkyne carbon, bonded to H): Forms one single bond to H and one triple bond to the adjacent C. Total electron domains = 2. Hybridisation = $\text{sp}$.
3. Carbon 2 (middle alkyne carbon): Forms one triple bond to C1 and one single bond to C3. Total electron domains = 2. Hybridisation = $\text{sp}$.
4. Carbon 3 (methyl carbon, $\text{CH}_3$): Forms three single bonds to H atoms and one single bond to C2. Total electron domains = 4. Hybridisation = $\text{sp}^3$.
5. The hybridisations present in the molecule are $\text{sp}$ (for the two alkyne carbons) and $\text{sp}^3$ (for the methyl carbon). $\text{sp}^2$ is not present anywhere in the molecule.
6. This matches option B.

Key Takeaways

• Count electron domains (not total bonds) to determine hybridisation. A triple bond is 1 domain, not 3.
• Alkynes contain $\text{sp}$ hybridised carbons, alkenes contain $\text{sp}^2$, and alkanes contain $\text{sp}^3$.
• A single molecule can contain carbon atoms with multiple different hybridisations; analyse each atom individually.

Common Mistakes

• Counting the total number of bonds instead of electron domains (e.g., thinking a triple bond counts as 3 domains, which incorrectly leads to $\text{sp}^3$ for the alkyne carbons).
• Assuming all carbon atoms in a molecule must have the same hybridisation.
• Confusing the hybridisation of an alkyne carbon ($\text{sp}$) with an alkene carbon ($\text{sp}^2$).

Things to Be Careful About

• Remember that a multiple bond (double or triple) counts as only ONE electron domain for the purpose of determining hybridisation and molecular geometry.
• Ensure you are counting domains around each carbon atom individually, not the molecule as a whole.

Background Concept

The boiling point of a substance is the temperature at which its vapour pressure equals atmospheric pressure — practically, the temperature at which the liquid turns to gas. To boil, molecules must escape the liquid phase, which means the intermolecular forces (attractions between molecules) holding them together must be overcome. The stronger these forces, the more energy (higher temperature) is needed, so the higher the boiling point.

There are several types of intermolecular force, in increasing strength:

1. London (dispersion / instantaneous dipole–induced dipole) forces — present in ALL molecules, and the only type in non-polar molecules.
2. Permanent dipole–dipole forces — in polar molecules.
3. Hydrogen bonding — the strongest, a special case of dipole–dipole interaction.

Hydrogen bonding requires a hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F) that also has a lone pair. The H becomes strongly $\delta+$ and is attracted to the lone pair on the N/O/F of a neighbouring molecule. For hydrogen bonding to occur, the H must be bonded to N, O, or F — S, Cl, etc. are not electronegative enough.

Understanding the Question

This is a "which statement explains" multiple-choice question. The fact given is that $\text{H}_2\text{O}$ boils at a higher temperature than $\text{H}_2\text{S}$. We must select the statement that correctly explains this difference. The trap is that several options mention properties of the covalent bonds (bond energy, bond length), which are intramolecular — they do not influence boiling point. The correct reasoning must be about intermolecular forces.

Approach

Step 1: Recall what determines boiling point — the strength of intermolecular forces.
Step 2: Identify the intermolecular forces present in each molecule.
Step 3: Compare their strengths.
Step 4: Eliminate options that refer to intramolecular properties or state the wrong relationship.

Step-by-Step Reasoning

• Boiling requires overcoming intermolecular forces. So the correct explanation must involve intermolecular forces, and the molecule with the higher boiling point must have the stronger intermolecular forces.
• Option A: "The bond energy of O–H is greater than the bond energy of S–H." Bond energy is the energy needed to break a covalent bond WITHIN a molecule — an intramolecular property. Boiling does not break covalent bonds; it separates whole molecules. Even if true, it is irrelevant to the boiling point. Wrong.
• Option B: "The intermolecular forces in $\text{H}_2\text{O}$ are weaker than the intermolecular forces in $\text{H}_2\text{S}$." If this were true, $\text{H}_2\text{O}$ would boil LOWER than $\text{H}_2\text{S}$, which contradicts the given fact. Wrong.
• Option C: "The S–H bond in $\text{H}_2\text{S}$ is longer than the O–H bond in $\text{H}_2\text{O}$." Bond length is again an intramolecular property, irrelevant to boiling point. Wrong.
• Option D: "There is significant intermolecular hydrogen bonding in $\text{H}_2\text{O}$ but not in $\text{H}_2\text{S}$." Correct. In $\text{H}_2\text{O}$, H is bonded to O (electronegativity $\approx 3.5$), so the O–H bond is very polar; the $\delta+$ hydrogen of one molecule is strongly attracted to a lone pair on the oxygen of a neighbouring molecule — hydrogen bonding. In $\text{H}_2\text{S}$, sulfur has electronegativity $\approx 2.5$, similar to hydrogen ($\approx 2.2$), so the S–H bond is barely polar and sulfur does not form hydrogen bonds; $\text{H}_2\text{S}$ has only weak London (van der Waals) forces. Hydrogen bonds are much stronger than London forces, so much more energy is needed to boil water — hence its higher boiling point.

Key Takeaways

• Boiling point (and melting point) is governed by intermolecular forces, never by intramolecular bond strength or bond length.
• Hydrogen bonding requires H covalently bonded to N, O, or F (highly electronegative atoms that also carry lone pairs).
• $\text{H}_2\text{S}$ does not hydrogen bond because S is not electronegative enough.
• Stronger intermolecular forces → higher boiling point.

Common Mistakes

• Choosing A or C: confusing intramolecular bond energy/length with intermolecular forces. Breaking a covalent bond is not what happens when a liquid boils.
• Thinking $\text{H}_2\text{S}$ exhibits hydrogen bonding: S is not electronegative enough; only N, O, and F form hydrogen bonds.
• Choosing B: it states the wrong direction — weaker forces would mean a lower boiling point, contradicting the question.

Things to Be Careful About

• The distinction between "intramolecular" (within a molecule) and "intermolecular" (between molecules) is the whole point of this question.
• Hydrogen bonding is an intermolecular force, not a covalent bond — it is much weaker than a covalent bond but much stronger than ordinary dipole–dipole or London forces.
• Electronegativity values: O $\approx 3.5$, S $\approx 2.5$, H $\approx 2.2$ — the O–H bond is far more polar than the S–H bond, which is why only water can hydrogen bond.

Background Concept

Silicon dioxide, $\text{SiO}_2$, is a giant covalent (macromolecular) substance. Each silicon atom forms four single covalent bonds to oxygen atoms in a tetrahedral arrangement, and each oxygen atom bridges two silicon atoms, so the whole solid is one huge three-dimensional network of atoms held together by strong covalent bonds. This network structure is what determines its physical properties: a very high melting point, great hardness, no electrical conductivity, and insolubility in water. It is completely different from a simple molecular compound, even though its empirical formula looks simple.

Understanding the Question

This one-mark multiple-choice question asks you to select the row that correctly describes two properties of silicon dioxide: its electrical conductivity in the liquid state and its solubility in water. It is testing whether you can predict physical properties directly from the type of structure and bonding, rather than memorising isolated facts. The four rows give every combination of conductor/non-conductor and soluble/insoluble, so you need to be certain about both properties to lock in the correct answer.

Approach

The key is to first recognise the structure of $\text{SiO}_2$: a giant covalent (macromolecular) lattice. Then apply two structure-property principles:

1. Electrical conductivity requires mobile charge carriers — free (delocalised) electrons or free-moving ions. A giant covalent network has neither, so it cannot conduct.
2. Solubility in water requires water molecules to break apart the structure. Water can overcome weak intermolecular forces, but it cannot break strong covalent bonds throughout a giant lattice, so the substance is insoluble.

Applying both principles at once eliminates three rows and leaves only the correct one.

Step-by-Step Reasoning

Property 1 — electrical conductivity in the liquid state.

In a giant covalent lattice every valence electron of silicon is localised in a covalent bond (silicon has four valence electrons, all shared with four oxygens). There are therefore no delocalised electrons available to carry charge. In the liquid state, even though the substance has melted, the liquid is still made of covalently bonded network fragments — there are no mobile ions either, because the substance is not ionic. With neither free electrons nor mobile ions, liquid $\text{SiO}_2$ is a non-conductor. This immediately rules out rows C and D, which say conductor.

Property 2 — solubility in water.

Water dissolves substances by surrounding individual molecules or ions and pulling them apart, which works when the forces between particles are weak (intermolecular forces) or when hydration energy can compensate for the energy needed to separate ions. In $\text{SiO}_2$, the particles are held together by strong covalent bonds extending throughout the entire lattice. There are no discrete molecules for water to surround, and water cannot break the Si–O covalent bonds. Hence $\text{SiO}_2$ is insoluble in water, ruling out row B as well.

Only row A remains: non-conductor and insoluble, which is correct.

Key Takeaways

• $\text{SiO}_2$ (quartz/sand) is a giant covalent (macromolecular) lattice, not a simple molecular solid.
• Giant covalent substances are hard, have very high melting points, do not conduct electricity in any state (no mobile electrons or ions), and are insoluble in water.
• Conductivity always requires mobile charge carriers; solubility requires that water can break apart the structure. Checking both criteria in turn is a reliable way to answer this type of question.

Common Mistakes

• Thinking $\text{SiO}_2$ conducts like graphite. Graphite conducts because it has delocalised electrons between its layers. $\text{SiO}_2$ has no such electrons — every electron is locked in a covalent bond — so it is a non-conductor.
• Treating $\text{SiO}_2$ as a simple molecular substance. The formula looks like a molecule, but the structure is an infinite network. This leads students to guess "soluble" or to expect a low melting point.
• Assuming that because it is solid at room temperature it won't conduct, but that molten ionic behaviour applies. Molten ionic compounds conduct due to mobile ions; molten $\text{SiO}_2$ still has no ions, so it does not conduct either.

Things to Be Careful About

• The question says "in the liquid state" — the reasoning must hold even though $\text{SiO}_2$ only melts at about 1700°C. The absence of mobile charge carriers is the point, not the temperature.
• "Insoluble" refers to water at normal conditions. $\text{SiO}_2$ does react with hydrofluoric acid, but that is a chemical reaction, not dissolution, and is outside the scope of this question.
• State clearly why each property holds (no free electrons, no mobile ions; water cannot break strong covalent bonds) — in an explanation question these reasons, not just the facts, are what earn marks.

Background Concept

Standard enthalpy change of reaction, $\Delta H^\ominus$, can be found from standard enthalpies of formation using Hess's law. Hess's law states that the enthalpy change for a reaction is independent of the route taken, because enthalpy is a state function. For any reaction:

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

Each term must be multiplied by the stoichiometric coefficient of that substance in the balanced equation. The standard enthalpy of formation of an element in its standard state is zero by definition.

Understanding the Question

The question gives three standard enthalpies of formation: $\text{P}_4\text{O}_{10}(\text{s})$, $\text{H}_2\text{O}(\text{l})$ and $\text{H}_3\text{PO}_4(\text{s})$. It asks for the standard enthalpy change of the reaction:

$$\text{P}_4\text{O}_{10}(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{H}_3\text{PO}_4(\text{s})$$

The command word is implicit: calculate the enthalpy change from formation data. This is a direct application of Hess's law.

Approach

1. Identify the products and reactants in the balanced equation.
2. Multiply each enthalpy of formation by its stoichiometric coefficient.
3. Apply the formula: products minus reactants.
4. Pay attention to signs: subtracting a negative number is equivalent to adding its magnitude.

Step-by-Step Reasoning

For the products, there is only one product, $\text{H}_3\text{PO}_4(\text{s})$, with coefficient 4:

$$4 \times (-1279) = -5116 \text{ kJ mol}^{-1}$$

For the reactants:

$$(-3012) + 6(-286) = -3012 - 1716 = -4728 \text{ kJ mol}^{-1}$$

Therefore:

$$\Delta H^\ominus = (-5116) - (-4728) = -388 \text{ kJ mol}^{-1}$$

The negative sign shows the reaction is exothermic. This matches option B.

The other options are traps: A comes from adding all the formation enthalpies without the products-minus-reactants sign convention; C comes from a mis-scaled calculation; D has the wrong sign and arithmetic.

Key Takeaways

• Always use $\Delta H^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants})$.
• Multiply each formation enthalpy by its stoichiometric coefficient.
• Watch negative signs carefully; subtracting a negative is adding.
• A negative $\Delta H^\ominus$ indicates an exothermic reaction.

Common Mistakes

• Adding the formation enthalpies of reactants and products together instead of subtracting reactants from products.
• Forgetting to multiply by the stoichiometric coefficients, especially the 4 for $\text{H}_3\text{PO}_4$ and the 6 for $\text{H}_2\text{O}$.
• Sign errors when subtracting the reactant sum: $(-5116) - (-4728)$ must become $-5116 + 4728$.
• Confusing $\Delta H_f^\ominus$ of a compound with $\Delta H^\ominus$ of the reaction.

Things to Be Careful About

• Standard enthalpy of formation data are usually given in $\text{kJ mol}^{-1}$; keep the unit in the final answer.
• The formula uses the balanced equation coefficients exactly as written.
• If the question gives $\Delta H_c^\ominus$ instead, the sign convention is reversed: reactants minus products.
• In an MCQ, once you obtain $-388 \text{ kJ mol}^{-1}$, check that the sign and magnitude match one of the options before selecting the letter.

Background Concept

The enthalpy change of neutralisation is the heat energy released when one mole of water is formed from the reaction of an acid with a base. For a strong acid and a strong base it is approximately constant (about $-57\text{ kJ mol}^{-1}$ for $\text{HCl}$ and $\text{NaOH}$).

Calorimetry is the measurement of heat changes. When a reaction occurs in a solution, the heat released (or absorbed) changes the temperature of the solution. The heat change is given by:

$$q = mc\Delta T$$

where $m$ is the mass of the solution, $c$ is its specific heat capacity, and $\Delta T$ is the temperature rise. In this question the data are given as a heat capacity per unit volume ($4.2\text{ J K}^{-1}\text{ cm}^{-3}$), so the formula becomes:

$$q = V \times C_V \times \Delta T$$

where $V$ is the total volume of the solution and $C_V$ is the heat capacity per unit volume. This is just a rearrangement of the same idea — the heat capacity of the whole solution is the volume times the heat capacity per unit volume.

The enthalpy change per mole is then:

$$\Delta H = \frac{q}{n}$$

where $n$ is the number of moles of the limiting reagent (the reactant that is used up completely).

Understanding the Question

We have $20\text{ cm}^3$ of $\text{HCl}$ solution containing $0.04\text{ mol}$ of $\text{HCl}$, and $20\text{ cm}^3$ of $\text{NaOH}$ solution containing $0.04\text{ mol}$ of $\text{NaOH}$. The two solutions are at the same initial temperature, and when they are mixed the temperature rises by $15\text{ K}$. The heat capacity per unit volume of the final solution is $4.2\text{ J K}^{-1}\text{ cm}^{-3}$.

The question asks for the enthalpy change of neutralisation of hydrochloric acid, i.e. the heat released per mole of $\text{HCl}$ neutralised. The four options are different algebraic expressions; we need to identify which one correctly combines the total volume, the heat capacity per unit volume, the temperature rise, and the number of moles.

The key points to notice:

• The total volume of the final solution is $40\text{ cm}^3$, not $20\text{ cm}^3$.
• The number of moles of $\text{HCl}$ is $0.04\text{ mol}$, and the reaction is equimolar, so exactly $0.04\text{ mol}$ of $\text{HCl}$ is neutralised.

Approach

1. Determine the total volume of the final solution: add the two volumes.
2. Calculate the total heat released using the heat capacity per unit volume and the temperature rise.
3. Divide the heat by the number of moles of $\text{HCl}$ to get the enthalpy change per mole.
4. Compare the resulting expression with the options.

Step-by-Step Reasoning

Step 1: Total volume

The two solutions are mixed, so the final volume is:

$$V = 20 + 20 = 40\text{ cm}^3$$

Step 2: Heat released

The heat released is the heat capacity per unit volume times the total volume times the temperature rise:

$$q = V \times C_V \times \Delta T = 40 \times 4.2 \times 15 = 2520\text{ J}$$

Step 3: Moles of HCl neutralised

The reaction is:

$$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$

There is $0.04\text{ mol}$ of $\text{HCl}$ and $0.04\text{ mol}$ of $\text{NaOH}$. Since the stoichiometry is 1:1, both are used up completely — neither is in excess. So $0.04\text{ mol}$ of $\text{HCl}$ is neutralised.

Step 4: Enthalpy change per mole

$$\Delta H = \frac{q}{n} = \frac{2520}{0.04} = 63000\text{ J mol}^{-1} = 63\text{ kJ mol}^{-1}$$

The reaction is exothermic, so the enthalpy change is negative:

$$\Delta H = -63\text{ kJ mol}^{-1}$$

The options give the magnitude (positive), and the expression that matches is:

$$\frac{40 \times 4.2 \times 15}{0.04}\text{ J mol}^{-1}$$

which is option C.

Why the other options are wrong:

• Option A uses $20\text{ cm}^3$ as the volume — this is only the volume of the acid, not the total volume of the solution. The heat is absorbed by the whole $40\text{ cm}^3$ of solution.
• Option B multiplies by $0.08$ instead of dividing by $0.04$. This is dimensionally wrong: multiplying by moles gives $\text{J mol}$, not $\text{J mol}^{-1}$, and $0.08$ is not the number of moles of $\text{HCl}$ (it would be the total moles of acid plus base, which is not what we divide by).
• Option D uses $20\text{ cm}^3$ for the volume and divides by $0.08$ mol — both errors combined.

Key Takeaways

• When mixing two solutions, the total volume is the sum of the individual volumes, and this total volume is what absorbs or releases the heat.
• Heat capacity per unit volume ($\text{J K}^{-1}\text{ cm}^{-3}$) is used in the same way as specific heat capacity: multiply by volume and temperature change.
• The enthalpy change per mole is the heat divided by the number of moles of the limiting reagent.
• Neutralisation is exothermic, so the enthalpy change is negative even though the options present positive magnitudes.

Common Mistakes

• Using the volume of only one solution (20 cm³) instead of the total (40 cm³). The heat is absorbed by the entire mixture, so the total volume must be used.
• Dividing by the wrong number of moles. The correct denominator is $0.04\text{ mol}$ of $\text{HCl}$, not $0.08\text{ mol}$ (which would be the total moles of acid and base combined).
• Forgetting the sign. The enthalpy change of neutralisation is negative because heat is released. The options give positive magnitudes, but a full answer should include the negative sign.
• Confusing heat capacity per unit volume with specific heat capacity. Here the units are $\text{J K}^{-1}\text{ cm}^{-3}$, so we multiply by volume, not mass.

Things to Be Careful About

• Always check the units of the heat capacity given. If it is per unit volume, use volume; if it is per unit mass, use mass.
• Ensure the temperature change is in kelvin (here it is already given as $15\text{ K}$, so no conversion is needed).
• The final enthalpy change should be expressed in $\text{J mol}^{-1}$ or $\text{kJ mol}^{-1}$ with the correct sign for an exothermic process.

Background Concept

Redox reactions involve the transfer of electrons. In an oxidation half-equation, electrons appear as products; in a reduction half-equation, electrons appear as reactants. The total number of electrons lost by the reducing agent(s) must equal the total number of electrons gained by the oxidising agent.

Iron(II) ethanedioate, $\text{FeC}_2\text{O}_4$, is an ionic compound containing both $\text{Fe}^{2+}$ and $\text{C}_2\text{O}_4^{2-}$ ions. In acidified potassium manganate(VII), the $\text{MnO}_4^-$ ion is a strong oxidising agent and can oxidise both of these ions. The stoichiometric ratio is therefore found by balancing the electrons released by both ions against the electrons accepted by one $\text{MnO}_4^-$ ion.

Understanding the Question

The question supplies three half-equations:

$$\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$ $$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$$ $$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-$$

We are asked how many moles of $\text{FeC}_2\text{O}_4$ react with one mole of $\text{MnO}_4^-$. Since one formula unit of $\text{FeC}_2\text{O}_4$ contains exactly one $\text{Fe}^{2+}$ ion and one $\text{C}_2\text{O}_4^{2-}$ ion, both oxidation half-equations contribute to the total electron release.

Approach

1. Determine how many electrons one mole of $\text{MnO}_4^-$ accepts from the reduction half-equation.
2. Determine how many electrons one mole of $\text{FeC}_2\text{O}_4$ releases by adding the two oxidation half-equations.
3. Set the total electrons released equal to the total electrons accepted and solve for the mole ratio.

Step-by-Step Reasoning

The reduction half-equation shows:

$$\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$

So one mole of $\text{MnO}_4^-$ accepts 5 moles of electrons.

The oxidation half-equation for iron(II):

$$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$$

So one mole of $\text{Fe}^{2+}$ releases 1 mole of electrons.

The oxidation half-equation for ethanedioate:

$$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-$$

So one mole of $\text{C}_2\text{O}_4^{2-}$ releases 2 moles of electrons.

Because one mole of $\text{FeC}_2\text{O}_4$ contains one mole of $\text{Fe}^{2+}$ and one mole of $\text{C}_2\text{O}_4^{2-}$, it releases:

$$1 + 2 = 3 \text{ mol e}^-$$

Let $n$ be the number of moles of $\text{FeC}_2\text{O}_4$ that react with one mole of $\text{MnO}_4^-$. The electron balance is:

$$3n = 5$$ $$n = \frac{5}{3} = 1.67$$

Therefore the correct option is B.

The other options arise from common mistakes:

• A, 0.60, is the inverse ratio $3/5$.
• C, 2.50, is $5/2$, which would be correct if only the ethanedioate ion were oxidised and the iron(II) were ignored.
• D, 5.00, is $5/1$, which would be correct if only the iron(II) ion were oxidised and the ethanedioate were ignored.

Key Takeaways

• Redox stoichiometry is governed by electron conservation.
• When a compound contains more than one reducing ion, add the electrons released by each ion per formula unit.
• The mole ratio is obtained by dividing the electrons accepted by the oxidising agent by the total electrons released per formula unit of the reducing compound.

Common Mistakes

• Using only the $\text{Fe}^{2+}$ oxidation and ignoring the $\text{C}_2\text{O}_4^{2-}$ oxidation, giving 5.00.
• Using only the $\text{C}_2\text{O}_4^{2-}$ oxidation and ignoring the $\text{Fe}^{2+}$, giving 2.50.
• Inverting the ratio and giving 0.60 instead of 1.67.
• Forgetting that $\text{FeC}_2\text{O}_4$ contains one $\text{Fe}^{2+}$ and one $\text{C}_2\text{O}_4^{2-}$, not two reducing ions of the same kind.

Things to Be Careful About

• Count electrons carefully: $\text{MnO}_4^-$ gains 5 electrons, not 1 or 2.
• The formula unit $\text{FeC}_2\text{O}_4$ provides 3 electrons in total, so the ratio is $5/3$.
• The answer is a mole ratio, so no units are needed, but the numerical value must be 1.67, not 1.7 unless the mark scheme allows rounding.

Background Concept

For a homogeneous equilibrium such as

$$2\text{NO}_2(\text{g}) \rightleftharpoons 2\text{NO}(\text{g}) + \text{O}_2(\text{g}),$$

the equilibrium constant $K_c$ is defined as

$$K_c = \frac{[\text{NO}]^2[\text{O}_2]}{[\text{NO}_2]^2}.$$

Here square brackets mean concentration in $\text{mol dm}^{-3}$. The coefficients in the balanced equation become powers in the $K_c$ expression: $\text{NO}$ is squared because its coefficient is 2, and $\text{NO}_2$ is squared for the same reason.

Because all species are gases in the same container, each concentration is simply:

$$[\text{X}] = \frac{\text{amount of X in mol}}{\text{volume in dm}^3}.$$

When the volume is exactly $1\text{ dm}^3$, the numerical value of the concentration is the same as the number of moles.

Equilibrium problems are often solved with an initial-change-equilibrium (ICE) table. The balanced equation fixes the ratios in which amounts change: if $x$ mol of $\text{O}_2$ is formed, then $2x$ mol of $\text{NO}$ is formed and $2x$ mol of $\text{NO}_2$ is consumed.

Understanding the Question

The question gives the initial amount of $\text{NO}_2$ ($4\text{ mol}$ in $1\text{ dm}^3$) and the equilibrium amount of $\text{O}_2$ ($0.8\text{ mol}$). It asks for the value of $K_c$ at that temperature. The four options are algebraic expressions, so the task is to identify the correct equilibrium concentrations and the correct $K_c$ expression.

The key is that $K_c$ uses equilibrium concentrations only, not initial amounts, and the stoichiometric coefficients must be respected.

Approach

1. Use the equilibrium amount of $\text{O}_2$ as the anchor.
2. Apply the stoichiometric ratios from the equation to find the amounts of $\text{NO}$ formed and $\text{NO}_2$ consumed.
3. Subtract the consumed $\text{NO}_2$ from the initial $4\text{ mol}$ to get the equilibrium amount of $\text{NO}_2$.
4. Since the volume is $1\text{ dm}^3$, use these amounts directly as concentrations.
5. Write the $K_c$ expression and substitute.

Step-by-Step Reasoning

Start with an ICE table in moles:

The equilibrium amount of $\text{O}_2$ is given as $0.8\text{ mol}$, so $x = 0.8$.

• $\text{NO}$ formed: $2x = 2 \times 0.8 = 1.6\text{ mol}$.
• $\text{NO}_2$ consumed: $2x = 1.6\text{ mol}$.
• $\text{NO}_2$ remaining: $4 - 1.6 = 2.4\text{ mol}$.

Because the container is $1\text{ dm}^3$, the equilibrium concentrations are:

$$[\text{NO}_2] = 2.4\text{ mol dm}^{-3}, \quad [\text{NO}] = 1.6\text{ mol dm}^{-3}, \quad [\text{O}_2] = 0.8\text{ mol dm}^{-3}.$$

Substitute into the $K_c$ expression:

$$K_c = \frac{(1.6)^2(0.8)}{(2.4)^2}.$$

This matches option D. Numerically, $K_c = \frac{2.56 \times 0.8}{5.76} \approx 0.356\text{ mol dm}^{-3}$.

Why the other options are wrong:

• Option A uses $0.8^2$ for $[\text{NO}]$ and $4^2$ for $[\text{NO}_2]^2$; it treats the amount of $\text{NO}$ as equal to the amount of $\text{O}_2$ and uses the initial $\text{NO}_2$ amount.
• Option B has $1.6 \times 0.8$ in the numerator, missing the square on $[\text{NO}]$.
• Option C squares $1.6$ correctly but uses the initial $4\text{ mol}$ of $\text{NO}_2$ instead of the equilibrium $2.4\text{ mol}$.

Key Takeaways

• $K_c$ is always written as products over reactants, with each concentration raised to its stoichiometric coefficient.
• Only equilibrium amounts (or concentrations) may be substituted into $K_c$; initial amounts are not used directly.
• The balanced equation gives the exact ratios in which amounts change, so knowing one equilibrium amount lets you find all the others.
• In a $1\text{ dm}^3$ container, amount in mol and concentration in $\text{mol dm}^{-3}$ are numerically equal, which simplifies the calculation.

Common Mistakes

• Using the initial amount of $\text{NO}_2$ ($4\text{ mol}$) in the denominator instead of the equilibrium amount ($2.4\text{ mol}$). This is the error in options A and C.
• Forgetting the coefficient 2 on $\text{NO}$ and writing $[\text{NO}]$ instead of $[\text{NO}]^2$, or using $0.8$ instead of $1.6$ for the equilibrium amount of $\text{NO}$.
• Confusing the change in $\text{NO}_2$ with the equilibrium amount: $2x$ mol is consumed, so the equilibrium amount is $4 - 2x$, not $4 - x$.
• Assuming concentrations equal moles when the volume is not $1\text{ dm}^3$. Here it is, but in general each amount must be divided by the volume.

Things to Be Careful About

• The coefficients in the balanced equation become powers in the $K_c$ expression. Missing a square changes the answer completely.
• All species are gases, so all are included in $K_c$. If a solid or pure liquid were present, it would be omitted.
• $K_c$ is constant only at a fixed temperature; the question specifies a constant temperature for this reason.
• If units were required, the expression gives $(\text{mol dm}^{-3})^3/(\text{mol dm}^{-3})^2 = \text{mol dm}^{-3}$, but the MCQ only asks for the algebraic form.

Background Concept

This question combines two ideas: the energy profile of an exothermic reaction, and what exactly becomes equal at equilibrium.

Activation energy and enthalpy change. For any reaction, the activation energy ($E_a$) is the minimum energy that colliding particles must possess for a reaction to occur. For an exothermic reaction, the products sit at a lower energy than the reactants, so the energy "hill" from reactants to the transition state is smaller than the hill from products back to the transition state. In other words, for an exothermic reaction $E_a(\text{forward}) < E_a(\text{reverse})$; the two are equal only when $\Delta H = 0$.

Dynamic equilibrium. At equilibrium, the forward and reverse reactions are still occurring, but at the same rate, so concentrations no longer change. The rate of a reaction is proportional to the frequency of effective (successful) collisions — collisions that have both sufficient energy and the correct orientation. The total frequency of collisions is much larger than the frequency of effective collisions, and only the effective ones contribute to the rate.

Understanding the Question

The question gives a single-step, exothermic, reversible reaction $\text{X} + \text{Y} \rightleftharpoons 2\text{Z}$ and asks which of three statements about the forward and reverse reactions are correct. Statement 1 compares activation energies; statements 2 and 3 compare collision frequencies at equilibrium. The key distinction between statements 2 and 3 is the word "effective" — this is the trap the question is built around.

Approach

Evaluate each statement on its own merits:

1. Use the exothermic nature of the reaction to compare the forward and reverse activation energies.
2. Recall that at equilibrium, forward rate = reverse rate, and that rate is governed by effective collisions, not total collisions.
3. Apply the same reasoning to statement 3, which correctly refers to effective collisions.

Step-by-Step Reasoning

Statement 1 — incorrect. The reaction is exothermic, so $\Delta H < 0$. On an energy profile, the products ($2\text{Z}$) are at a lower energy than the reactants ($\text{X} + \text{Y}$). The forward activation energy is the energy rise from reactants to the transition state; the reverse activation energy is the energy rise from products to the transition state. Since the products are lower, the reverse "hill" is taller: $E_a(\text{forward}) < E_a(\text{reverse})$. They are not equal. Statement 1 is false.

Statement 2 — incorrect. At equilibrium, the forward and reverse rates are equal. But the rate is proportional to the frequency of effective collisions, not to the total frequency of collisions. The total collision frequency between X and Y particles is generally not equal to the total collision frequency between Z particles (the concentrations, and hence collision frequencies, of the species differ). Even if they were, total collision frequency is not what determines rate. So statement 2 is false.

Statement 3 — correct. At equilibrium, forward rate = reverse rate. Since rate is proportional to the frequency of effective collisions, the frequency of effective collisions between X and Y must equal the frequency of effective collisions between Z particles. This is exactly the condition that maintains dynamic equilibrium. Statement 3 is true.

Only statement 3 is correct, so the answer is D.

Key Takeaways

• For an exothermic reaction, $E_a(\text{forward}) < E_a(\text{reverse})$; equality of activation energies only holds for $\Delta H = 0$.
• At equilibrium, it is the rates that are equal, and rate depends on the frequency of effective collisions, not total collisions.
• The word "effective" is chemically significant: it means collisions with sufficient energy and correct orientation.

Common Mistakes

• Choosing statement 1 as correct: forgetting that exothermic means products are lower in energy, so the reverse activation energy is larger.
• Choosing statement 2 as correct: confusing total collision frequency with effective collision frequency, or thinking equilibrium means equal collision frequencies rather than equal rates.
• Misreading statement 2 and 3 as identical: the presence of "effective" in statement 3 is the entire point of the question.

Things to Be Careful About

• Note that the reaction is described as "single-step"; this ensures there is a single transition state and the simple activation-energy comparison applies, but it does not change the equilibrium reasoning.
• "Effective" collisions are those with sufficient energy and correct orientation — both conditions matter for a successful reaction.
• At equilibrium, concentrations are constant but not necessarily equal; likewise collision frequencies are not necessarily equal — only the rates (and hence effective collision frequencies) are equal.