Chemistry 9701/13 — October/November 2024

Background Concept

This question tests two separate periodic trends:

1. First ionisation energy down Group 2: As you go down Group 2 (Be → Mg → Ca → Sr → Ba), the first ionisation energy decreases. The valence electrons are in higher principal shells, further from the nucleus, with more shielding from inner electrons, so they are more easily removed.

2. Halogen–halogen (Q–Q) bond energy: The bond dissociation energy of the diatomic halogens X–X follows the order Cl–Cl > Br–Br > I–I, with F–F anomalously low. Fluorine's small size causes lone-pair–lone-pair repulsion between the two atoms, weakening the bond. Chlorine, being larger, avoids this repulsion and has the strongest halogen–halogen bond.

Understanding the Question

The question gives two independent criteria and asks for the compound that satisfies both:

• The Group 2 metal must be the one with the higher first ionisation energy.
• The halogen must be the one with the higher Q–Q bond energy.

The four options are the four possible combinations of Mg/Ca with Cl/Br.

Approach

1. Identify which of Mg and Ca has the higher first ionisation energy.
2. Identify which of Cl and Br has the higher halogen–halogen bond energy.
3. Combine the two winners to write the formula.

Step-by-Step Reasoning

Step 1 — Group 2 element
First ionisation energy decreases down Group 2. Mg is above Ca, so Mg has the higher first ionisation energy. This rules out options C and D (which contain Ca).

Step 2 — Halogen element
The halogen–halogen bond energy is highest for chlorine. Cl–Cl ≈ 242 kJ mol⁻¹, while Br–Br ≈ 193 kJ mol⁻¹. This rules out options B and D (which contain Br).

Step 3 — Combine
The element that satisfies both is Mg with Cl, giving the formula MgCl₂. This is option A.

Key Takeaways

• First ionisation energy decreases down a group.
• The halogen–halogen bond energy is highest for Cl₂; F₂ is anomalously weak due to lone-pair repulsion.
• When a question gives two independent criteria, evaluate each separately and then combine the results.

Common Mistakes

• Assuming fluorine has the strongest F–F bond because it is the most electronegative halogen. Bond energy is not the same as electronegativity; F–F is actually the weakest of the common halogens (apart from astatine).
• Thinking first ionisation energy increases down a group. It decreases because of increased shell number and shielding.
• Mixing up the two trends and picking CaBr₂.

Things to Be Careful About

• Read "higher first ionisation energy" and "higher Q–Q bond energy" as two separate filters.
• Remember that the question uses Q as a generic halogen symbol; the actual halogen is either Cl or Br.
• The formula of a Group 2 chloride/bromide is MX₂ because the metal forms a 2+ ion and the halogen forms a 1− ion.

Background Concept

The electron shell of an atom is divided into sub-shells: s, p, d and f. The p sub-shell consists of three orbitals. According to Hund's rule, electrons occupy empty orbitals singly before pairing in an orbital. Therefore the occupancy of the p sub-shell reveals how many p electrons the element has in its outer shell, and this, together with the Period number, identifies the element.

Understanding the Question

The question gives clues about the electron occupancy of the p sub-shells of two elements, Y and Z, from Period 2 and Period 3. We must identify Y and Z, then select the formula of the compound formed between them from the options.

Approach

Use the sentence about p-orbital occupancy to count the number of p electrons in the outer shell. Period 2 means the outer shell is n = 2; Period 3 means the outer shell is n = 3. Once the element is identified, use its valency to write the formula of the binary compound.

Step-by-Step Reasoning

For Y (Period 2): a Period 2 element has the outer configuration $2s^2\,2p^x$. The p sub-shell has three orbitals. "All three orbitals occupied; only one is fully occupied" means two orbitals contain one electron each and one orbital contains two electrons. Total p electrons = $1 + 1 + 2 = 4$, so the configuration is $2s^2\,2p^4$. This is oxygen, O.

For Z (Period 3): a Period 3 element has outer configuration $3s^2\,3p^x$. "Only two orbitals occupied" means exactly two p electrons, giving $3s^2\,3p^2$. This is silicon, Si.

Now consider the compound of oxygen and silicon. Oxygen has six outer electrons and needs two more to complete an octet; silicon has four outer electrons and can form four covalent bonds. The stable common oxide of silicon is $\text{SiO}_2$.

Checking the options: $\text{CCl}_4$ contains neither Y nor Z; $\text{SiCl}_4$ contains Si but Cl has five 3p electrons, so all three orbitals would be occupied; $\text{SO}_2$ contains S, whose Period 3 p sub-shell is $3p^4$ and hence all three orbitals occupied. Only $\text{SiO}_2$ matches both clues.

Key Takeaways

The key skill is translating orbital-occupancy wording into electron configurations. Remember: each p sub-shell has three orbitals; "occupied" means containing at least one electron; pairing only occurs once no empty orbital remains. The Period number fixes the principal quantum number, so the p count then identifies the element.

Common Mistakes

• Thinking that "occupied" means "full". An orbital can be occupied by one electron.
• Concluding that Y is sulfur because $p^4$ appears in Period 3 too; the Period given rules this out.
• Giving the formula incorrectly. Silicon is in Group 14 and oxygen in Group 16, so the oxide is $\text{SiO}_2$.

Things to Be Careful About

• Count electrons, not just orbitals. Three occupied p orbitals could mean 3, 4, 5 or 6 p electrons, and the phrase "only one fully occupied" is the extra clue that fixes the count at 4.
• Interpret the sub-shell occupancy for the ground state, following Hund's rule.
• When writing the answer, make sure the element symbols and subscript are correct: option C is $\text{SiO}_2$, not $\text{Si}_2\text{O}$ or $\text{SiO}$.

Background Concept

Glauber's salt is a hydrated salt: each formula unit of sodium sulfate carries $x$ molecules of water of crystallisation, written $\text{Na}_2\text{SO}_4 \cdot x\text{H}_2\text{O}$. On heating, the water is driven off as steam, leaving the anhydrous salt behind. The value of $x$ is simply the mole ratio of water to anhydrous salt in the crystal: for every one mole of $\text{Na}_2\text{SO}_4$ there are $x$ moles of $\text{H}_2\text{O}$. Finding $x$ therefore requires converting each given mass into moles using the appropriate molar mass, then dividing.

The molar mass of anhydrous sodium sulfate is $M_r(\text{Na}_2\text{SO}_4) = 2(23) + 32 + 4(16) = 142 \text{ g mol}^{-1}$. The molar mass of water is $M_r(\text{H}_2\text{O}) = 2(1) + 16 = 18 \text{ g mol}^{-1}$.

Understanding the Question

The question gives the masses of the two components released by heating: $1.91 \text{ g}$ of water and $1.51 \text{ g}$ of anhydrous $\text{Na}_2\text{SO}_4$. It asks for $x$, the number of moles of water per mole of salt in the hydrated crystal. The command is effectively a calculation: no theory beyond the mole concept is needed.

Approach

1. Work out the molar mass of $\text{Na}_2\text{SO}_4$.
2. Convert the mass of anhydrous salt to moles.
3. Convert the mass of water to moles.
4. Divide the moles of water by the moles of salt to get $x$.
5. Round to the nearest whole number, since $x$ must be an integer.

Step-by-Step Reasoning

First, the molar mass of the anhydrous salt:

$$M_r(\text{Na}_2\text{SO}_4) = 2(23) + 32 + 4(16) = 142 \text{ g mol}^{-1}$$

Moles of anhydrous salt:

$$n(\text{Na}_2\text{SO}_4) = \frac{1.51}{142} = 0.01063 \text{ mol}$$

Moles of water:

$$n(\text{H}_2\text{O}) = \frac{1.91}{18.0} = 0.1061 \text{ mol}$$

Ratio:

$$x = \frac{0.1061}{0.01063} = 9.98 \approx 10$$

So the formula is $\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O}$, which matches option C.

Distractor analysis: option D ($11.25$) results from using an incorrect molar mass for sodium sulfate (treating it as $160 \text{ g mol}^{-1}$ rather than $142$). Option B ($8.85$) and option A ($1$) arise from arithmetic slips or from ignoring the mole ratio entirely.

Key Takeaways

• A hydrated salt formula $\text{Salt} \cdot x\text{H}_2\text{O}$ means $x$ moles of water per mole of anhydrous salt.
• To find $x$, convert both masses to moles and divide: $x = n(\text{H}_2\text{O}) / n(\text{salt})$.
• Always use the molar mass of the anhydrous salt, not the hydrated salt.

Common Mistakes

• Using the total mass of the hydrated sample instead of the separate masses of salt and water.
• Using the wrong molar mass for $\text{Na}_2\text{SO}_4$ (e.g. $160$ instead of $142$), which produces option D.
• Forgetting to convert masses to moles and simply comparing masses (giving $1.91/1.51 \approx 1.27$, which is not $x$).
• Not rounding the ratio to a whole number.

Things to Be Careful About

• Use $M_r(\text{Na}_2\text{SO}_4) = 142 \text{ g mol}^{-1}$ and $M_r(\text{H}_2\text{O}) = 18 \text{ g mol}^{-1}$.
• Keep the units consistent; both masses are in grams.
• The ratio $0.1061 / 0.01063$ is very close to 10, so rounding to the nearest integer is safe.
• State the final answer with the option letter in an MCQ.

Background Concept

The amount of substance is measured in moles. One mole of any substance contains the same number of particles — the Avogadro constant, $6.02 \times 10^{23}$ mol⁻¹. Therefore, to compare the number of particles in different samples, we compare the number of moles in each. Moles are obtained by dividing mass by molar mass (for solids and liquids) or by dividing gas volume by the molar volume (24 dm³ mol⁻¹ at room temperature and pressure, r.t.p.).

Understanding the Question

This is a multiple-choice comparison question. Each option gives a 6.0 quantity — either a gas volume or a mass — and names the particle to count: argon atoms, carbon dioxide molecules, magnesium atoms, water molecules. The task is to find which sample has the largest number of those named particles. The key is to bring all four options to a common basis — moles.

Approach

Convert each option to moles:

• For a gas at r.t.p., moles = volume / 24 dm³ mol⁻¹.
• For a solid or liquid, moles = mass / molar mass.
  Then compare the four mole values; the largest corresponds to the greatest number of particles, since all share the same Avogadro constant.

Step-by-Step Reasoning

A: argon is a gas at room conditions. Molar volume at r.t.p. = 24 dm³ mol⁻¹. moles = 6.0/24 = 0.25 mol of Ar atoms.

B: CO₂ molar mass = 12 + 2(16) = 44 g mol⁻¹. moles = 6.0/44 = 0.136 mol.

C: Mg molar mass = 24.3 g mol⁻¹. moles = 6.0/24.3 = 0.247 mol.

D: H₂O molar mass = 2(1) + 16 = 18 g mol⁻¹. moles = 6.0/18 = 0.333 mol.

Comparing: 0.333 > 0.25 > 0.247 > 0.136, so D contains the greatest number of water molecules.

Why the distractors fail: A is tempting because gases have large molar volumes, but 6.0 dm³ is only 0.25 mol. B fails because CO₂ has a large molar mass (44). C is close, but Mg's molar mass (24.3) is larger than water's (18), so 6.0 g of Mg gives fewer moles than 6.0 g of water.

Key Takeaways

• To compare particle numbers, always convert to moles.
• Molar volume of a gas at r.t.p. = 24 dm³ mol⁻¹ (at s.t.p. it is 22.4 dm³ mol⁻¹).
• moles = mass / molar mass.
• The named particle matters: an element like argon exists as atoms, while compounds like CO₂ and H₂O exist as molecules.

Common Mistakes

• Using 22.4 dm³ (s.t.p.) instead of 24 dm³ (r.t.p.) — even then A gives 0.268 mol, still less than D.
• Using the wrong molar mass: CO₂ = 44 (not 28), H₂O = 18 (not 20).
• Forgetting to divide mass by molar mass and instead comparing masses directly (all are 6.0, which would be wrong).
• Confusing atoms and molecules: argon is monatomic, so one mole of Ar atoms equals one mole of Ar gas.

Things to Be Careful About

• Units: gas volume in dm³, molar volume 24 dm³ mol⁻¹.
• The Avogadro constant is identical for all samples, so comparing moles is sufficient — no need to multiply by 6.02 × 10²³.
• State the particle type in the answer: D is "water molecules".

Background Concept

A dative covalent (coordinate) bond forms when one atom or ion supplies both electrons of a shared pair to another atom or ion that has an empty orbital. The donor must have a lone pair; the acceptor must have a vacant orbital. Once formed, a dative covalent bond is indistinguishable from an ordinary covalent bond in the species.

This is different from:

• an ionic bond, which is the electrostatic attraction between oppositely charged ions after electron transfer;
• a hydrogen bond, which is a special dipole–dipole interaction involving a hydrogen atom covalently bonded to N, O or F and a lone pair on N, O or F;
• dipole–dipole forces, which are intermolecular attractions between polar molecules, not chemical bonds.

Understanding the Question

The question gives phosphine, $\text{PH}_3$, and asks what type of interaction occurs when it reacts with a hydrogen ion, $\text{H}^+$. Phosphorus has five valence electrons: three are used in the three $\text{P}-\text{H}$ bonds, leaving one lone pair. A hydrogen ion is a bare proton with no electrons and an empty 1s orbital. When $\text{PH}_3$ and $\text{H}^+$ combine, the lone pair on phosphorus is shared with the hydrogen ion, forming a new covalent bond. The correct classification of this bond is a dative covalent bond.

Approach

Look at the electron availability on each species. $\text{PH}_3$ has a lone pair, and $\text{H}^+$ has an empty orbital. Whenever a lone pair is donated into an empty orbital, the bond formed is coordinate (dative) covalent. The other options describe either intermolecular forces or a completely different type of bonding, so they can be eliminated once the dative bond is recognised.

Step-by-Step Reasoning

1. Phosphorus has the electron configuration $1s^2\,2s^2\,2p^6\,3s^2\,3p^3$. In $\text{PH}_3$, three of the five valence electrons form $\text{P}-\text{H}$ bonds, leaving one lone pair on phosphorus.
2. $\text{H}^+$ has no electrons; it is simply a proton with an empty 1s orbital.
3. The lone pair on phosphorus is donated into the empty orbital of $\text{H}^+$, forming a $\text{P}-\text{H}$ bond in which both electrons come from phosphorus.
4. This is exactly the definition of a dative covalent (coordinate) bond.
5. The product is the phosphonium ion, $\text{PH}_4^+$, in which all four $\text{P}-\text{H}$ bonds are equivalent once formed.
6. The other options are incorrect because:
   • hydrogen bonding requires H bonded to N, O or F, and a lone pair on N, O or F; $\text{PH}_3$ does not satisfy this;
   • dipole–dipole forces are intermolecular, not a bond formed between $\text{PH}_3$ and $\text{H}^+$;
   • an ionic bond would require electron transfer and the formation of separate ions, not the sharing of a lone pair.

Key Takeaways

• A dative covalent bond forms when a lone pair is donated into an empty orbital.
• The donor must have a lone pair; the acceptor must have an empty orbital.
• Once formed, a dative covalent bond is identical to an ordinary covalent bond.
• Distinguish between intramolecular bonds (covalent, ionic, dative covalent) and intermolecular forces (dipole–dipole, hydrogen bonding, van der Waals forces).

Common Mistakes

• Choosing hydrogen bond because $\text{H}^+$ contains hydrogen. Hydrogen bonding requires a hydrogen atom covalently bonded to N, O or F, and a lone pair on N, O or F. $\text{PH}_3$ does not meet this requirement.
• Choosing ionic bond because $\text{H}^+$ is a charged ion. No electron transfer occurs; instead, a lone pair is shared.
• Choosing dipole–dipole forces because $\text{PH}_3$ is a polar molecule. Dipole–dipole forces are intermolecular attractions, not a bond formed between $\text{PH}_3$ and $\text{H}^+$.

Things to Be Careful About

• A dative covalent bond is still a covalent bond; do not describe it as a separate type of bond that remains different after formation.
• In mechanisms, use an arrow from the lone pair on the donor to the acceptor to show dative bond formation.
• Remember that $\text{H}^+$ has no electrons and an empty orbital, while $\text{PH}_3$ has a lone pair on phosphorus.
• Do not confuse coordinate bonding with hydrogen bonding just because a hydrogen ion is involved.

Background Concept

Across Period 3 of the periodic table (Na, Mg, Al, Si, P, S, Cl, excluding Ar), several physical properties exhibit distinct trends due to the increasing nuclear charge and decreasing atomic radius, while the inner electron shielding remains relatively constant.

• Atomic radius decreases steadily across the period.
• First ionisation energy generally increases but has notable dips at aluminium (Al) and sulfur (S). The dip at Al occurs because the outermost electron is in a higher-energy 3p orbital, which is easier to remove than a 3s electron. The dip at S occurs due to electron-electron repulsion between the paired electrons in the 3p orbital.
• Electronegativity increases steadily and smoothly across the period because the increasing nuclear charge and decreasing radius draw bonding electrons more strongly. There are no dips or peaks.
• Melting point increases from Na to Si (due to increasing metallic bonding strength and the giant covalent structure of Si), then drops sharply for P, S, and Cl (which have simple molecular structures with weak van der Waals forces).

Understanding the Question

The question asks to identify which of the four graphs (A, B, C, or D) represents the trend in electronegativity across Period 3 elements (Na to Cl). The x-axis for all graphs is the element identity, and the y-axis is the magnitude of the physical property.

Approach

Recall the specific shape of the electronegativity trend across Period 3. Then, identify what the other three graphs represent by matching their shapes to known periodic trends. This process of elimination confirms the correct graph.

Step-by-Step Reasoning

1. Electronegativity trend: Electronegativity is the ability of an atom to attract shared electrons in a covalent bond. Across Period 3, the nuclear charge increases (11 to 17) while the atomic radius decreases. Shielding by inner electrons is roughly constant. Consequently, the attraction for bonding electrons increases steadily from Na to Cl. The graph must show a smooth, monotonic increase. Graph D matches this perfectly.
2. Graph A (peak at Si): This represents melting point. Na, Mg, and Al have metallic bonding that strengthens across the period. Si has a giant covalent (macromolecular) structure with very strong covalent bonds, giving it the highest melting point. P, S, and Cl exist as simple molecules with weak intermolecular forces, so their melting points drop sharply.
3. Graph B (steady decrease): This represents atomic radius. The increasing nuclear charge pulls the electron shells closer, reducing the radius steadily from Na to Cl.
4. Graph C (general increase with dips at Al and S): This represents first ionisation energy. The general increase is due to higher nuclear charge and smaller radius. The dip at Al is because the 3p electron is higher in energy and shielded by the 3s electrons. The dip at S is due to repulsion between the paired electrons in one of the 3p orbitals, making it easier to remove one.

Key Takeaways

• Electronegativity increases smoothly across a period.
• First ionisation energy has a jagged trend with dips at Group 13 (e.g., Al) and Group 16 (e.g., S).
• Atomic radius decreases smoothly.
• Melting point peaks at Group 14 (e.g., Si) due to the change from metallic to giant covalent to simple molecular structures.

Common Mistakes

• Confusing the ionisation energy trend (with dips at Al and S) with the electronegativity trend (which is a smooth increase). Students often memorise the "jagged" IE graph and apply it to electronegativity.
• Mistaking the melting point graph (peak at Si) for electronegativity, perhaps because Si is the most "central" or important element in Period 3 bonding.

Things to Be Careful About

• Electronegativity values are not measured directly but are derived scales (like Pauling scale), so the graph is a smooth theoretical trend without the quantum-mechanical dips seen in ionisation energies.
• Always exclude argon (Ar) when discussing Period 3 trends for properties like electronegativity and ionisation energy, as noble gases have a stable full shell and do not typically form bonds, making their electronegativity undefined or irrelevant in this context.

Background Concept

Elements can exist in different structural forms called allotropes. The physical properties of an element depend on the type of bonding and structure present in that allotrope.

• Giant covalent structures consist of a network of atoms joined by strong covalent bonds. They are usually hard but brittle, have very high melting/boiling points, and do not conduct electricity because the electrons are localised in covalent bonds.
• Giant metallic structures consist of positive ions in a sea of delocalised electrons. The delocalised electrons allow electrical conductivity in both the solid and liquid states, and the layers of ions can slide over each other, giving malleability and ductility.
• Giant ionic structures consist of oppositely charged ions held by strong electrostatic forces. They conduct electricity only when molten or in aqueous solution, not in the solid state, and are typically hard but brittle.

Tin is a metal, but its grey allotrope behaves like a non-metal because it adopts a giant covalent diamond-like structure.

Understanding the Question

The question gives a table of properties for grey tin and white tin:

• boiling point: similar for both (about 2540 K)
• electrical conductivity: grey tin has none in solid or liquid; white tin conducts in both solid and liquid
• malleability: grey tin is brittle; white tin is malleable

We are asked which structural change might take place when grey tin changes to white tin. The options are transitions between giant covalent, giant ionic, and giant metallic structures.

Approach

Use the physical properties as evidence for the type of bonding:

1. Electrical conductivity in the solid state is the key clue. A solid that conducts electricity must have delocalised electrons or mobile ions. Among the options, only a giant metallic structure conducts in the solid state.
2. Brittleness and lack of conductivity point to a giant covalent structure, not a giant ionic structure. Although ionic solids are brittle, they do not conduct in the solid state and usually have different melting/boiling behaviour.
3. Compare the two allotropes and select the option that matches grey tin to a non-conducting brittle structure and white tin to a conducting malleable structure.

Step-by-Step Reasoning

1. Grey tin: it has no electrical conductivity in the solid or liquid state. This rules out metallic bonding (which would conduct in both) and ionic bonding (which would conduct when liquid). The brittleness is consistent with a giant covalent network, where strong directional bonds resist deformation but break suddenly under stress.
2. White tin: it conducts electricity in both the solid and liquid states. This is characteristic of metallic bonding, where delocalised electrons remain available to carry charge even when the lattice is disrupted by melting. Its malleability also supports a metallic structure, because layers of metal ions can slide without breaking the metallic bonding.
3. Therefore, grey tin is giant covalent and white tin is giant metallic. The change is from giant covalent to giant metallic.

Checking the options:

• A giant covalent to giant ionic: incorrect because white tin is not ionic; it conducts in the solid state, which ionic solids do not.
• B giant covalent to giant metallic: correct.
• C giant ionic to giant covalent: incorrect because grey tin is not ionic; it does not conduct when liquid.
• D giant ionic to giant metallic: incorrect because grey tin is not ionic.

Key Takeaways

• Electrical conductivity in the solid state is a strong indicator of metallic bonding.
• Brittleness and lack of conductivity in all states suggest a giant covalent structure.
• Ionic structures conduct only when molten or in aqueous solution, not in the solid state.
• Allotropes of the same element can have completely different structures and therefore different physical properties.

Common Mistakes

• Assuming tin is always metallic because it is a metal. Grey tin is an allotrope with a giant covalent structure.
• Confusing ionic and covalent structures: both can be brittle, but ionic solids conduct when molten, whereas giant covalent substances generally do not.
• Choosing an option that makes grey tin ionic because it is brittle, without checking the conductivity evidence.

Things to Be Careful About

• Read the table carefully: the conductivity entry for grey tin says "none in solid or liquid", which eliminates both ionic and metallic possibilities.
• Malleability is a property of metals, so it strongly supports the metallic structure of white tin.
• The boiling points are similar and do not help distinguish the structures, so do not use them as the deciding evidence.

Background Concept

Solids can be classified by the type of particles they contain and the forces holding them together. Four common lattice types are:

• Ionic lattice: oppositely charged ions held by strong electrostatic forces (e.g. calcium fluoride, sodium chloride).
• Metallic lattice: positive ions in a sea of delocalised electrons (e.g. nickel, iron).
• Giant covalent (macromolecular) lattice: atoms held by a continuous network of covalent bonds (e.g. silicon(IV) oxide, diamond, graphite).
• Simple molecular lattice: discrete molecules held together by weak intermolecular forces such as van der Waals forces, dipole-dipole interactions, or hydrogen bonds (e.g. sulfur, ice, iodine).

Understanding the Question

The question asks which solid has a simple molecular lattice. We must identify the solid whose structure consists of small, discrete molecules rather than a continuous network of ions, atoms, or metallic bonding.

Approach

Recall the structure of each solid in the options:

• Calcium fluoride (CaF$_2$) — ionic lattice.
• Nickel (Ni) — metallic lattice.
• Silicon(IV) oxide (SiO$_2$) — giant covalent lattice.
• Sulfur (S$_8$) — simple molecular lattice.

Only sulfur matches the description.

Step-by-Step Reasoning

1. Calcium fluoride is an ionic compound of Ca$^{2+}$ and F$^-$ ions arranged in an ionic lattice — not molecular.
2. Nickel is a metal, so its structure is a metallic lattice of cations in a sea of delocalised electrons — not molecular.
3. Silicon(IV) oxide is a covalent network solid in which each silicon is bonded to four oxygens in a giant three-dimensional framework — a giant covalent lattice, not molecular.
4. Sulfur exists as discrete S$_8$ ring molecules held together only by weak van der Waals forces — a simple molecular lattice.

Therefore, the correct answer is D.

Key Takeaways

• Sulfur is the classic example of a simple molecular solid among the elements.
• Simple molecular lattices have low melting and boiling points because only weak intermolecular forces need to be overcome, not covalent or ionic bonds.

Common Mistakes

• Confusing silicon(IV) oxide with a molecular solid because it is a covalent compound. In reality, SiO$_2$ is a giant covalent (macromolecular) lattice, not a simple molecular structure.
• Assuming that all covalent compounds are molecular — many, such as diamond, graphite, and silicon(IV) oxide, are giant covalent lattices.

Things to Be Careful About

• Distinguish between "covalent compound" and "simple molecular structure". A compound can be covalent yet still form a giant covalent lattice.
• Remember that elements can adopt different lattice types; sulfur specifically forms discrete molecules, whereas metals and many non-metal oxides do not.

Background Concept

Hess's law states that the total enthalpy change of a reaction is independent of the route taken, provided the initial and final states are the same. This allows us to calculate the enthalpy change of a reaction by combining the enthalpy changes of other reactions whose sum equals the target reaction.

The standard enthalpy change of combustion, $\Delta H_c$, is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions (298 K, 1 atm). The standard enthalpy change of formation, $\Delta H_f$, is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

For butane, $\text{C}_4\text{H}_{10}$, complete combustion produces carbon dioxide and water:

$$\text{C}_4\text{H}_{10}(\text{g}) + \frac{13}{2}\text{O}_2(\text{g}) \rightarrow 4\text{CO}_2(\text{g}) + 5\text{H}_2\text{O}(\text{l})$$

The critical insight is that the combustion of carbon, $\text{C}(\text{s}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g})$, is exactly the formation of $\text{CO}_2$ from its elements. Therefore:

$$\Delta H_f(\text{CO}_2) = \Delta H_c(\text{C}) = -394 \text{ kJ mol}^{-1}$$

Similarly, the combustion of hydrogen, $\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l})$, is the formation of water:

$$\Delta H_f(\text{H}_2\text{O}) = \Delta H_c(\text{H}_2) = -286 \text{ kJ mol}^{-1}$$

Understanding the Question

The question provides three enthalpy values: the standard combustion enthalpies of carbon and hydrogen, and the standard formation enthalpy of butane. It asks us to determine the standard combustion enthalpy of butane from these values, selecting the correct option among four.

This is a Hess's law problem. The key is to recognise that the combustion enthalpies of carbon and hydrogen give us the formation enthalpies of $\text{CO}_2$ and $\text{H}_2\text{O}$, which are the products of butane combustion. The formation enthalpy of butane is the enthalpy of the reactant from its elements.

Approach

We construct a Hess cycle with two routes from the elements (carbon, hydrogen, oxygen) to the combustion products ($\text{CO}_2$ and $\text{H}_2\text{O}$):

Route 1: Elements $\rightarrow$ butane $\rightarrow$ combustion products.
The enthalpy change is: $\Delta H_f(\text{butane}) + \Delta H_c(\text{butane})$.

Route 2: Elements $\rightarrow$ combustion products directly.
The enthalpy change is: $4 \times \Delta H_f(\text{CO}_2) + 5 \times \Delta H_f(\text{H}_2\text{O}) = 4 \times \Delta H_c(\text{C}) + 5 \times \Delta H_c(\text{H}_2)$.

By Hess's law, both routes give the same total enthalpy change:

$$\Delta H_f(\text{butane}) + \Delta H_c(\text{butane}) = 4 \times \Delta H_c(\text{C}) + 5 \times \Delta H_c(\text{H}_2)$$

Rearranging:

$$\Delta H_c(\text{butane}) = 4 \times \Delta H_c(\text{C}) + 5 \times \Delta H_c(\text{H}_2) - \Delta H_f(\text{butane})$$

Step-by-Step Reasoning

1. Write the balanced combustion equation for butane:
   $\text{C}_4\text{H}_{10}(\text{g}) + \frac{13}{2}\text{O}_2(\text{g}) \rightarrow 4\text{CO}_2(\text{g}) + 5\text{H}_2\text{O}(\text{l})$.
   This tells us the stoichiometric coefficients: 4 mol of $\text{CO}_2$ and 5 mol of $\text{H}_2\text{O}$ are produced per mole of butane.

2. Identify formation enthalpies from combustion enthalpies:

   • $\Delta H_f(\text{CO}_2) = \Delta H_c(\text{C}) = -394 \text{ kJ mol}^{-1}$
   • $\Delta H_f(\text{H}_2\text{O}) = \Delta H_c(\text{H}_2) = -286 \text{ kJ mol}^{-1}$

3. Apply Hess's law:

   $$\Delta H_c(\text{butane}) = 4 \times (-394) + 5 \times (-286) - (-129)$$

4. Calculate step by step:

   • $4 \times (-394) = -1576 \text{ kJ mol}^{-1}$
   • $5 \times (-286) = -1430 \text{ kJ mol}^{-1}$
   • Sum: $-1576 + (-1430) = -3006 \text{ kJ mol}^{-1}$
   • Subtract the formation enthalpy: $-3006 - (-129) = -3006 + 129 = -2877 \text{ kJ mol}^{-1}$

5. Select the correct option: B, $-2877 \text{ kJ mol}^{-1}$.

Distractor analysis:

• A ($-551$ kJ mol$^{-1}$): This comes from $394 + 286 - 129 = 551$, i.e., ignoring the stoichiometric coefficients entirely. The candidate added the two combustion enthalpies and subtracted the formation enthalpy without multiplying by 4 and 5.
• C ($-3135$ kJ mol$^{-1}$): This comes from $4 \times 394 + 5 \times 286 + 129 = 3135$, i.e., adding the formation enthalpy instead of subtracting it. The sign of the formation term is wrong.
• D ($-4307$ kJ mol$^{-1}$): A more complex error, likely involving incorrect stoichiometric ratios or multiple sign errors.

Key Takeaways

• The combustion of an element is the same as the formation of its oxide. This lets us convert between $\Delta H_c$ and $\Delta H_f$ values.
• Hess's law lets us combine enthalpy changes along different routes to find an unknown enthalpy change.
• Always multiply enthalpy values by the stoichiometric coefficients from the balanced equation.
• Watch the signs: the formation enthalpy of the reactant is subtracted in this type of Hess calculation.

Common Mistakes

• Ignoring stoichiometric coefficients: Not multiplying $\Delta H_c(\text{C})$ by 4 and $\Delta H_c(\text{H}_2)$ by 5 leads to option A.
• Wrong sign on the formation term: Adding $\Delta H_f(\text{butane})$ instead of subtracting it leads to option C.
• Using the wrong number of water molecules: Butane has 10 hydrogen atoms, so combustion produces 5 mol of $\text{H}_2\text{O}$, not 4 or 6.
• Forgetting the negative signs: The given values are all negative; forgetting a minus sign changes the result significantly.

Things to Be Careful About

• The balanced equation for butane combustion has a fractional coefficient: $\frac{13}{2}\text{O}_2$. This is normal for combustion of a hydrocarbon.
• State symbols matter: $\text{H}_2\text{O}$ is liquid in standard combustion enthalpy terms.
• The answer must carry the negative sign: butane combustion is exothermic.
• Units: $\text{kJ mol}^{-1}$. The value is per mole of butane.
• When subtracting a negative number, the minus sign becomes a plus: $-3006 - (-129) = -3006 + 129$.

Background Concept

An enthalpy change, $\Delta H$, is positive for an endothermic process (heat absorbed from the surroundings) and negative for an exothermic process (heat released to the surroundings).

Understanding the Question

We are given three processes and asked which statement correctly describes their enthalpy changes. The key is to decide for each process whether it releases or absorbs energy.

Approach

Classify each process by its energy change:

• Neutralisation is always exothermic.
• Combustion is always exothermic.
• Condensation (gas to liquid) is exothermic because intermolecular bonds form.

Step-by-Step Reasoning

1. $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})$: This is neutralisation, which releases heat. $\Delta H$ is negative.
2. $\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})$: This is combustion, which releases heat. $\Delta H$ is negative.
3. $\text{NH}_3(\text{g}) \rightarrow \text{NH}_3(\text{l})$: This is condensation, a gas changing to a liquid. Intermolecular forces form, releasing energy. $\Delta H$ is negative.

Since all three processes are exothermic, none has a positive enthalpy change. Therefore, statement A is correct.

Key Takeaways

• Exothermic processes have negative $\Delta H$; endothermic processes have positive $\Delta H$.
• Combustion, neutralisation, and condensation are all exothermic.
• Phase changes from gas to liquid or solid release energy; the reverse changes absorb energy.

Common Mistakes

• Thinking condensation is endothermic because it involves cooling. In fact, forming intermolecular bonds releases energy.
• Assuming combustion must be endothermic because it requires ignition. The activation energy is separate from the overall enthalpy change, which is negative.

Things to Be Careful About

• The sign convention: a negative $\Delta H$ means energy is released by the system.
• Process 3 is a physical change, not a chemical reaction, but it still has an enthalpy change and is exothermic.

Background Concept

Redox (reduction–oxidation) reactions involve the transfer of electrons from one species to another. The species that loses electrons is oxidised (its oxidation number increases) and acts as the reducing agent; the species that gains electrons is reduced (its oxidation number decreases) and acts as the oxidising agent.

To balance a redox equation, chemists split the reaction into two half-equations — one for reduction and one for oxidation — balance atoms and charge in each, then combine them so that the total number of electrons lost equals the total number gained.

The medium matters: in acidic solution, $\text{H}^+$ and $\text{H}_2\text{O}$ are used to balance H and O; in alkaline solution, $\text{OH}^-$ and $\text{H}_2\text{O}$ are used. This question specifies alkaline solution, so $\text{OH}^-$ must appear in the balanced half-equations.

Understanding the Question

We are told that in alkaline solution, $\text{MnO}_4^-$ oxidises $\text{SO}_3^{2-}$ to $\text{SO}_4^{2-}$, while $\text{MnO}_4^-$ itself is reduced to $\text{MnO}_2$. The question asks for the ratio of $\text{MnO}_4^-$ to $\text{SO}_3^{2-}$ in the balanced chemical equation — i.e., how many $\text{MnO}_4^-$ ions react with how many $\text{SO}_3^{2-}$ ions.

This is a classic redox-balancing problem. The answer is found by:

1. Determining the electron change for each species.
2. Writing balanced half-equations.
3. Combining them so electrons cancel.

Approach

1. Assign oxidation numbers: Mn in $\text{MnO}_4^-$ is +7, in $\text{MnO}_2$ is +4; S in $\text{SO}_3^{2-}$ is +4, in $\text{SO}_4^{2-}$ is +6.
2. Write the reduction half-equation ($\text{MnO}_4^- \rightarrow \text{MnO}_2$) and oxidation half-equation ($\text{SO}_3^{2-} \rightarrow \text{SO}_4^{2-}$), balancing O with $\text{H}_2\text{O}$ and H with $\text{OH}^-$ (alkaline conditions), then balancing charge with electrons.
3. Find the lowest common multiple of the electron changes (3 and 2 $\rightarrow$ 6).
4. Multiply each half-equation by the factor that gives 6 electrons, add them, cancel spectator $\text{H}_2\text{O}$/$\text{OH}^-$, and read off the coefficients.

Step-by-Step Reasoning

Step 1 — Oxidation numbers

In $\text{MnO}_4^-$, each O is -2, so Mn must be +7 (since -1 = Mn + 4(-2) $\rightarrow$ Mn = +7). In $\text{MnO}_2$, Mn = +4. The change is +7 $\rightarrow$ +4, a gain of 3 electrons per Mn (reduction).

In $\text{SO}_3^{2-}$, S = +4 (since -2 = S + 3(-2) $\rightarrow$ S = +4). In $\text{SO}_4^{2-}$, S = +6. The change is +4 $\rightarrow$ +6, a loss of 2 electrons per S (oxidation).

Step 2 — Reduction half-equation

Start with $\text{MnO}_4^- \rightarrow \text{MnO}_2$.

Balance O: left has 4 O, right has 2 O. Add $2\text{H}_2\text{O}$ to the right:
$\text{MnO}_4^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$

Balance H: right has 4 H. Add $4\text{H}^+$ to the left (acidic method):
$\text{MnO}_4^- + 4\text{H}^+ \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$

Convert to alkaline: add $4\text{OH}^-$ to both sides:
$\text{MnO}_4^- + 4\text{H}^+ + 4\text{OH}^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} + 4\text{OH}^-$
$\text{MnO}_4^- + 4\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} + 4\text{OH}^-$
$\text{MnO}_4^- + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{OH}^-$

Balance charge: left charge is -1, right is -4. Add $3\text{e}^-$ to the left:
$\text{MnO}_4^- + 2\text{H}_2\text{O} + 3\text{e}^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$

Check: atoms — Mn 1=1, O 6=6, H 4=4; charge — (-1)+(-3) = -4 = -4 $\checkmark$

Step 3 — Oxidation half-equation

Start with $\text{SO}_3^{2-} \rightarrow \text{SO}_4^{2-}$.

Balance O: left has 3 O, right has 4 O. Add $1\text{H}_2\text{O}$ to the left:
$\text{SO}_3^{2-} + \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-}$

Balance H: left has 2 H. Add $2\text{H}^+$ to the right:
$\text{SO}_3^{2-} + \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 2\text{H}^+$

Convert to alkaline: add $2\text{OH}^-$ to both sides:
$\text{SO}_3^{2-} + \text{H}_2\text{O} + 2\text{OH}^- \rightarrow \text{SO}_4^{2-} + 2\text{H}^+ + 2\text{OH}^-$
$\text{SO}_3^{2-} + \text{H}_2\text{O} + 2\text{OH}^- \rightarrow \text{SO}_4^{2-} + 2\text{H}_2\text{O}$
$\text{SO}_3^{2-} + 2\text{OH}^- \rightarrow \text{SO}_4^{2-} + \text{H}_2\text{O}$

Balance charge: left is -4, right is -2. Add $2\text{e}^-$ to the right:
$\text{SO}_3^{2-} + 2\text{OH}^- \rightarrow \text{SO}_4^{2-} + \text{H}_2\text{O} + 2\text{e}^-$

Check: atoms — S 1=1, O 5=5, H 2=2; charge — (-2)+(-2) = -4 = (-2)+0+(-2) $\checkmark$

Step 4 — Combine

Reduction: 3 e$^-$ per $\text{MnO}_4^-$; Oxidation: 2 e$^-$ per $\text{SO}_3^{2-}$.
LCM = 6. Multiply reduction by 2, oxidation by 3:

$2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6\text{e}^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^-$

$3\text{SO}_3^{2-} + 6\text{OH}^- \rightarrow 3\text{SO}_4^{2-} + 3\text{H}_2\text{O} + 6\text{e}^-$

Add:
$2\text{MnO}_4^- + 4\text{H}_2\text{O} + 3\text{SO}_3^{2-} + 6\text{OH}^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^- + 3\text{SO}_4^{2-} + 3\text{H}_2\text{O}$

Cancel $3\text{H}_2\text{O}$ from both sides (4 - 3 = 1 $\text{H}_2\text{O}$ left) and $6\text{OH}^-$ from both sides (8 - 6 = 2 $\text{OH}^-$ right):

$2\text{MnO}_4^- + \text{H}_2\text{O} + 3\text{SO}_3^{2-} \rightarrow 2\text{MnO}_2 + 2\text{OH}^- + 3\text{SO}_4^{2-}$

Step 5 — Read the ratio

Coefficients: 2 $\text{MnO}_4^-$ : 3 $\text{SO}_3^{2-}$ $\rightarrow$ ratio 2 : 3 $\rightarrow$ option A.

Why the distractors are wrong:

• B (3:2) reverses the correct ratio.
• C (4:7) and D (7:4) do not correspond to any combination that balances electrons (they would imply electron transfers of 12 and 14 respectively, which don't match the 3 e$^-$/2 e$^-$ changes per ion).

Key Takeaways

• Redox equations must balance both atoms and charge.
• In alkaline solution, use $\text{OH}^-$ and $\text{H}_2\text{O}$ (not $\text{H}^+$) to balance.
• The stoichiometric ratio comes from the coefficients after cancelling electrons.
• Always verify the final equation balances for atoms and charge.

Common Mistakes

• Using $\text{H}^+$ instead of $\text{OH}^-$ in alkaline solution — the half-equations will be wrong.
• Forgetting to balance charge with electrons — the electron count will be off.
• Not simplifying the final equation (leaving $\text{H}_2\text{O}$/$\text{OH}^-$ on both sides).
• Reversing the ratio (option B) — the ratio is $\text{MnO}_4^-$ : $\text{SO}_3^{2-}$, not $\text{SO}_3^{2-}$ : $\text{MnO}_4^-$.

Things to Be Careful About

• Oxidation number of Mn in $\text{MnO}_4^-$ is +7 (not +8 — the overall charge is -1 with four O at -2 each).
• Oxidation number of S in $\text{SO}_3^{2-}$ is +4, in $\text{SO}_4^{2-}$ is +6.
• The LCM step: 3 and 2 $\rightarrow$ 6, so coefficients 2 and 3.
• Always check atoms and charge balance in the final equation.

Background Concept

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. For gas-phase equilibria, changes in volume or pressure (by changing volume) shift the equilibrium towards the side with fewer (or more) moles of gas. Removing a reactant or product also shifts the equilibrium to oppose the change (e.g., removing a reactant shifts the equilibrium to produce more of it).

Understanding the Question

We have a central vessel at equilibrium containing $\text{N}_2$, $\text{H}_2$, and $\text{NH}_3$. It is connected to an evacuated vessel (left) and a vessel with solid lithium (right). We must determine which action (opening taps or keeping them closed) results in the highest amount of ammonia at equilibrium. The stem tells us lithium reacts with nitrogen to form solid $\text{Li}_3\text{N}$.

Approach

Evaluate the effect of each option on the equilibrium $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$:

1. Opening tap 1 changes the volume/pressure. Apply the moles-of-gas rule.
2. Opening tap 2 introduces a chemical reaction that removes a component. Apply the concentration change rule.
3. Compare the outcomes to find which leaves the most $\text{NH}_3$.

Step-by-Step Reasoning

• Option A (Keep both closed): The system is isolated. The equilibrium remains exactly as it was, with its maximum possible yield of ammonia under the given conditions.
• Option C (Open tap 1 only): The evacuated vessel is connected, so the gas mixture expands to fill a larger volume. The total pressure drops. The equilibrium has 4 moles of gas on the left ($\text{N}_2 + 3\text{H}_2$) and 2 moles on the right ($2\text{NH}_3$). To counteract the pressure drop, the equilibrium shifts to the left (towards more moles of gas). This consumes $\text{NH}_3$, reducing its amount.
• Option D (Open tap 2 only): Lithium reacts with nitrogen: $6\text{Li}(\text{s}) + \text{N}_2(\text{g}) \rightarrow 2\text{Li}_3\text{N}(\text{s})$. This continuously removes $\text{N}_2$ from the middle vessel. The decrease in $[\text{N}_2]$ causes the equilibrium to shift to the left to produce more $\text{N}_2$, which consumes $\text{NH}_3$ and $\text{H}_2$. The amount of $\text{NH}_3$ decreases.
• Option B (Open both taps): Both disturbances occur simultaneously. The pressure drops and $\text{N}_2$ is removed, both causing a leftward shift. The amount of $\text{NH}_3$ decreases the most.

Thus, keeping both taps closed (Option A) results in the most ammonia.

Key Takeaways

• Expanding a gas equilibrium into a larger volume (lowering pressure) shifts it towards the side with more gaseous moles.
• Removing a reactant via a secondary reaction shifts the primary equilibrium to produce more of that reactant, consuming the products.
• To maximize a product at equilibrium, avoid any disturbances that shift the equilibrium away from the product side.

Common Mistakes

• Assuming that opening a tap to an evacuated vessel increases the amount of product because "more space" is available. (Volume increase lowers pressure, shifting to more moles of gas).
• Forgetting that lithium reacts with nitrogen gas, not ammonia. (If it reacted with ammonia, the reasoning would differ, but the stem explicitly states it reacts with nitrogen).
• Miscounting the moles of gas on each side of the ammonia synthesis equation (4 moles on the left, 2 on the right).

Things to Be Careful About

• Always count the moles of gaseous species when applying Le Chatelier's principle to pressure/volume changes. Solids and liquids do not contribute to the gas mole count.
• Read the stem carefully for secondary reactions. The information about lithium reacting with nitrogen is the key to understanding what happens when tap 2 is opened.
• "Most ammonia" means we are looking for the option that causes the least leftward shift (or no shift at all). Any disturbance that removes reactants or lowers pressure will shift this specific equilibrium to the left.

Background Concept

For a general equilibrium $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$, the equilibrium constant in terms of concentrations is

$$K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$$

where square brackets denote equilibrium concentrations in $\text{mol dm}^{-3}$. For this reaction:

$$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$$

Since all species are in the same vessel of volume $V$, $[\text{species}] = n(\text{species})/V$. Here the total number of moles of gas on each side is 2, so $V$ cancels and mole amounts may be used directly in the expression.

Understanding the Question

The question gives initial amounts of $\text{H}_2$ and $\text{I}_2$ and the equilibrium amount of $\text{HI}$. It asks which expression correctly gives $K_c$. The key is that $K_c$ must use equilibrium amounts of all three species, so we must first work out how much $\text{H}_2$ and $\text{I}_2$ remain. The volume is not given, but because it cancels, the expression can be written using mole amounts.

Approach

Use the stoichiometry of the equation to find the change in amount. Since 2 mol $\text{HI}$ are produced from 1 mol $\text{H}_2$ and 1 mol $\text{I}_2$, forming 0.26 mol $\text{HI}$ consumes 0.13 mol of each reactant. Subtract these from the initial amounts to get equilibrium amounts, then substitute into the $K_c$ expression and compare with the options.

Step-by-Step Reasoning

1. Initial: $n(\text{H}_2) = 0.20\text{ mol}$, $n(\text{I}_2) = 0.15\text{ mol}$, $n(\text{HI}) = 0$.
2. Equilibrium: $n(\text{HI}) = 0.26\text{ mol}$.
3. From the equation, 2 mol $\text{HI}$ are formed per 1 mol $\text{H}_2$ consumed and 1 mol $\text{I}_2$ consumed. Therefore consumed $\text{H}_2$ = consumed $\text{I}_2$ = $0.26/2 = 0.13\text{ mol}$.
4. Equilibrium amounts:

$$n(\text{H}_2) = 0.20 - 0.13 = 0.07\text{ mol}$$ $$n(\text{I}_2) = 0.15 - 0.13 = 0.02\text{ mol}$$

1. Substitute into $K_c$:

$$K_c = \frac{(0.26/V)^2}{(0.07/V)(0.02/V)} = \frac{(0.26)^2}{0.07 \times 0.02}$$

This matches option C.

Distractor analysis:

• A uses initial amounts and has no square on $[\text{HI}]$.
• B incorrectly uses $2 \times 0.26$ and squares it; 2 is not a concentration factor and initial amounts are not equilibrium amounts.
• D uses 0.13 for both $\text{H}_2$ and $\text{I}_2$, but 0.13 is the amount consumed, not the amount remaining.

Key Takeaways

• $K_c$ is always written with equilibrium concentrations, not initial amounts.
• The stoichiometric coefficients become powers in the $K_c$ expression.
• Use the reaction stoichiometry to convert the amount of product formed into amounts of reactants consumed.
• When the total moles of gas are the same on both sides, the volume cancels and mole amounts can be used directly in $K_c$.

Common Mistakes

• Substituting initial moles into $K_c$: this ignores the reaction that has occurred.
• Forgetting to square $[\text{HI}]$, because the coefficient of $\text{HI}$ is 2.
• Thinking that 0.26 mol $\text{HI}$ requires 0.26 mol of each reactant; the 2:1 stoichiometry means only 0.13 mol of each is consumed.
• Using 0.13 mol as the equilibrium amount of $\text{H}_2$ or $\text{I}_2$; 0.13 is the change, not the amount remaining.

Things to Be Careful About

• Check that the expression uses products over reactants, with coefficients as powers.
• If the volume does not cancel, convert moles to concentrations before writing $K_c$.
• Here the volume cancels, but it is safer to show concentrations as $n/V$ before cancelling.
• The correct denominator is $0.07 \times 0.02$, not $0.13 \times 0.13$ or $0.20 \times 0.15$.