Chemistry 9701/13 — October/November 2025

Background Concept

An atom is described by its proton number $Z$ and mass number $A$. For a neutral atom, the number of electrons equals $Z$. A positive ion has lost electrons, so its electron count is $Z$ minus the charge. Electrons occupy orbitals in order of increasing energy: $1\text{s}$, $2\text{s}$, $2\text{p}$, $3\text{s}$, $3\text{p}$, ... with maximum occupancies 2, 2, 6, 2, 6 respectively.

Understanding the Question

The particle is written $^{27}_{13}\text{Al}^{+}$: the lower number 13 is the proton number, and the upper number 27 is the mass number. The $+$ charge means one electron has been removed from the neutral aluminium atom. The question asks for the ground-state electronic configuration of this ion.

Approach

First find how many electrons the ion contains. Then fill the orbitals in the correct energy order until all electrons are placed. Compare the resulting configuration with the options.

Step-by-Step Reasoning

1. Neutral Al has $Z = 13$, so it has 13 electrons.
2. $\text{Al}^{+}$ has lost one electron: $13 - 1 = 12$ electrons.
3. Fill orbitals in order:
   • $1\text{s}^2$ (2 electrons)
   • $2\text{s}^2$ (2, total 4)
   • $2\text{p}^6$ (6, total 10)
   • $3\text{s}^2$ (2, total 12)
4. The configuration is $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2$, which is option B.

Option A places the last two electrons as $3\text{s}^1 3\text{p}^1$; this is not the ground state because $3\text{s}$ must fill before $3\text{p}$. Option C has 14 electrons, which would be a neutral silicon atom, not $\text{Al}^{+}$. Option D has far too many electrons and incorrectly includes $3\text{d}$ and $4\text{s}$ orbitals.

Key Takeaways

For an ion, always adjust the electron count for the charge before writing the configuration. The mass number is not needed for electron configuration. Fill subshells in order of increasing energy.

Common Mistakes

• Using the mass number 27 as the number of electrons.
• Forgetting the $+$ charge and writing the neutral Al configuration $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^1$.
• Filling $3\text{p}$ before $3\text{s}$ is fully occupied.
• Including $3\text{d}$ electrons for a species with only 12 electrons.

Things to Be Careful About

• The lower number in $^{A}_{Z}\text{X}$ is the proton number; the upper number is the mass number.
• A $+$ charge means electrons have been removed; a $-$ charge means electrons have been added.
• Check the total number of electrons by summing the superscripts in the configuration.
• Use the Aufbau order $1\text{s} < 2\text{s} < 2\text{p} < 3\text{s} < 3\text{p}$ for this element.

Background Concept

Ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms or ions, producing one mole of gaseous ions with one more positive charge. The first ionisation energy removes the first (outermost, most loosely held) electron; each successive ionisation energy removes the next electron from an increasingly positive ion.

Successive ionisation energies always increase because each electron is removed from a more positively charged ion, so the remaining electrons are held more tightly. The increase, however, is not smooth. When the electron being removed comes from a different, inner shell — closer to the nucleus and less shielded by other electrons — there is a dramatic step-like jump in ionisation energy. This jump is the key signature: it marks the transition from removing outer-shell (valence) electrons to removing inner-shell electrons.

For a main-group element, the number of electrons removed before the first large jump equals the number of outer-shell electrons, which equals the group number.

All Period 3 elements have their valence electrons in the third shell:

• Na: $3s^1$ (1 outer electron)
• Mg: $3s^2$ (2)
• Al: $3s^2 3p^1$ (3)
• Si: $3s^2 3p^2$ (4)
• P: $3s^2 3p^3$ (5)
• S: $3s^2 3p^4$ (6)
• Cl: $3s^2 3p^5$ (7)
• Ar: $3s^2 3p^6$ (8)

For phosphorus, removing five electrons gives $P^{5+}$ with the configuration $1s^2 2s^2 2p^6$ — the neon noble-gas core. The sixth electron must come from the 2p subshell, which is much closer to the nucleus and far less shielded, so the sixth ionisation energy is dramatically larger than the fifth.

Understanding the Question

The table supplies the 5th through 10th ionisation energies for three unknown Period 3 elements, X, Y and Z. The task is to identify each element by finding where the large jump occurs in its series of ionisation energies.

The command is "What are the correct identities" — a deduction. For each element we must count how many electrons are removed before the big jump; that number tells us the number of outer-shell electrons and hence the group and identity of the element.

Approach

For each element:

1. Examine the successive differences (or ratios) between consecutive ionisation energies.
2. Identify the largest jump — the biggest relative increase.
3. The number of electrons removed before that jump equals the number of outer-shell electrons.
4. Match that number to a Period 3 element.

Step-by-Step Reasoning

Element X: 5th = 6274, 6th = 21 269, 7th = 25 398, 8th = 29 855, 9th = 35 868, 10th = 40 960.

The jump from 6274 to 21 269 is roughly a factor of 3.4, while the subsequent jumps (25 398, 29 855, 35 868, 40 960) are all modest increases. The large jump is therefore between the 5th and 6th ionisation energies. This means the 6th electron is removed from the inner 2p subshell, so X has 5 outer-shell electrons. A Period 3 element with 5 outer electrons is phosphorus (P), Group 15.

Element Y: 5th = 7012, 6th = 8496, 7th = 27 107, ...

The jump from 8496 to 27 107 is roughly a factor of 3.2 — the large jump is between the 6th and 7th ionisation energies. Y therefore has 6 outer-shell electrons, identifying it as sulfur (S), Group 16.

Element Z: 5th = 6542, 6th = 9362, 7th = 11 018, 8th = 33 606, ...

The jumps 6542 → 9362 and 9362 → 11 018 are small; the large jump is from 11 018 to 33 606, between the 7th and 8th ionisation energies. Z therefore has 7 outer-shell electrons, identifying it as chlorine (Cl), Group 17.

So X = P, Y = S, Z = Cl, which is option C.

Why the distractors are wrong:

• A (Na, Mg, Al): these elements have 1, 2 and 3 outer-shell electrons, so their large jumps would appear between the 1st–2nd, 2nd–3rd and 3rd–4th ionisation energies — all before the 5th ionisation energy. None of the data shown would contain a jump, so A cannot be correct.
• B (Mg, Al, Si): 2, 3 and 4 outer electrons; again the jumps would occur before the 5th ionisation energy. Incorrect.
• D (S, Cl, Ar): S and Cl correctly have 6 and 7 outer electrons, matching Y and Z, but Ar has 8 outer-shell electrons, so its jump would be between the 8th and 9th ionisation energies. Element X shows a jump between the 5th and 6th, so X cannot be S. Incorrect.

Key Takeaways

• The position of the large jump in successive ionisation energies reveals the number of outer-shell electrons and hence the group of a main-group element.
• The jump occurs because the next electron is removed from an inner, more tightly held shell (here the 2p subshell).
• This reasoning is a powerful tool for identifying unknown elements from ionisation-energy data.

Common Mistakes

• Miscounting the jump position: the jump between the nth and (n+1)th ionisation energy means there are n outer-shell electrons, not n+1.
• Using absolute differences instead of relative jumps: at high ionisation energies even a "small" step is a large absolute number; always compare the ratio or relative increase to find the true shell-change jump.
• Partially matching options: option D correctly places S and Cl for Y and Z but fails on X; each element must be checked independently.

Things to Be Careful About

• Identify the jump as the largest relative increase, not merely the largest absolute increase.
• Remember that Period 3 elements have their valence electrons in the third shell (3s, 3p); the jump signals entry into the 2p subshell.
• Work each element independently, then match the complete set to the options.

Background Concept

The mole is the amount of substance that contains $6.02 \times 10^{23}$ particles (the Avogadro constant, $L$ or $N_A$). The number of particles is related to the amount in moles by:

$N = n \times N_A$

A key subtlety in this type of question is that "oxygen atoms" refers to individual atoms, not molecules. The number of oxygen atoms in a sample depends on how many oxygen atoms appear in each formula unit of the compound.

Understanding the Question

The question asks which given amount of a compound contains exactly $9.03 \times 10^{23}$ oxygen atoms. This is a mole-conversion problem: first convert the given particle count into moles of oxygen atoms, then check which option produces that many oxygen atoms.

Approach

1. Convert $9.03 \times 10^{23}$ atoms into moles using the Avogadro constant.
2. For each option, multiply the given moles of compound by the number of oxygen atoms per formula unit.
3. Identify the option whose product equals the target moles of oxygen atoms.

Step-by-Step Reasoning

Step 1 — Find the target amount of oxygen atoms.

$n = \frac{9.03 \times 10^{23}}{6.02 \times 10^{23}} = 1.5\text{ mol}$

So we need a sample containing 1.5 mol of oxygen atoms.

Step 2 — Check each option.

• A Aluminium oxide is $\mathrm{Al_2O_3}$, with 3 oxygen atoms per formula unit:
  $0.25 \times 3 = 0.75\text{ mol O} \quad (\times)$
• B Sulfur dioxide is $\mathrm{SO_2}$, with 2 oxygen atoms per molecule:
  $0.75 \times 2 = 1.5\text{ mol O} \quad (\checkmark)$
• C Sulfur trioxide is $\mathrm{SO_3}$, with 3 oxygen atoms per molecule:
  $1.5 \times 3 = 4.5\text{ mol O} \quad (\times)$
• D Water is $\mathrm{H_2O}$, with 1 oxygen atom per molecule:
  $3.0 \times 1 = 3.0\text{ mol O} \quad (\times)$

Only B gives exactly 1.5 mol of oxygen atoms.

Key Takeaways

• Always divide the number of particles by $6.02 \times 10^{23}$ to get moles.
• Multiply the moles of compound by the number of atoms of interest per formula unit.
• The formula of the compound determines how many oxygen atoms each particle contributes.

Common Mistakes

• Confusing molecules with atoms: e.g. treating $\mathrm{SO_2}$ as contributing 1 oxygen atom instead of 2.
• Using the wrong formula: e.g. writing aluminium oxide as $\mathrm{AlO}$ instead of $\mathrm{Al_2O_3}$.
• Forgetting to divide the particle count by $N_A$ before comparing.

Things to Be Careful About

• Check the subscript on oxygen in each formula carefully.
• Remember that $9.03 \times 10^{23}$ is deliberately chosen as $1.5 \times N_A$, so the arithmetic should come out cleanly.
• Read the question precisely: it asks for oxygen atoms, not oxygen molecules ($\mathrm{O_2}$).

Background Concept

This question tests the mole concept and stoichiometry. The amount of any substance in moles is found from its mass and molar mass:

A balanced chemical equation gives the mole ratio in which reactants combine and products form. When two reactants are supplied, one may be in excess and the other is the limiting reagent — the one that is completely used up. The limiting reagent determines the maximum amount of product that can form. Finally, the volume of a gas at room conditions (about 25 °C and 1 atm) is found using the molar gas volume of $24\ \text{dm}^3\ \text{mol}^{-1}$:

Understanding the Question

Methane and steam react according to the equation:

We are told that $0.80\ \text{g}$ of methane and $1.35\ \text{g}$ of steam react, and that one reactant is used up completely. We must find the volume of hydrogen produced at room conditions. The phrase "one of the reactants is used up" signals a limiting-reagent problem: we must decide which reactant runs out first and base the product amount on it.

Approach

1. Convert each reactant mass to moles using its molar mass.
2. Compare the available mole ratio with the required stoichiometric ratio ($1\ :\ 2$) to identify the limiting reagent.
3. Use the limiting reagent and the stoichiometric ratio to find the moles of hydrogen produced.
4. Convert moles of hydrogen to volume using the molar gas volume at room conditions ($24\ \text{dm}^3\ \text{mol}^{-1}$).

Step-by-Step Reasoning

Step 1 — Moles of methane.

The molar mass of methane, $\text{CH}_4$, is $12 + 4(1) = 16\ \text{g}\ \text{mol}^{-1}$.

Step 2 — Moles of steam.

The molar mass of water, $\text{H}_2\text{O}$, is $2(1) + 16 = 18\ \text{g}\ \text{mol}^{-1}$.

Step 3 — Identify the limiting reagent.

The equation shows that $1\ \text{mol}$ of methane reacts with $2\ \text{mol}$ of steam. So $0.050\ \text{mol}$ of methane would need:

Only $0.075\ \text{mol}$ of steam is available, which is less than $0.100\ \text{mol}$. Therefore steam is the limiting reagent and methane is in excess. All product amounts must be calculated from the steam.

Step 4 — Moles of hydrogen produced.

From the equation, $2\ \text{mol}$ of steam produces $4\ \text{mol}$ of hydrogen, i.e. a $1 : 2$ ratio. So:

Step 5 — Volume of hydrogen at room conditions.

Using the molar gas volume of $24\ \text{dm}^3\ \text{mol}^{-1}$:

This corresponds to option B.

Key Takeaways

• Always convert masses to moles before applying a stoichiometric ratio.
• In a limiting-reagent problem, calculate how much of each reactant is needed and identify which runs out first.
• The limiting reagent, not the excess reagent, controls the amount of product formed.
• At room conditions, one mole of any gas occupies $24\ \text{dm}^3$.

Common Mistakes

• Using methane as the limiting reagent without checking: methane needs $0.100\ \text{mol}$ of steam, but only $0.075\ \text{mol}$ is present, so steam is limiting. Using methane would give $0.050 \times 4 = 0.20\ \text{mol}$ of hydrogen and a volume of $4.8\ \text{dm}^3$ — option C, a common trap.
• Forgetting the stoichiometric ratio and assuming $1\ \text{mol}$ of steam gives $1\ \text{mol}$ of hydrogen. The ratio is $2\ \text{H}_2\text{O} : 4\ \text{H}_2$, i.e. $1 : 2$.
• Using the wrong molar gas volume: the question specifies room conditions, so $24\ \text{dm}^3\ \text{mol}^{-1}$ applies, not $22.4\ \text{dm}^3\ \text{mol}^{-1}$ (standard conditions) or $22.7\ \text{dm}^3\ \text{mol}^{-1}$.

Things to Be Careful About

• Check the units of mass carefully; here they are already in grams.
• Molar masses must be calculated correctly: $\text{CH}_4 = 16$, $\text{H}_2\text{O} = 18$.
• In the final volume, keep the correct unit ($\text{dm}^3$) and the correct number of significant figures consistent with the data given.
• The balanced equation must be read correctly: $2\ \text{mol}$ of steam produce $4\ \text{mol}$ of hydrogen, so the ratio is $1:2$, not $1:1$ or $1:4$.

Background Concept

Sodium and potassium are both Group 1 metals. In the metallic bonding model, each atom releases its outer (valence) electron into a shared pool, forming positive cations (Na⁺, K⁺) arranged in a lattice, surrounded by a mobile 'sea' of delocalised electrons. The metal is held together by the electrostatic attraction between these cations and the delocalised electrons. This attraction is what gives metals their strength, melting point, and electrical conductivity. The stronger the attraction, the more energy is needed to break the lattice apart, so the higher the melting point.

The strength of the metallic bond depends on two main factors: the charge on the cations and the size of the cations. For Group 1 metals, the charge is always +1, so the key difference is size. As you go down Group 1, atomic radius increases because each element has an additional electron shell. The larger the cation, the farther the delocalised electrons are from the nucleus, and the weaker the attraction. This explains why sodium (smaller cation) has a higher melting point than potassium (larger cation).

Understanding the Question

This is a multiple-choice question asking which statement correctly explains why sodium and potassium have different melting points. The question tests whether you understand (1) what type of bonding holds metals together, and (2) what factor determines the strength of that bond. Each option describes a different type of attraction, and you must identify which one correctly describes metallic bonding and correctly explains the difference between Na and K.

The command word is 'explain' — you need to identify the correct scientific reason, not just observe that they differ. The correct option must both describe the correct bonding model and correctly compare sodium and potassium.

Approach

1. Recognise that Na and K are metals, so their melting points are governed by metallic bonding.
2. Recall the metallic bonding model: cations + sea of delocalised electrons.
3. Identify the factor that controls metallic bond strength: the attraction between cations and delocalised electrons.
4. Compare Na and K: Na has a smaller atomic radius, so the delocalised electrons are closer to the nucleus and more strongly attracted — hence stronger metallic bond — hence higher melting point.
5. Eliminate the other options by identifying what bonding model they actually describe.

Step-by-Step Reasoning

Option A — Correct. 'The attraction between cations and delocalised electrons is stronger in sodium.' This correctly describes metallic bonding and correctly compares Na to K. Sodium has a smaller atomic radius (fewer electron shells), so its delocalised valence electrons are closer to the nucleus and more strongly attracted. The stronger metallic bond means more energy is needed to separate the ions, giving sodium a higher melting point (98°C) than potassium (63°C).

Option B — Incorrect. 'The attraction between cations and anions is stronger in sodium.' This describes ionic bonding, not metallic bonding. Metals do not contain anions — there are no negatively charged ions in a metal lattice. The delocalised electrons are not anions; they are mobile electrons.

Option C — Incorrect. 'The attraction between atoms is stronger in sodium.' This describes a molecular or covalent solid where discrete atoms are attracted to each other (e.g. by van der Waals forces). Metals are not held together by attractions between individual atoms; they are held by the attraction between cations and the delocalised electron sea.

Option D — Incorrect. 'The attraction between nuclei and shared electron pairs is stronger in sodium.' This describes covalent bonding, where two nuclei share a localised pair of electrons. In a metal, the electrons are not shared between specific pairs of atoms — they are delocalised across the entire lattice. This option confuses metallic bonding with covalent bonding.

Therefore, option A is the only statement that correctly describes metallic bonding and correctly explains the melting point difference.

Key Takeaways

• Metals are held together by metallic bonding: cations in a sea of delocalised electrons.
• The strength of the metallic bond depends on the charge of the cations and their size.
• Smaller cations → stronger attraction to delocalised electrons → higher melting point.
• Going down Group 1, atomic radius increases, metallic bond weakens, melting point decreases.
• Be able to distinguish between metallic, ionic, and covalent bonding descriptions.

Common Mistakes

• Choosing B — confusing metallic bonding with ionic bonding. Metals have no anions; the delocalised electrons are not anions.
• Choosing C — thinking of metals as atoms attracted to each other. Metals are not molecular; there are no discrete atoms bonded together.
• Choosing D — confusing metallic bonding with covalent bonding. In metals, electrons are delocalised, not shared between specific pairs of nuclei.
• Misidentifying the factor — thinking that more protons alone means stronger bonding, without considering atomic radius. Both Na and K have a +1 cation charge, so size is the deciding factor.

Things to Be Careful About

• Remember that the melting point is a macroscopic property that reflects the strength of the bonding — stronger bonds need more energy to break.
• In Group 1, the cation charge is the same (+1) for all elements, so the trend in metallic bond strength is determined by atomic/cation radius, not charge.
• Do not confuse 'delocalised electrons' with 'anions' — they are fundamentally different species.
• The term 'shared electron pairs' (option D) is a covalent bonding concept and should never be used to describe metallic bonding.

Background Concept

NH4CN is an ionic compound containing the ammonium ion, NH4+, and the cyanide ion, CN-. The ammonium ion is formed when ammonia, NH3, accepts a proton, H+. The cyanide ion has a carbon-nitrogen triple bond.

A coordinate (dative covalent) bond is a covalent bond in which both shared electrons come from one atom. In NH4+, the nitrogen atom already has a lone pair in NH3, and this lone pair is used to form the bond to H+.

A triple bond consists of one sigma bond and two pi bonds. A double bond consists of one sigma bond and one pi bond, and a single bond is one sigma bond.

Understanding the Question

The question asks for two separate counts in one formula unit of NH4CN:

1. The number of coordinate bonds.
2. The number of pi bonds.

The formula NH4CN must first be recognised as containing two separate ions: NH4+ and CN-. The bonds inside each ion must then be counted.

Approach

1. Identify the ions present in NH4CN.
2. Analyse the bonding in NH4+.
3. Analyse the bonding in CN-.
4. Add the counts for coordinate bonds and pi bonds.

Step-by-Step Reasoning

1. NH4CN dissociates into NH4+ and CN-.
2. In NH4+, nitrogen forms four N-H bonds. Three of these come from the three unpaired electrons of nitrogen combining with three hydrogen atoms. The fourth bond is formed using nitrogen's lone pair, so it is a coordinate bond.
   • Coordinate bonds in NH4+ = 1.
3. In CN-, carbon and nitrogen share three pairs of electrons, forming a triple bond.
   • A triple bond has one sigma bond and two pi bonds.
   • Pi bonds in CN- = 2.
4. Total coordinate bonds = 1; total pi bonds = 2.

This matches option B.

Key Takeaways

• NH4+ always contains one coordinate bond because NH3 donates its lone pair to H+.
• A triple bond always contributes two pi bonds.
• When counting bonds in an ionic compound, first identify the ions present, then analyse the bonding within each ion.

Common Mistakes

• Counting all four N-H bonds in NH4+ as ordinary covalent bonds, forgetting that one is dative.
• Counting the C≡N triple bond as three pi bonds instead of one sigma and two pi bonds.
• Treating NH4CN as a molecule with direct N-C bonding rather than as separate NH4+ and CN- ions.

Things to Be Careful About

• The word "coordinate" means the same as "dative covalent".
• A coordinate bond is still a covalent bond; it is counted separately only because both electrons come from one atom.
• The pi bond count depends on the bond order: single = 0 pi, double = 1 pi, triple = 2 pi.

Background Concept

In any chemical reaction, bonds in the reactants must first be broken and new bonds must then be formed in the products. Bond breaking requires energy input, so it is endothermic. Bond forming releases energy, so it is exothermic. Whether the overall reaction is exothermic or endothermic depends on the balance between these two energy changes:

If more energy is released in forming bonds than is absorbed in breaking bonds, the reaction is exothermic and $\Delta H$ is negative. If more energy is absorbed in breaking bonds than is released in forming bonds, the reaction is endothermic and $\Delta H$ is positive.

Understanding the Question

The question describes the neutralisation of ethanoic acid with an excess of aqueous sodium hydroxide, releasing $55\text{ kJ mol}^{-1}$ of energy. It asks which statement correctly explains why this reaction is exothermic. The question is not asking for the enthalpy value itself; it is testing the general principle of why reactions release energy in terms of bond breaking and bond forming.

Approach

The key is to recall two facts:

1. Breaking bonds always absorbs energy.
2. Forming bonds always releases energy.

Then decide which energy change must be larger for the overall process to release energy. The exothermic condition is that the energy released on forming product bonds exceeds the energy needed to break reactant bonds. Apply this to each option and select the one that states, in the correct order, that bond forming releases more energy than bond breaking absorbs.

Step-by-Step Reasoning

• Bond breaking: energy is taken in. So it can never be described as giving out energy.
• Bond forming: energy is given out. So an exothermic reaction must involve more energy given out in bond forming than taken in in bond breaking.

Now examine the options:

• A: It says only bond breaking takes place. This is chemically wrong: products must form bonds too, and bond breaking alone would absorb energy, not release it.
• B: It says only bond forming takes place. This is also incomplete: reactant bonds must be broken before new bonds can form.
• C: It says more energy is given out in breaking bonds than is taken in to form bonds. This reverses the endothermic/exothermic nature of both processes. Bond breaking does not give out energy, so this is incorrect.
• D: It says more energy is given out in forming bonds than is taken in to break bonds. This is the correct statement: bond forming releases energy, and if this release is greater than the energy absorbed in breaking bonds, the overall reaction is exothermic.

In the neutralisation reaction, new bonds such as O–H bonds in water are formed, and this energy release outweighs the energy needed to break the existing bonds in the acid, base, and water.

Key Takeaways

• Bond breaking is endothermic; bond forming is exothermic.
• A reaction is exothermic when the energy released in forming new bonds is greater than the energy absorbed in breaking old bonds.
• A reaction is endothermic when the energy absorbed in breaking old bonds is greater than the energy released in forming new bonds.
• When reading options that compare energy changes, pay attention to the direction: breaking bonds takes energy in, forming bonds gives energy out.

Common Mistakes

• Choosing C because it sounds like a plausible energy comparison, but it reverses the roles of bond breaking and bond forming.
• Assuming that bond breaking can release energy because an overall reaction is exothermic. Bond breaking always requires energy; the exothermicity comes from bond forming.
• Thinking that only bond forming matters, or only bond breaking matters. Both must be considered; the sign of $\Delta H$ is determined by their relative sizes.

Things to Be Careful About

• The phrase “given out” must be paired with bond forming, and “taken in” must be paired with bond breaking.
• The enthalpy change of neutralisation is negative, so the reaction is exothermic. The magnitude, $55\text{ kJ mol}^{-1}$, is not used to choose the option, but it confirms that neutralisation is an exothermic process.
• Make sure the comparison in the answer is between “energy released in forming bonds” and “energy absorbed in breaking bonds”, not the reverse.

Background Concept

When a fuel burns, the chemical energy stored in its bonds is released as heat. In a simple calorimetry experiment the heat released is transferred to a known mass of water, and the temperature rise of the water is measured. The heat gained by the water is given by

where $m$ is the mass of water, $c$ is its specific heat capacity ($4.18\text{ J g}^{-1}\text{K}^{-1}$ for water) and $\Delta T$ is the temperature rise. The experiment is not perfectly efficient: only 45.0% of the energy released by the fuel is absorbed by the water. Therefore the measured $q$ is only 45.0% of the total energy released. The question also asks for the energy per gram of fuel, so after finding the total energy we divide by the mass of fuel burnt.

Understanding the Question

This is a one-mark multiple-choice question about calorimetry and energy efficiency. The stem gives:

• mass of fuel = $1.60\text{ g}$
• mass of water = $200\text{ g}$
• temperature rise from $18.0\,^\circ\text{C}$ to $66.0\,^\circ\text{C}$
• only 45.0% of the energy released is absorbed by the water.

The command is a calculation: find the total energy released per gram of fuel burnt. We must not stop at the energy absorbed by the water, nor at the total energy released; we must also divide by the mass of fuel.

Approach

1. Calculate the heat absorbed by the water using $q = mc\Delta T$.
2. Recognise that this heat is 45.0% of the total energy released, so divide $q$ by 0.450 to obtain the total energy released by the fuel.
3. Divide the total energy by the mass of fuel ($1.60\text{ g}$) to obtain the energy released per gram.
4. Compare with the options.

Step-by-Step Reasoning

First find the temperature rise:

Then calculate the heat absorbed by the water:

This is the energy that actually warmed the water. Since this is 45.0% of the total energy released by the fuel:

Finally, express this per gram of fuel:

So the correct option is B.

Why the other options are wrong:

• A, $25\,100\text{ J}$, is the heat absorbed by the water divided by the mass of fuel, i.e. it uses the absorbed energy instead of the total energy.
• C, $89\,200\text{ J}$, is the total energy released by the whole sample, but the question asks for the energy per gram.
• D, $143\,000\text{ J}$, comes from incorrectly multiplying the total energy by 1.60 (or otherwise mishandling the percentage/mass conversion).

Key Takeaways

• The heat gained by water in a calorimetry experiment is $q = mc\Delta T$.
• A percentage efficiency means the measured heat is a fraction of the total energy released; to recover the total, divide by the fraction as a decimal.
• Always check the final unit requested: here "per gram" requires dividing by the mass of fuel.
• A temperature difference has the same numerical value in kelvin and in degrees Celsius.

Common Mistakes

• Choosing C because it is the total energy released by the whole sample, not the energy per gram.
• Choosing A because the absorbed energy is divided by 1.60 before converting to total energy.
• Choosing D because the total energy is multiplied by 1.60 or the percentage is applied incorrectly.
• Using 45 instead of 0.450 in the division, or multiplying by 0.45 instead of dividing by it.
• Forgetting that $\Delta T$ is a difference, so using $18.0$ or $66.0$ alone is wrong.

Things to Be Careful About

• Use $c = 4.18\text{ J g}^{-1}\text{K}^{-1}$ for water unless another value is given.
• Convert 45.0% to 0.450 before using it in a calculation.
• Keep track of units: the final answer is in $\text{J g}^{-1}$, not just J.
• The temperature difference in kelvin is numerically equal to the difference in degrees Celsius, so no extra conversion is needed for $\Delta T$.
• Read the question wording carefully: "per gram of fuel" means the total energy must be divided by 1.60 g.

Background Concept

Redox reactions involve electron transfer. To balance a redox equation, split it into oxidation and reduction half-equations. First assign oxidation numbers to every element. Oxidation number is the charge an atom would have if the electrons in each bond were assigned to the more electronegative atom; in an ion, the sum of the oxidation numbers equals the charge on the ion.

In this question:

• S in $\text{SO}_3^{2-}$: O is $-2$, three O give $-6$, ion charge is $-2$, so S is $+4$.
• S in $\text{S}_2\text{O}_4^{2-}$: four O give $-8$, ion charge $-2$, so two S total $+6$, each S is $+3$.
• S in $\text{S}_2\text{O}_6^{2-}$: six O give $-12$, ion charge $-2$, so two S total $+10$, each S is $+5$.
• Zn metal is 0; $\text{Zn}^{2+}$ is $+2$.
• Mn in $\text{MnO}_2$ is $+4$; $\text{Mn}^{2+}$ is $+2$.

Once the oxidation number changes are known, write half-equations. Balance atoms other than O and H, then balance O and H using the species appropriate to the medium: in acid use $\text{H}_2\text{O}$ and $\text{H}^+$; in alkaline solution use $\text{OH}^-$ and $\text{H}_2\text{O}$. Add electrons to balance charge, then multiply half-equations so that electrons cancel.

Understanding the Question

The question gives two skeleton equations with unknown whole-number coefficients. The first is the reduction of sulfite ions by zinc metal, producing hydroxide ions, so it is balanced in alkaline conditions. The second is the oxidation of sulfite ions by manganese dioxide in acidic conditions, with $\text{H}^+$ as a reactant. You are asked for the coefficients $a$, $b$, $w$ and $x$. The four options give different combinations, so the task is to balance both equations fully.

Approach

For each equation, identify which element is oxidised and which is reduced. Then write the two half-equations, balance atoms and charge, equalise the number of electrons, and add the half-equations. Finally read off the required coefficients.

For the first equation, Zn is oxidised and S in sulfite is reduced. For the second, Mn is reduced and S in sulfite is oxidised. In both cases the sulfur product contains two sulfur atoms, so two sulfite ions are needed for each product ion.

Step-by-Step Reasoning

First equation

Oxidation half-equation:

Reduction half-equation, in alkaline conditions:

Check the reduction half-equation:

• S: 2 on each side.
• O: $6 + 2 = 8$ on the left; $4 + 4 = 8$ on the right.
• H: 4 on each side.
• Charge: left $-4 + 0 - 2 = -6$; right $-2 - 4 = -6$.

Adding the two half-equations cancels the electrons:

So $a = 2$, $b = 2$, $c = 4$, $d = 1$. This eliminates options A and C, which have $a = 1$.

Second equation

Reduction half-equation, in acid:

Check: Mn 1 = 1; O 2 = 2; H 4 = 4; charge $+4 - 2 = +2$ on the left and $+2$ on the right.

Oxidation half-equation:

Check: S 2 = 2; O 6 = 6; charge $-4 = -2 - 2$.

Adding:

So $w = 2$, $x = 4$, $y = 1$, $z = 2$.

Therefore the correct option is B: $a = 2$, $b = 2$, $w = 2$, $x = 4$.

Key Takeaways

• Oxidation number changes identify which species is oxidised and which is reduced.
• Half-equations must balance both atoms and charge.
• The medium matters: use $\text{H}^+$ and $\text{H}_2\text{O}$ in acid, and $\text{OH}^-$ and $\text{H}_2\text{O}$ in alkali.
• The coefficients in a redox equation come from the electron balance, not from trial and error.

Common Mistakes

• Forgetting that $\text{S}_2\text{O}_4^{2-}$ and $\text{S}_2\text{O}_6^{2-}$ each contain two sulfur atoms, so two sulfite ions are needed per product ion.
• Using $\text{H}^+$ in the first equation, which is alkaline; the correct species is $\text{OH}^-$.
• Balancing oxygen with water but forgetting to balance hydrogen afterwards.
• Not checking that charge is balanced in each half-equation.
• Choosing coefficients by inspection instead of using half-equations.

Things to Be Careful About

• The oxidation number of S is $+4$ in $\text{SO}_3^{2-}$, $+3$ in $\text{S}_2\text{O}_4^{2-}$, and $+5$ in $\text{S}_2\text{O}_6^{2-}$.
• In acidic half-equations, add $\text{H}^+$ and $\text{H}_2\text{O}$; in alkaline half-equations, add $\text{OH}^-$ and $\text{H}_2\text{O}$.
• Electrons must cancel exactly when the half-equations are added.
• State symbols and ionic charges are part of a correctly balanced equation.
• The correct option is B: $a = 2$, $b = 2$, $w = 2$, $x = 4$.

Background Concept

An oxidising agent is a substance that oxidises another species, meaning it causes another species to lose electrons. In doing so, the oxidising agent itself gains electrons and is reduced. Conversely, a reducing agent causes another species to gain electrons and is itself oxidised.

The key tool for identifying redox behaviour is the oxidation number (oxidation state). For an element in its elemental form, the oxidation number is 0. For hydrogen in most compounds, it is +1, but in metal hydrides (like KH, NaH), hydrogen is -1 because the metal is more electropositive (less electronegative) than hydrogen.

To identify whether hydrogen acts as an oxidising agent, we track its oxidation number: if it decreases from 0 to -1, hydrogen is reduced and acts as an oxidising agent; if it increases from 0 to +1, hydrogen is oxidised and acts as a reducing agent.

Understanding the Question

The question presents four chemical equations and asks us to identify the one in which hydrogen acts as an oxidising agent. This requires us to determine, for each reaction, whether hydrogen is reduced (gaining electrons) or oxidised (losing electrons). The correct option is the one where hydrogen's oxidation number decreases.

Approach

1. Assign oxidation numbers to hydrogen in the reactants and products of each equation.
2. Determine whether hydrogen's oxidation number increases (oxidised → reducing agent) or decreases (reduced → oxidising agent).
3. Select the equation where hydrogen is reduced.

Step-by-Step Reasoning

Let's go through each option:

Option A: $\text{H}_2 + 2\text{K} \rightarrow 2\text{K}^+\text{H}^-$

• In $\text{H}_2$, hydrogen has oxidation number 0.
• In $\text{K}^+\text{H}^-$, hydrogen has oxidation number -1 (potassium is a metal, more electropositive than hydrogen, so hydrogen takes the negative oxidation state as hydride, H^-).
• Hydrogen goes from 0 to -1: it gains an electron, so it is reduced. Potassium goes from 0 to +1: it loses an electron, so it is oxidised.
• Since hydrogen is reduced, it acts as the oxidising agent. ✓

Option B: $\text{H}_2 + \text{I}_2 \rightarrow 2\text{HI}$

• In $\text{H}_2$, hydrogen has oxidation number 0.
• In $\text{HI}$, hydrogen has oxidation number +1 (iodine is more electronegative than hydrogen).
• Hydrogen goes from 0 to +1: it loses an electron, so it is oxidised. Iodine goes from 0 to -1: it gains electrons, so it is reduced.
• Hydrogen is oxidised, so it acts as the reducing agent, not the oxidising agent. ✗

Option C: $\text{H}_2 + \text{Cu}_2\text{S} + \text{ZnO} \rightarrow 2\text{Cu} + \text{ZnS} + \text{H}_2\text{O}$

• In $\text{H}_2$, hydrogen has oxidation number 0.
• In $\text{H}_2\text{O}$, hydrogen has oxidation number +1.
• Hydrogen goes from 0 to +1: it is oxidised, so it acts as the reducing agent. ✗
• (Copper in Cu2S: S is -2, so Cu is +1; in elemental Cu, it's 0 — copper is reduced.)

Option D: $3\text{H}_2 + \text{N}_2 \rightarrow 2\text{NH}_3$

• In $\text{H}_2$, hydrogen has oxidation number 0.
• In $\text{NH}_3$, hydrogen has oxidation number +1 (nitrogen is more electronegative than hydrogen).
• Hydrogen goes from 0 to +1: it is oxidised, so it acts as the reducing agent. ✗

Therefore, only option A shows hydrogen acting as an oxidising agent.

Key Takeaways

• An oxidising agent is itself reduced (gains electrons, oxidation number decreases).
• A reducing agent is itself oxidised (loses electrons, oxidation number increases).
• Hydrogen's oxidation number is 0 in the elemental form, +1 in most compounds, and -1 in metal hydrides.
• To determine whether a species acts as an oxidising or reducing agent, track its oxidation number change.

Common Mistakes

• Confusing oxidising and reducing agents: an oxidising agent is reduced, not oxidised.
• Forgetting that hydrogen can have oxidation number -1 in metal hydrides. In KH, hydrogen is -1, not +1.
• Assuming hydrogen always has oxidation number +1 in compounds — this is true for most compounds but not for metal hydrides.

Things to Be Careful About

• Always assign oxidation numbers carefully, considering electronegativity. In KH (a metal hydride), the metal is less electronegative than hydrogen, so hydrogen carries the negative oxidation state.
• Remember that the oxidising agent is the species that is reduced (gains electrons), and the reducing agent is the species that is oxidised (loses electrons).
• In option C, note that hydrogen is oxidised, so hydrogen acts as the reducing agent, and copper is reduced, so Cu2S (or more precisely, Cu+ in Cu2S) acts as the oxidising agent.

Background Concept

The ideal gas equation, $pV = nRT$, links the pressure, volume, amount and temperature of an ideal gas. Here $p$ is pressure in Pa, $V$ is volume in $\text{m}^3$, $n$ is amount in mol, $T$ is temperature in K, and $R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$. Nitrogen exists as diatomic molecules, $\text{N}_2$, so its molar mass is $2 \times 14.0 = 28.0\text{ g mol}^{-1}$.

Understanding the Question

This is a one-mark multiple-choice question asking for the pressure exerted by a known mass of nitrogen in a fixed volume at a given temperature. The data are given in convenient laboratory units: grams, $\text{dm}^3$ and Celsius. To use the ideal gas equation directly, all quantities must be in SI base units: moles, $\text{m}^3$ and kelvin.

Approach

1. Convert the mass of nitrogen into moles using $n = m/M_r$.
2. Convert the volume from $\text{dm}^3$ to $\text{m}^3$.
3. Convert the temperature from Celsius to kelvin.
4. Rearrange $pV = nRT$ to $p = nRT/V$ and substitute.

Step-by-Step Reasoning

Moles of nitrogen.

Using $M_r = 28.0$ is essential because nitrogen gas is $\text{N}_2$, not N.

Volume in SI units.

$1\text{ dm}^3 = 10^{-3}\text{ m}^3$

so

$V = 2.00\text{ dm}^3 = 2.00 \times 10^{-3}\text{ m}^3$

Temperature in kelvin.

$T = 40.0 + 273.15 = 313.15\text{ K}$

Pressure.

The numerator is about $92.9\text{ J}$ (since $\text{Pa m}^3 = \text{J}$), and dividing by $2.00 \times 10^{-3}\text{ m}^3$ gives

$p \approx 4.64 \times 10^4\text{ Pa} = 46400\text{ Pa}$

This matches option C.

Option D is double this value, which is what you get if you use $M_r = 14.0\text{ g mol}^{-1}$ (atomic nitrogen) instead of $28.0$. Option A is close to the result obtained by using $T = 40\text{ K}$ instead of $313.15\text{ K}$. Option B is an arithmetic or unit slip. The correct answer is therefore C.

Key Takeaways

• Always use SI units in $pV = nRT$: pressure in Pa, volume in $\text{m}^3$, temperature in K.
• Remember common gases such as $\text{N}_2$, $\text{O}_2$, $\text{H}_2$ and $\text{Cl}_2$ are diatomic; their molar mass is twice the atomic mass.
• A quick magnitude check helps: about 0.036 mol of gas at 313 K in 2 L gives a pressure of tens of thousands of Pa, not thousands.

Common Mistakes

• Using $M_r = 14.0$ for nitrogen instead of $28.0$, giving option D.
• Leaving the temperature in Celsius, giving a much smaller pressure (option A).
• Forgetting to convert $\text{dm}^3$ to $\text{m}^3$; this makes the calculated pressure 1000 times too large.
• Mixing up units of $R$; $R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$ only works with Pa, $\text{m}^3$, mol and K.

Things to Be Careful About

• $2.00\text{ dm}^3 = 2.00 \times 10^{-3}\text{ m}^3$, not $0.200\text{ m}^3$.
• Add 273.15 (or at least 273) to Celsius temperatures; never use the Celsius value directly.
• Give the final answer to three significant figures, matching the data.
• The unit is Pa; if you accidentally work in kPa, the numerical value would be 46.4, not 46400.

Background Concept

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

at a fixed temperature. Each concentration is raised to the power of its stoichiometric coefficient in the balanced equation. Here, the coefficient 4 in front of $\text{O}_2$ means $[\text{O}_2]$ appears raised to the fourth power — a very common trap. Since $[\text{O}_2] = 7.6 \times 10^{-6}\,\text{mol dm}^{-3}$ is small, $[\text{O}_2]^4$ is astronomically smaller (about $3.3 \times 10^{-21}$), so $K_c$ becomes very large, telling us this equilibrium strongly favours the bound form $\text{Hb}(\text{O}_2)_4$.

Understanding the Question

The stem gives the equilibrium $\text{Hb}(\text{aq}) + 4\text{O}_2(\text{aq}) \rightleftharpoons \text{Hb}(\text{O}_2)_4(\text{aq})$ and states that at equilibrium $[\text{O}_2] = 7.6 \times 10^{-6}\,\text{mol dm}^{-3}$ and that $[\text{Hb}] = [\text{Hb}(\text{O}_2)_4]$. The task is to find the numerical value of $K_c$. The equal concentrations are deliberately provided so that they cancel from the expression — the answer depends only on the oxygen concentration. Recognising this cancellation is the heart of the question.

Approach

Write the $K_c$ expression; replace the two equal concentrations with the same symbol and cancel; substitute the given $[\text{O}_2]$; compute the fourth power and take the reciprocal; compare with the options. As a sanity check, because the product is strongly favoured at these tiny concentrations, $K_c$ should be far greater than 1 — only option A satisfies this.

Step-by-Step Reasoning

1. Write the expression.

1. Cancel the equal concentrations. Let $[\text{Hb}] = [\text{Hb}(\text{O}_2)_4] = x$. Then

1. Substitute the oxygen concentration.

1. Evaluate the fourth power.

1. Take the reciprocal.

This matches option A.

Now the distractors: option B, $1.3 \times 10^{5}$, is $1/[\text{O}_2]$ — what you get by forgetting the fourth power. Option C, $7.6 \times 10^{-6}$, is just the given $[\text{O}_2]$ itself. Option D, $3.3 \times 10^{-21}$, is precisely $[\text{O}_2]^4$ — the denominator that must be inverted, not quoted as $K_c$.

Key Takeaways

• Stoichiometric coefficients become exponents in $K_c$ expressions — the 4 in front of $\text{O}_2$ means $[\text{O}_2]^4$, not $4[\text{O}_2]$.
• Where two concentrations are stated to be equal, they cancel from a ratio — use that before substituting numbers.
• Magnitude checks are powerful: a tiny reactant concentration raised to a power gives an enormous $K_c$ when products are favoured.
• Always compare your computed value with the options to catch arithmetic slips.

Common Mistakes

• Forgetting the fourth power: using $[\text{O}_2]^1$ gives $1.3 \times 10^{5}$ (option B).
• Quoting the fourth-power value itself, $3.3 \times 10^{-21}$ (option D), instead of its reciprocal.
• Misidentifying which species belong in the numerator and denominator of $K_c$ (products over reactants).
• Units of $K_c$: this reaction has $\Delta n = 1 - 1 - 4 = -4$, so the units would be $(\text{mol dm}^{-3})^{-4}$; the question asks only the numerical value, so no units are needed.