Chemistry 9701/11 — October/November 2025

Background Concept

Electronic configuration describes how electrons are arranged in orbitals around the nucleus. For an ion, the number of electrons differs from the neutral atom: a positive ion has lost electrons, while a negative ion has gained electrons. The proton number (atomic number) never changes — only the electron count changes.

Understanding the Question

The particle is written as $^{27}_{13}\text{Al}^+$. The superscript 27 is the mass number (protons + neutrons), and the subscript 13 is the proton number (number of protons). The + charge means one electron has been removed from the neutral atom. So we must write the configuration for the 12 electrons that remain.

Approach

1. Identify the number of electrons in the neutral atom — this equals the proton number.
2. Adjust for the charge: subtract electrons for a positive charge, add electrons for a negative charge.
3. Write the configuration in order of increasing orbital energy.

Step-by-Step Reasoning

• Neutral Al has 13 electrons: $1s^2 2s^2 2p^6 3s^2 3p^1$.
• For Al⁺, remove one electron. Electrons are removed from the highest-energy occupied orbital first, which is 3p.
• Removing that 3p electron leaves: $1s^2 2s^2 2p^6 3s^2$.
• Check the total: 2 + 2 + 6 + 2 = 12 electrons. ✓

Key Takeaways

• For positive ions, remove electrons from the highest-energy occupied orbital (the outermost shell).
• The proton number stays the same; only the electron count changes.
• Always verify the total electron count after writing the configuration.

Common Mistakes

• Confusing the mass number (27) with the proton number (13).
• Removing an electron from an inner shell (e.g., 2p) instead of the valence shell (3p).
• Counting 13 electrons for Al⁺ instead of 12.

Things to Be Careful About

The charge tells you how many electrons were lost (positive) or gained (negative). Also remember that 3p is higher in energy than 3s, so it is emptied first when forming a positive ion.

Background Concept

Successive ionisation energies are the energies needed to remove, one by one, each electron from a gaseous atom or ion. Removing the first electron is the first ionisation energy; removing the second is the second, and so on.

Electrons are arranged in shells. Electrons in the same shell have broadly similar energies, so a small, steady increase is seen as successive electrons are removed from the same shell (each removal leaves a more positive ion, so the next electron is held a little more tightly). However, once all the valence electrons of the outermost shell have been removed, the next electron must come from the next shell in — the noble-gas core — which is at a much lower energy level and much closer to the nucleus. Removing that electron requires a dramatically larger amount of energy, producing a characteristically large jump in the succession of ionisation energies.

The position of this jump tells you how many electrons were in the outermost shell: for these Period 3 elements (Na through Ar), the number of electrons in the outermost shell equals the group number. For example, sulfur (group 16) has 6 valence electrons, so the 6th electron is removed reasonably easily, but the 7th comes from the inner shells — hence a jump between IE6 and IE7.

Understanding the Question

This question presents a table of the 5th to the 10th ionisation energies for three Period 3 elements, labelled X, Y and Z. The values are in kJ mol⁻¹. You must decide which element each of X, Y and Z is, choosing from the four answer options.

The data only show ionisation energies from the 5th to the 10th, so you cannot see jumps at the very start of the sequence (e.g. between IE1 and IE2, which would distinguish Na from Mg). Instead, you must recognise that the jumps visible in this window reveal how many electrons belong to the valence shell of each element.

Approach

For each element in turn, look down its column of ionisation energies and find where a large increase occurs between two consecutive values. The position of that leap indicates the end of the valence shell: if the jump is between the nth and the (n+1)th ionisation energy, the element has n valence electrons. Since all three elements are in Period 3, the number of valence electrons equals the group number, and the element can be named directly.

The steady small increases within the same shell do not matter — only the large leaps identify the shell boundary.

Step-by-Step Reasoning

Element X:
the values rise steadily from 6274 to 21269? Let's check carefully. The 5th value is 6274, the 6th is 21269. The jump from 6274 to 21269 is enormous (roughly trebling), while the following values (25398, 29855, 35868, 40960) rise only modestly. The large jump therefore occurs between the 5th and the 6th ionisation energises. This means the 5 valence electrons have been removed, and the 6th electron must come from an inner shell. So X has 5 valence electrons, placing it in group 15 — phosphorus, P.

Element Y:
the 5th, 6th, and 7th values are 7012, 8496 and 27107. The increase from 7012 to 8496 is small — both electrons come from the same (outermost) shell. The jump between the 6th and the 7th (8496 → 27107) is large, so the 7th electron is being removed from the inner core. Thus Y has 6 valence electrons, placing it in group 16 — sulfur, S.

Element Z:
the values rise steadily and slowly through 6542, 9362, 11018, then leap to 33606 between the 7th and the 8th values. So the 8th electron is the first one pulled from the inner shells, meaning Z has 7 valence electrons and is in group 17 — chlorine, Cl.

Matching these to the options:

• X = P
• Y = S
• Z = Cl

This corresponds to option C.

Key Takeaways

• A large jump between successive ionisation energies signals the boundary between the valence shell and the inner (core) shell.
• The position of the jump reveals the number of valence electrons, and hence the group of the element.
• Within the same shell, successive ionisation energies increase only slowly; the inner electrons demand far higher energies because they are closer to the nucleus and less shielded.
• In an MCQ, the same reasoning is applied repeatedly — identify each element from its own jump position before looking at the option list.

Common Mistakes

• Misidentifying where the jump occurs. For X, the jump is between the 5th and 6th values; for Y between the 6th and 7th; for Z between the 7th and 8th. Counting the electron number of rather than the boundary position causes wrong choices.
• Confusing the electron being removed with the number of electrons removed so far. The nth electron removed is drawn from the ion that already lost n−1 electrons; the jump tells you the identity of the electron being removed, which reveals the count of valence electrons.
• Forgetting that the elements are Period 3, so the count of valence electrons exactly equals the group number — if the elements were in another period, the same reasoning still works, but the group mapping is unchanged.n
• Picking options that pair the elements in the wrong order (e.g. swapping P and S). Checking each jump independently prevents this.

Things to Be Careful About

• The unit is kJ mol⁻¹; the absolute values are not needed, only their relative sizes.
• Ensure you compare consecutive pairs only — every element in the table shows a large jump somewhere, and it is the location that matters.
• The question asks for the identity of all three elements; the correct option must satisfy all three letters simultaneously.
• There is no need to calculate anything — this is a pattern-recognition question based on the trend in successive ionisation energies.
• In an MCQ, reasoning one element at a time and checking against the options is the most reliable approach; a single correct element often narrows the choice to one or two options.

Background Concept

The mole is the SI unit for the amount of substance. One mole contains exactly the Avogadro constant, $6.02 \times 10^{23}$, of particles (atoms, molecules, ions, or electrons). The relationship is:

When a question asks how many atoms of a particular element are present in a given amount of a compound, you must multiply the amount of compound by the subscript of that element in the chemical formula. For example, one formula unit of $\text{Al}_2\text{O}_3$ contains three O atoms, so 1 mol of $\text{Al}_2\text{O}_3$ contains 3 mol of O atoms.

Understanding the Question

The question gives a specific number of oxygen atoms, $9.03 \times 10^{23}$, and asks which of four amounts of compounds contains exactly that many oxygen atoms. The key steps are: (1) convert the given number of atoms into moles of atoms using the Avogadro constant, and (2) check each option by multiplying the amount of compound by the number of O atoms per formula unit.

Approach

1. Convert the target number of O atoms to moles of O atoms.
2. For each option, compute the moles of O atoms = (moles of compound) $\times$ (subscript of O in the formula).
3. Compare each result with the target and select the matching option.

Step-by-Step Reasoning

Step 1: Convert the target.

Step 2: Evaluate each option.

• Option A: $\text{Al}_2\text{O}_3$ contains 3 O atoms per formula unit. $0.25 \times 3 = 0.75$ mol O atoms. This is $0.75 \times 6.02 \times 10^{23} = 4.52 \times 10^{23}$ O atoms — too few.
• Option B: $\text{SO}_2$ contains 2 O atoms per molecule. $0.75 \times 2 = 1.5$ mol O atoms. This is $1.5 \times 6.02 \times 10^{23} = 9.03 \times 10^{23}$ O atoms — exactly the target.
• Option C: $\text{SO}_3$ contains 3 O atoms per molecule. $1.5 \times 3 = 4.5$ mol O atoms. This is $4.5 \times 6.02 \times 10^{23} = 2.71 \times 10^{24}$ O atoms — too many.
• Option D: $\text{H}_2\text{O}$ contains 1 O atom per molecule. $3.0 \times 1 = 3.0$ mol O atoms. This is $3.0 \times 6.02 \times 10^{23} = 1.81 \times 10^{24}$ O atoms — too many.

Only option B matches the target of $9.03 \times 10^{23}$ O atoms.

Key Takeaways

• The Avogadro constant links the number of particles to the amount in moles.
• Always multiply the amount of compound by the subscript of the element in the formula to find the amount of that element's atoms.
• The value $9.03 \times 10^{23}$ is exactly $1.5 \times 6.02 \times 10^{23}$, a common exam-friendly value.

Common Mistakes

• Forgetting to multiply by the subscript: e.g., taking 0.75 mol $\text{SO}_2$ as 0.75 mol of O atoms instead of $0.75 \times 2 = 1.5$ mol.
• Confusing $\text{SO}_2$ (2 O atoms) with $\text{SO}_3$ (3 O atoms).
• Using the wrong value of the Avogadro constant or dividing instead of multiplying.

Things to Be Careful About

• Read each formula carefully and note the subscript of oxygen.
• Keep track of units: "mol of compound" vs "mol of atoms".
• The answer must be expressed as the option letter, not the amount.

Background Concept

This question tests stoichiometry: converting masses to moles, identifying the limiting reagent, and using the molar gas volume at room temperature and pressure (24 dm³ mol⁻¹).

Understanding the Question

Methane and steam react in a 1:2 mole ratio. Masses of both reactants are given, so one is in excess and one is used up (the limiting reagent). We must find which reactant is limiting, then calculate the volume of hydrogen produced.

Approach

1. Convert both masses to moles.
2. Compare the available mole ratio with the stoichiometric ratio to find the limiting reactant.
3. Use the limiting reactant to calculate moles of H₂.
4. Convert moles of H₂ to volume using the molar volume 24 dm³ mol⁻¹.

Step-by-Step Reasoning

1. Moles of CH₄ = 0.80 / 16 = 0.050 mol
2. Moles of H₂O = 1.35 / 18 = 0.075 mol
3. The reaction needs 2 mol H₂O per 1 mol CH₄. 0.050 mol CH₄ needs 0.100 mol H₂O. Only 0.075 mol is present, so H₂O is the limiting reactant.
4. Ratio H₂O : H₂ = 2 : 4 = 1 : 2, so 0.075 mol H₂O gives 0.150 mol H₂.
5. Volume = 0.150 × 24 = 3.60 dm³, which is option B.

Key Takeaways

• Always identify the limiting reagent when masses of both reactants are given.
• Molar volume at room conditions is 24 dm³ mol⁻¹ (25°C, 1 atm).
• Use the stoichiometric ratio from the balanced equation.

Common Mistakes

• Assuming methane is limiting just because it has the smaller mass.
• Using the wrong mole ratio (e.g., 4 mol H₂ per 1 mol CH₄ without checking the limiting reagent).
• Using 22.4 dm³ mol⁻¹ (STP) instead of 24 dm³ mol⁻¹ (room conditions).

Things to Be Careful About

• Molar masses: CH₄ = 16 g mol⁻¹, H₂O = 18 g mol⁻¹.
• The ratio H₂O : H₂ is 1 : 2, not 1 : 4.
• Room conditions vs STP molar volumes.

Background Concept

Metals consist of a giant lattice of positive ions surrounded by a sea of delocalised electrons. The metallic bond is the electrostatic attraction between these positive cations and the delocalised electrons. The strength of this attraction depends on the charge on the cation and its size: a smaller cation with the same charge attracts the delocalised electrons more strongly.

Understanding the Question

The question asks why sodium and potassium have different melting points. Both are Group 1 metals, so both have the same type of metallic bonding and the same cation charge (+1). The difference must come from the strength of the metallic bond, which is controlled by the size of the cation.

Approach

Compare the two metals in terms of the metallic bonding model. Identify which factor differs between sodium and potassium, then decide which option correctly describes the attraction responsible for metallic bonding.

Step-by-Step Reasoning

1. Sodium and potassium are both metals, so their atoms lose one outer electron to form +1 cations and a sea of delocalised electrons.
2. Sodium is above potassium in Group 1, so the sodium cation has a smaller ionic radius than the potassium cation.
3. With the same charge but a smaller radius, the sodium cation has a higher charge density and attracts the delocalised electrons more strongly.
4. A stronger metallic bond means more energy is needed to break the lattice, so sodium has a higher melting point than potassium.
5. Option A correctly identifies this attraction between cations and delocalised electrons.

Key Takeaways

• Metallic bonding is the attraction between cations and delocalised electrons, not between atoms or between cations and anions.
• Down Group 1, the cation becomes larger, the metallic bond weakens, and melting points decrease.
• Charge density is key: smaller ions with the same charge form stronger metallic bonds.

Common Mistakes

• Choosing B: this describes ionic bonding between cations and anions, which does not apply to metals.
• Choosing C: metals do not contain discrete atoms held together by interatomic attractions; the metallic bond involves cations and delocalised electrons.
• Choosing D: this describes covalent bonding, where nuclei are attracted to shared electron pairs, not metallic bonding.

Things to Be Careful About

• Use the correct metallic bonding model: cations plus delocalised electrons.
• Remember that sodium and potassium have the same cation charge, so the deciding factor is cation size.
• Melting point trends in Group 1 are the opposite of trends in some other groups because the metallic bond weakens as the cation gets larger.

Background Concept

A coordinate (dative) bond is a covalent bond in which the shared pair of electrons is supplied by only one of the bonded atoms. It is often formed when a lone pair on one atom is donated to an atom or ion with an empty orbital, as when $\text{NH}_3$ accepts $\text{H}^+$ to form $\text{NH}_4^+$.

A pi bond is formed by sideways overlap of p orbitals above and below the internuclear axis. A single bond is one sigma bond; a double bond is one sigma + one pi bond; a triple bond is one sigma + two pi bonds.

$\text{NH}_4\text{CN}$ is ionic and is composed of one $\text{NH}_4^+$ ion and one $\text{CN}^-$ ion, so the total number of bonds in the formula unit is found by adding the numbers for each ion separately.

Understanding the Question

The question asks you to count twice: the number of coordinate bonds and the number of pi bonds in one formula unit of $\text{NH}_4\text{CN}$. The options test whether you recognise (a) that $\text{NH}_4^+$ contains a dative bond, and (b) that the cyanide ion $\text{CN}^-$ contains a triple bond, which has two pi bonds.

Approach

Because the compound is ionic, treat the cation and anion separately.

1. Count the coordinate bonds in $\text{NH}_4^+$.
2. Count the pi bonds in $\text{CN}^-$.
3. Add the two answers and select the row.

Step-by-Step Reasoning

For the ammonium ion:

• Ammonia, $\text{NH}_3$, has a lone pair on nitrogen. When it reacts with $\text{H}^+$, the lone pair is donated into the empty 1s orbital of the proton to form a $\text{N}{-}\text{H}$ bond.
• This bond is therefore a coordinate (dative) bond, while the other three $\text{N}{-}\text{H}$ bonds in $\text{NH}_4^+$ are ordinary covalent bonds.
• Even though all four bonds are identical after formation, the ion was formed using one lone pair donation, so there is exactly one coordinate bond.

For the cyanide ion:

• In $\text{CN}^-$, carbon and nitrogen are joined by a triple bond, $\text{C}\equiv\text{N}$.
• A triple bond contains one sigma bond and two pi bonds.
• No other multiple bonds are present, so the number of pi bonds is 2.

Combining these counts gives:

• coordinate bonds = 1
• pi bonds = 2

This matches option B.

The distractors can be understood as follows:

• Row A says 0 coordinate bonds, ignoring the dative bond in $\text{NH}_4^+$.
• Row C repeats the same error and also counts the pi bonds as 3.
• Row D has the correct coordinate bond count but incorrectly treats the triple bond as contributing three pi bonds rather than one sigma and two pi.

Key Takeaways

• Coordinate bonds are common in polyatomic ions such as $\text{NH}_4^+$, $\text{H}_3\text{O}^+$ and complex ions.
• Multiple bonds always consist of one sigma bond plus one or two pi bonds.
• In an ionic compound, count structural features per ion and then combine them per formula unit.

Common Mistakes

• Counting the $\text{C}\equiv\text{N}$ triple bond as three pi bonds. A triple bond has one sigma and two pi, so the pi count is 2, not 3.
• Forgetting that $\text{NH}_4^+$ contains a dative bond. Three of the $\text{N}{-}\text{H}$ bonds are ordinary covalent; the fourth is coordinate.
• Thinking coordinate bonds are pi bonds. A dative bond is a sigma-type bond, not counted separately as a pi bond.

Things to Be Careful About

• Use the correct terminolgy: "coordinate bond" and "pi bond" are counted independently.
• Read "one formula unit" carefully: it means one $\text{NH}_4^+$ and one $\text{CN}^-$, not a molecule of covalent $\text{NH}_4\text{CN}$.
• Remember that the dative bond is counted once, even though the four $\text{N}{-}\text{H}$ bonds in the ammonium ion are indistinguishable once formed.

Background Concept

In all chemical reactions, the energy change comes from two opposing processes:

• Bond breaking: energy must be supplied to break existing bonds — this is an endothermic process (energy absorbed).
• Bond forming: energy is released when new bonds form — this is an exothermic process (energy released).

The overall enthalpy change ($\Delta H$) of a reaction is the difference between the energy absorbed in breaking bonds and the energy released in forming bonds:

• If more energy is released in forming bonds than absorbed in breaking bonds → exothermic ($\Delta H$ negative).
• If more energy is absorbed in breaking bonds than released in forming bonds → endothermic ($\Delta H$ positive).

Understanding the Question

The neutralisation of ethanoic acid (CH3COOH) with excess aqueous NaOH releases 55 kJ mol⁻¹ of energy. The question asks us to identify which statement correctly explains why this reaction is exothermic. All four options agree the reaction is exothermic; they differ only in their explanation of why, in terms of bond breaking and bond forming.

Approach

Evaluate each statement against the fundamental rule: bond breaking absorbs energy, bond forming releases energy. An exothermic reaction must have more energy released in bond forming than absorbed in bond breaking.

Step-by-Step Reasoning

• Option A — "only bond breaking takes place": incorrect. Every chemical reaction involves both breaking old bonds and forming new ones. Also, bond breaking absorbs energy, it does not release it.
• Option B — "only bond forming takes place": incorrect. Bonds must also be broken for a reaction to occur; bond forming alone is not a complete description.
• Option C — "more energy is given out in breaking bonds than is taken in to form bonds": incorrect. Bond breaking takes in (absorbs) energy; bond forming gives out (releases) energy. This statement has the energy flow backwards.
• Option D — "more energy is given out in forming bonds than is taken in to break bonds": correct. This correctly states that the energy released during bond formation exceeds the energy absorbed during bond breaking, which is exactly why the reaction is exothermic.

Key Takeaways

• Bond breaking is always endothermic (absorbs energy); bond forming is always exothermic (releases energy).
• An exothermic reaction has a net release of energy because bond forming releases more energy than bond breaking absorbs.

Common Mistakes

• Confusing which process absorbs vs releases energy (bond breaking absorbs; bond forming releases).
• Thinking a reaction involves only bond breaking or only bond forming — both always occur.

Things to Be Careful About

• Read the direction of energy flow carefully in each option — options C and D are mirror images, and only D has it correct.
• The amount of energy released (55 kJ mol⁻¹) is not needed to answer; the question is purely about the qualitative explanation.

Background Concept

This question tests calorimetry — the measurement of heat changes during a chemical reaction. When a fuel burns, the energy released heats the surrounding water. The key equation is $q = mc\Delta T$, where $q$ is the heat absorbed, $m$ is the mass of water, $c$ is the specific heat capacity of water (4.18 J g⁻¹ K⁻¹), and $\Delta T$ is the temperature change.

Understanding the Question

The question gives:

• Mass of fuel = 1.60 g
• Percentage of energy absorbed by water = 45.0%
• Mass of water = 200 g
• Initial temperature = 18.0°C
• Final temperature = 66.0°C

We need to find the total energy released per gram of fuel.

Approach

1. Calculate the heat absorbed by the water using $q = mc\Delta T$.
2. Since only 45.0% of the energy released is absorbed, divide by 0.450 to find the total energy released.
3. Divide by the mass of fuel to get energy per gram.

Step-by-Step Reasoning

1. Temperature change: $\Delta T = 66.0 - 18.0 = 48.0$ K (temperature difference is the same in °C and K).
2. Heat absorbed by water: $q = 200 \times 4.18 \times 48.0 = 40\,128$ J.
3. Total energy released: Since 45.0% is absorbed, $\text{total} = \frac{40\,128}{0.450} = 89\,173$ J.
4. Energy per gram: $\frac{89\,173}{1.60} = 55\,733$ J/g ≈ 55,700 J/g.

This matches option B.

Key Takeaways

• Always use $q = mc\Delta T$ for calorimetry, with $c = 4.18$ J g⁻¹ K⁻¹ for water.
• Temperature differences in °C equal those in K.
• When a percentage of energy is absorbed, divide by the decimal fraction (0.450) to get the total.

Common Mistakes

• Forgetting to divide by the fuel mass: This gives 89,200 J (option C), which is the total energy, not per gram.
• Using 100% instead of 45%: This gives 40,128 J, which is not an option.
• Multiplying instead of dividing by 0.45: Would give a smaller number.

Things to Be Careful About

• The question asks "per gram of fuel", so the final step must divide by 1.60 g.
• The options are in J, but the answer is J/g — check the units carefully.
• Use the correct specific heat capacity of water (4.18 J g⁻¹ K⁻¹).

Background Concept

This question tests redox balancing. In a redox reaction, oxidation and reduction happen together; the total number of electrons lost in oxidation must equal the total number gained in reduction. Oxidation numbers are bookkeeping charges assigned to atoms to track electron transfer.

Understanding the Question

Two separate reactions are shown. In the first, sulfite ions react with zinc metal. In the second, sulfite ions react with manganese dioxide in acidic conditions. The coefficients $a, b, w, x$ are the coefficients of $\text{SO}_3^{2-}$, $\text{H}_2\text{O}$, $\text{SO}_3^{2-}$, and $\text{H}^+$ respectively. We need to balance each equation and match the coefficients to the options.

Approach

1. Assign oxidation numbers to all species, especially S and the metal/Mn.
2. Identify which element is oxidised and which is reduced.
3. Balance the electron transfer using coefficients.
4. Balance remaining atoms (O, H) and check charge balance.
5. Match the coefficients to the options.

Step-by-Step Reasoning

First equation

• In $\text{SO}_3^{2-}$, O is -2, so S is +4.
• In $\text{S}_2\text{O}_4^{2-}$, O is -2, total O = -8, ion charge -2, so 2S = +6, S = +3.
• Zn goes from 0 to +2: oxidation, losing 2e.
• Each S goes from +4 to +3: reduction, gaining 1e per S. Since the product contains two S atoms, two sulfite ions are reduced while gaining 2e.
• Therefore 1 Zn supplies the 2e needed by 2 $\text{SO}_3^{2-}$: $a = 2$, $d = 1$.
• Balance charge: left side has charge $-4$ (from 2 sulfite ions). Right side has $+2$ (from $\text{Zn}^{2+}$) $-c$ (from $c\,\text{OH}^-$) $-2$ (from $\text{S}_2\text{O}_4^{2-}$) $= -c$. Thus $-4 = -c$, so $c = 4$.
• Balance H: $2b = c = 4$, so $b = 2$.
• Check O: left $6 + 2 = 8$; right $4 + 4 = 8$. Balanced.

Second equation

• In $\text{MnO}_2$, Mn is +4; in $\text{Mn}^{2+}$, Mn is +2. Reduction: gain of 2e.
• In $\text{SO}_3^{2-}$, S is +4. In $\text{S}_2\text{O}_6^{2-}$, O is -2, total O = -12, ion charge -2, so 2S = +10, S = +5. Oxidation: loss of 1e per S, or 2e per $\text{S}_2\text{O}_6^{2-}$.
• One $\text{MnO}_2$ gains 2e; two $\text{SO}_3^{2-}$ lose 2e. Therefore $w = 2$, $y = 1$.
• Balance O: left has 2 (from $\text{MnO}_2$) + 6 (from 2 sulfite) = 8 O atoms. Right has 6 (from $\text{S}_2\text{O}_6^{2-}$) + $z$ (from water), so $z = 2$.
• Balance H: $x = 2z = 4$.
• Check charge: left $-4 + 4 = 0$; right $+2 - 2 = 0$. Balanced.

Thus $a = 2$, $b = 2$, $w = 2$, $x = 4$, which is option B.

Key Takeaways

• Always assign oxidation numbers before balancing redox equations.
• Balance electron transfer first, then atoms and charge.
• In oxoanions, the oxidation state of the central atom can be found from the total charge and the -2 oxidation state of oxygen.
• A coefficient in front of a formula applies to every atom in that formula.

Common Mistakes

• Incorrectly assigning the oxidation state of S in $\text{S}_2\text{O}_4^{2-}$ or $\text{S}_2\text{O}_6^{2-}$.
• Forgetting that the product contains two S atoms, so two sulfite ions are needed per redox unit.
• Balancing O and H without first balancing the electron transfer.
• Mixing up the coefficients for $\text{SO}_3^{2-}$ in the two different reactions.

Things to Be Careful About

• The options give only $a, b, w, x$; you do not need to find $c, d, y, z$ unless they help with balancing.
• Check both atom balance and charge balance; a balanced redox equation must satisfy both.
• In acidic conditions, $\text{H}^+$ and $\text{H}_2\text{O}$ are used to balance H and O; in basic conditions, $\text{OH}^-$ and $\text{H}_2\text{O}$ are used.

Background Concept

An oxidising agent (oxidant) is a species that oxidises another species, meaning it causes the other species to lose electrons. In doing so, the oxidising agent itself is reduced — it gains electrons. Conversely, a reducing agent is itself oxidised (loses electrons). The key to identifying an oxidising agent is to find the species whose oxidation number decreases during the reaction.

Hydrogen normally has oxidation number $+1$ in compounds such as $\text{H}_2\text{O}$, $\text{HCl}$ and $\text{NH}_3$. However, with very electropositive metals (Group 1 and Group 2 metals), hydrogen can have oxidation number $-1$, forming ionic hydrides such as $\text{KH}$, $\text{NaH}$ and $\text{CaH}_2$. In these compounds hydrogen has gained an electron to achieve a noble-gas-like electron configuration, so it exists as the hydride ion, $\text{H}^-$.

Understanding the Question

The question asks which of four equations shows hydrogen acting as an oxidising agent. This means we must find the reaction in which hydrogen is reduced — where its oxidation number decreases from 0 (in elemental $\text{H}_2$) to a negative value. The correct answer is the only reaction in which hydrogen gains electrons.

Approach

1. Assign oxidation numbers to hydrogen in each reaction, both in reactants and products.
2. Look for the reaction where hydrogen's oxidation number decreases (from 0 to $-1$).
3. That reaction shows hydrogen acting as an oxidising agent.

Step-by-Step Reasoning

Go through each option in turn.

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

• K: oxidation number goes from 0 (element) to $+1$ (in $\text{K}^+$). Potassium is oxidised (loses an electron).
• H: oxidation number goes from 0 (in $\text{H}_2$) to $-1$ (in $\text{H}^-$). Hydrogen is reduced (gains an electron).
• Since hydrogen gains electrons, it is the oxidising agent. ✓

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

• H: $0 \rightarrow +1$. Hydrogen is oxidised (loses an electron).
• I: $0 \rightarrow -1$. Iodine is reduced (gains an electron).
• Hydrogen is oxidised, so it acts as a reducing agent, not an oxidising agent. ✗

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

• Cu: $+1 \rightarrow 0$. Copper is reduced (gains electrons).
• H: $0 \rightarrow +1$. Hydrogen is oxidised (loses an electron).
• Hydrogen is oxidised, so it acts as a reducing agent. ✗

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

• H: $0 \rightarrow +1$. Hydrogen is oxidised.
• N: $0 \rightarrow -3$. Nitrogen is reduced.
• Hydrogen is oxidised, so it acts as a reducing agent. ✗

Only option A shows hydrogen being reduced, so hydrogen acts as an oxidising agent only in A.

Key Takeaways

• An oxidising agent is itself reduced (gains electrons); a reducing agent is itself oxidised (loses electrons).
• Hydrogen is usually $+1$ in compounds, but can be $-1$ in ionic hydrides formed with very electropositive metals.
• To identify redox behaviour, always assign oxidation numbers and track the changes.

Common Mistakes

• Confusing oxidising and reducing agents: an oxidising agent is itself reduced, not oxidised. Many students pick B or D thinking hydrogen "gives" electrons, but in those reactions hydrogen is oxidised and is the reducing agent.
• Assuming hydrogen always has oxidation number $+1$: in ionic hydrides such as $\text{KH}$, hydrogen is $-1$.
• Not checking all options: each option must be tested; the reaction where hydrogen is reduced is the only correct one.

Things to Be Careful About

• The oxidation number of an element in its elemental form ($\text{H}_2$) is 0.
• In ionic hydrides, hydrogen is $-1$, not $+1$.
• The oxidising agent is the species that is reduced — look for the decrease in oxidation number.

Background Concept

The ideal gas equation is:

where $p$ is pressure, $V$ is volume, $n$ is the amount of gas in moles, $R$ is the gas constant, and $T$ is temperature in kelvin. For this equation to give pressure in pascals, the volume must be in cubic metres ($\text{m}^3$) and temperature in kelvin (K). The gas constant is $R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$.

Nitrogen gas is diatomic, so its molecule is $\text{N}_2$. The molar mass is therefore:

Understanding the Question

We are given the mass of nitrogen gas, the volume of the vessel, and the temperature. We need to calculate the pressure. The question is a one-mark multiple-choice question, so the working must be quick but careful. Three conversions are needed before substituting into the ideal gas equation:

• mass of $\text{N}_2$ to moles,
• volume from $\text{dm}^3$ to $\text{m}^3$,
• temperature from $^\circ\text{C}$ to K.

Approach

1. Convert the mass of nitrogen to moles using $n = \frac{m}{M_r}$.
2. Convert the volume to $\text{m}^3$.
3. Convert the temperature to kelvin.
4. Rearrange $pV = nRT$ to give $p = \frac{nRT}{V}$.
5. Substitute the values and compare with the options.

Step-by-Step Reasoning

Step 1: Calculate moles of nitrogen.

Since nitrogen is $\text{N}_2$, its molar mass is $28.0\text{ g mol}^{-1}$.

Step 2: Convert volume to SI units.

$1\text{ dm}^3 = 1 \times 10^{-3}\text{ m}^3$, so:

Step 3: Convert temperature to kelvin.

Step 4: Substitute into the ideal gas equation.

Evaluating the numerator first:

Then:

This matches option C.

If a candidate used $M_r = 14.0$ instead of $28.0$, they would get double the number of moles and hence double the pressure, $92\,900\text{ Pa}$, which is option D. This is a common trap because nitrogen atoms have $A_r = 14.0$, but nitrogen gas exists as $\text{N}_2$ molecules.

Key Takeaways

• Always use the molar mass of the actual gas molecule: nitrogen is $\text{N}_2$, so $M_r = 28.0$.
• The ideal gas equation requires SI units: volume in $\text{m}^3$, temperature in K.
• $1\text{ dm}^3 = 1 \times 10^{-3}\text{ m}^3$.
• $T(\text{K}) = T(^\circ\text{C}) + 273$.

Common Mistakes

• Using $14.0\text{ g mol}^{-1}$ as the molar mass of nitrogen gas instead of $28.0\text{ g mol}^{-1}$. Nitrogen gas is diatomic.
• Forgetting to convert $\text{dm}^3$ to $\text{m}^3$. This would give a pressure in kPa rather than Pa, or a numerically incorrect value.
• Forgetting to add 273 to the Celsius temperature. Using $40.0\text{ K}$ would give a much smaller pressure.
• Using $R$ with inconsistent units. With $p$ in Pa, $V$ in $\text{m}^3$, and $T$ in K, $R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$.

Things to Be Careful About

• Check the units at every step: mass in g, molar mass in g mol$^{-1}$, volume in m$^3$, temperature in K.
• The answer should be reported to three significant figures because the data are given to three significant figures.
• In multiple-choice questions, a quick estimate can help: $n \approx 0.0357$, $RT \approx 2600$, so $p \approx \frac{0.0357 \times 2600}{0.002} \approx 46\,000\text{ Pa}$, clearly identifying option C.

Background Concept

For a general equilibrium $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$, the equilibrium constant is written as

where each concentration is raised to the power of its stoichiometric coefficient in the balanced equation. Solids and pure liquids are omitted; here every species is aqueous, so all four appear. The key feature of this question is the coefficient 4 in front of $\text{O}_2$: it becomes the exponent 4 on $[\text{O}_2]$ in the denominator. Getting this exponent right is the entire point of the problem.

Understanding the Question

The equation shows one haemoglobin molecule binding four oxygen molecules. We are told 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 command is to find the numerical value of $K_c$. This is a pure substitution-and-arithmetic task: write the correct expression, use the given equality to simplify it, then evaluate.

Approach

1. Write the $K_c$ expression from the balanced equation, remembering the exponent 4 on $[\text{O}_2]$.
2. Use the given equality $[\text{Hb}] = [\text{Hb}(\text{O}_2)_4]$ to cancel the numerator and denominator ratio.
3. Substitute the value of $[\text{O}_2]$ and raise it to the fourth power.
4. Take the reciprocal to obtain $K_c$ and compare with the options.

Step-by-Step Reasoning

The equilibrium constant expression is

The product is on top, the two reactants on the bottom, and the 4 from the equation becomes the power on $[\text{O}_2]$.

Because the question states that $[\text{Hb}]$ and $[\text{Hb}(\text{O}_2)_4]$ are equal, the ratio $[\text{Hb}(\text{O}_2)_4]/[\text{Hb}] = 1$. The expression therefore collapses to

Now substitute: $(7.6 \times 10^{-6})^4$. Work this out in two parts: $7.6^4 = 3336.2$ and $(10^{-6})^4 = 10^{-24}$, giving $3.34 \times 10^{-21}$. Then

This matches option A.

Why the distractors are wrong:

• B ($1.3 \times 10^5$) comes from using $1/[\text{O}_2]$ instead of $1/[\text{O}_2]^4$ — forgetting the exponent 4.
• C ($7.6 \times 10^{-6}$) just repeats the given concentration of oxygen, with no equilibrium expression at all.
• D ($3.3 \times 10^{-21}$) is $(7.6 \times 10^{-6})^4$ itself — the denominator before taking the reciprocal. It is the value you would get if you forgot that $K_c$ is products over reactants.

Key Takeaways

• The stoichiometric coefficient in a balanced equation becomes the exponent in the $K_c$ expression — never ignore it.
• When two equilibrium concentrations are equal, the ratio of their terms simplifies to 1, which can dramatically simplify the expression.
• Always check the magnitude: a very small equilibrium concentration of a reactant raised to a power gives a very large $K_c$ when it appears in the denominator.

Common Mistakes

• Using power 1 instead of power 4 on $[\text{O}_2]$ — this gives option B. The coefficient 4 in the equation is not optional.
• Forgetting the reciprocal — writing $K_c = [\text{O}_2]^4$ rather than $1/[\text{O}_2]^4$ gives option D. Remember products over reactants.
• Quoting the given concentration as $K_c$ — option C is simply the data, not a calculation.
• Arithmetic slip with powers of ten: $(10^{-6})^4 = 10^{-24}$, not $10^{-10}$ or $10^{-6}$. Keep the exponent bookkeeping separate from the coefficient.

Things to Be Careful About

• The exponent 4 applies to the whole concentration, including the power of ten: $(7.6 \times 10^{-6})^4$ is not $7.6 \times 10^{-24}$.
• Significant figures: the data has 2 s.f., so the answer is quoted to 2 s.f. as $3.0 \times 10^{20}$.
• Units: $K_c$ for this equilibrium would carry units, but the question asks only for the numerical value, so no unit is needed in the answer.
• Check the direction: $K_c$ is large here because the product is strongly favoured at equilibrium (the binding is effectively very favourable), which is consistent with a large value like $10^{20}$.

Background Concept

A pH titration curve plots the pH of a solution against the volume of titrant added. The shape of the curve reveals the strengths of the acid and base, and the volume at the equivalence point reveals their stoichiometric ratio.

• Initial pH indicates the strength and concentration of the acid. A pH of 1 for a 0.1 mol dm$^{-3}$ solution means it is a strong monoprotic acid (fully dissociated).
• Equivalence point volume depends on the stoichiometry. For equal concentrations, the volume ratio is the inverse of the ratio of H$^+$ to OH$^-$ provided per formula unit.
• Equivalence point pH is 7 for strong acid + strong base, $<$ 7 for strong acid + weak base, and $>$ 7 for weak acid + strong base.
• Final pH indicates the strength and concentration of the excess titrant. A final pH of 13 for a 0.1 mol dm$^{-3}$ solution indicates a strong diprotic base (e.g., [OH$^-$] = 0.2 mol dm$^{-3}$, pOH = 0.7, pH = 13.3).

Understanding the Question

We are given a titration curve for 20 cm$^3$ of acid P titrated with alkali Q of the same concentration. We must identify P and Q from the four options based on the curve's features: initial pH, equivalence point volume, equivalence point pH, and final pH.

Approach

We will systematically check the four key features of the titration curve against the properties of the options:

1. Initial pH to determine acid strength.
2. Equivalence point volume to determine the proton/hydroxide ratio.
3. Equivalence point pH to determine base strength.
4. Final pH to confirm base strength and concentration.

Step-by-Step Reasoning

1. Initial pH = 1: The acid P is 0.1 mol dm$^{-3}$ (since $-\log(0.1) = 1$). A pH of 1 means it is fully dissociated, so P is a strong acid. This eliminates D (CH$_3$COOH is a weak acid, its 0.1 mol dm$^{-3}$ solution would have an initial pH of about 2.8).

2. Equivalence point volume = 10 cm$^3$: 20 cm$^3$ of acid P is neutralised by 10 cm$^3$ of alkali Q. Since they have the same concentration (0.1 mol dm$^{-3}$), the moles of P are twice the moles of Q ($n_P = 2n_Q$). For complete neutralisation, the total moles of H$^+$ must equal the total moles of OH$^-$. Therefore, each mole of P must provide twice as many reactive particles as each mole of Q. This means P is monoprotic (1 H$^+$) and Q is diprotic (2 OH$^-$).

   • A (HCl and NaOH): both monoprotic. Equivalence would be at 20 cm$^3$. Incorrect.
   • B (H$_2$SO$_4$ and NH$_3$): H$_2$SO$_4$ is diprotic, NH$_3$ is monoprotic. Equivalence would be at 10 cm$^3$, but NH$_3$ is a weak base. Incorrect.
   • C (HCl and Ba(OH)$_2$): HCl is monoprotic, Ba(OH)$_2$ is diprotic. Equivalence is at 10 cm$^3$. Correct stoichiometry.
   • D (CH$_3$COOH and Sr(OH)$_2$): stoichiometry matches, but CH$_3$COOH is weak. Incorrect.

3. Equivalence point pH = 7: The sharp vertical section passes exactly through pH 7. This indicates the salt formed is neutral, which only happens when a strong acid is titrated with a strong base.

   • B involves NH$_3$, a weak base. The equivalence point would be acidic (pH $<$ 7) due to hydrolysis of the NH$_4^+$ ion. Incorrect.
   • D involves CH$_3$COOH, a weak acid. The equivalence point would be basic (pH $>$ 7) due to hydrolysis of the CH$_3$COO$^-$ ion. Incorrect.
   • C involves HCl (strong acid) and Ba(OH)$_2$ (strong base). The equivalence point is exactly at pH 7. Correct.

4. Final pH = 13: After the equivalence point, the pH levels off at 13. This is the pH of excess 0.1 mol dm$^{-3}$ Ba(OH)$_2$. Since Ba(OH)$_2$ is a strong diprotic base, it fully dissociates to give [OH$^-$] = $2 \times 0.1 = 0.2$ mol dm$^{-3}$.
   $pOH = -\log(0.2) \approx 0.7$
   $pH = 14 - 0.7 = 13.3 \approx 13$
   This matches the final pH on the curve perfectly, confirming C is correct.

Key Takeaways

• The initial pH and final pH of a titration curve immediately reveal whether the acid and base are strong or weak.
• The volume at the equivalence point relative to the initial volume of the analyte reveals the stoichiometric ratio of H$^+$ to OH$^-$, allowing you to deduce if an acid or base is monoprotic, diprotic, etc.
• The pH at the equivalence point (7, $<$ 7, or $>$ 7) distinguishes between strong/weak acid and strong/weak base combinations.

Common Mistakes

• Ignoring the equivalence point volume: Many students focus only on the pH at the equivalence point and forget to check the volume. If they only look at pH 7, they might incorrectly choose A, forgetting that A would require 20 cm$^3$ of NaOH, not 10 cm$^3$.
• Confusing weak base final pH: Choosing B because H$_2$SO$_4$ and NH$_3$ have the correct 2:1 volume ratio. However, NH$_3$ is a weak base; its solution will not reach pH 13, and the equivalence point will be below pH 7.
• Misreading the initial pH: Choosing D because Sr(OH)$_2$ is a strong diprotic base. However, CH$_3$COOH is a weak acid, so its initial pH would be around 2.8, not 1.

Things to Be Careful About

• State symbols and dissociation: Remember that Ba(OH)$_2$ and Sr(OH)$_2$ are strong bases and fully dissociate in water to give 2 OH$^-$ ions per formula unit. This doubles the [OH$^-$] and thus lowers the pOH by $\log(2) \approx 0.3$, raising the pH to $\approx 13.3$ for a 0.1 mol dm$^{-3}$ solution.
• Equivalence point vs. half-equivalence point: Do not confuse the pH at the equivalence point (where moles of H$^+$ = moles of OH$^-$) with the pH at the half-equivalence point (where pH = p$K_a$ for a weak acid). The graph clearly shows the vertical section centered at pH 7.
• Concentration assumptions: The problem states P and Q have the same concentration. Always use this to set up the mole ratio: $n = c \times V$. If $c_P = c_Q$, then $n_P / n_Q = V_P / V_Q = 20 / 10 = 2$. This directly gives the proton/hydroxide ratio.