Chemistry 9701/12 — October/November 2025

Background Concept

The electron configuration of transition metals and their ions depends on the filling order of orbitals and the order of electron removal. For neutral atoms, the 4s orbital fills before the 3d orbital (e.g., $[\text{Ar}] 4\text{s}^2 3\text{d}^n$). However, when transition metals form positive ions, electrons are removed from the 4s orbital before the 3d orbital, because in the ion, the 3d orbitals are lower in energy than the 4s orbital.

Hund's rule states that electrons will fill degenerate orbitals (orbitals of the same energy, like the five 3d orbitals) singly with parallel spins before they pair up. This minimises electron-electron repulsion and results in the most stable arrangement.

Understanding the Question

The question asks to identify the correct electrons-in-boxes (orbital diagram) notation for the $\text{Fe}^{3+}$ ion. We are given four options showing the arrangement of electrons in the 3d and 4s orbitals after the $[\text{Ar}]$ core. We need to determine the correct number of electrons in each subshell and their correct arrangement according to Hund's rule.

Approach

1. Determine the atomic number of iron and write the electron configuration of the neutral atom.
2. Determine the number of electrons removed to form the $\text{Fe}^{3+}$ ion and remove them in the correct order (4s first, then 3d).
3. Write the resulting electron configuration for $\text{Fe}^{3+}$.
4. Apply Hund's rule to distribute the 3d electrons among the five 3d boxes.
5. Match the result with the given options.

Step-by-Step Reasoning

1. Neutral Iron Atom: Iron (Fe) has a proton number of 26. The noble gas core is Argon ($[\text{Ar}]$), which accounts for 18 electrons. The remaining 8 electrons fill the 4s and 3d subshells. The neutral atom configuration is $[\text{Ar}] 4\text{s}^2 3\text{d}^6$.
2. Forming $\text{Fe}^{3+}$: The $\text{Fe}^{3+}$ ion has a charge of +3, meaning it has lost 3 electrons. Transition metals lose their 4s electrons before their 3d electrons.
   • Remove 2 electrons from the 4s subshell: $[\text{Ar}] 3\text{d}^6$.
   • Remove 1 electron from the 3d subshell: $[\text{Ar}] 3\text{d}^5$.
   • The configuration of $\text{Fe}^{3+}$ is $[\text{Ar}] 3\text{d}^5 4\text{s}^0$.
3. Electron Arrangement (Hund's Rule): The 3d subshell has five degenerate orbitals (boxes). There are 5 electrons to place in these 5 orbitals. According to Hund's rule, each orbital will receive one electron before any pairing occurs. Thus, all five 3d boxes will contain one electron (all with the same spin, typically represented as up arrows). The 4s box will be empty.
4. Evaluating Options:
   • Option A: Shows 3 electrons in 3d and 2 in 4s. Incorrect configuration.
   • Option B: Shows 3 electrons in 3d and 2 in 4s. Incorrect configuration.
   • Option C: Shows 5 electrons in 3d but one pair and three singles. This violates Hund's rule, which requires maximum unpaired electrons in degenerate orbitals.
   • Option D: Shows 5 electrons in 3d, all unpaired (one in each box), and the 4s box is empty. This matches our derived configuration and correctly applies Hund's rule.

Key Takeaways

• Transition metals lose 4s electrons before 3d electrons when forming cations.
• Hund's rule must be applied when drawing orbital diagrams for partially filled degenerate subshells (like d or f orbitals) to ensure maximum multiplicity (unpaired electrons).
• The electrons-in-boxes notation visually represents the quantum mechanical filling rules (Aufbau, Pauli exclusion, and Hund's rule).

Common Mistakes

• Removing 3d electrons first: A common mistake is to remove electrons from the 3d subshell before the 4s subshell when forming ions, leading to an incorrect configuration like $[\text{Ar}] 3\text{d}^3 4\text{s}^2$.
• Violating Hund's rule: Option C is a distractor that shows the correct number of electrons (5 in 3d, 0 in 4s) but pairs them up prematurely. Students might forget Hund's rule and just fill the first box with a pair.
• Confusing neutral atom and ion configurations: Forgetting that neutral Fe is $3\text{d}^6 4\text{s}^2$ and incorrectly assuming it is $3\text{d}^8$ or similar.

Things to Be Careful About

• Always remove electrons from the highest principal quantum number ($n$) first. For transition metals, $4\text{s}$ ($n=4$) is emptied before $3\text{d}$ ($n=3$), even though $4\text{s}$ fills first in the neutral atom.
• Ensure the orbital diagram strictly follows Hund's rule: single electrons in all degenerate orbitals before any pairing. Arrows must be parallel (same direction) for unpaired electrons.
• Pay attention to the core notation: $[\text{Ar}]$ means the inner 18 electrons are not shown. Only the valence/outermost subshells (3d and 4s) need to be evaluated.

Background Concept

First ionisation energy is defined as the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous singly positive ions. The three key features are: the starting material must be isolated gaseous atoms, one electron is removed per atom, and the product is a gaseous 1+ ion plus a free electron.

Understanding the Question

The question asks which of the four equations has an energy change equal to the first ionisation energy of bromine. We must check two things for each option: the starting state of bromine (must be a gaseous atom, not a molecule) and the direction of electron transfer (must be loss of an electron to form a +1 ion).

Approach

Test each option against the definition:

1. Does it start from a gaseous bromine atom, Br(g)?
2. Does it show loss of one electron to form a gaseous Br⁺ ion?
   Only an option satisfying both conditions can represent the first ionisation energy.

Step-by-Step Reasoning

1. First ionisation energy always starts from gaseous atoms — Br(g), never Br₂(g).
2. It removes one electron to form a +1 ion, so the electron appears as a product.
3. Option A: Br(g) → Br⁺(g) + e⁻ — starts from a gaseous atom and removes one electron to form a gaseous 1+ ion. This matches the definition exactly.
4. Option B: Br(g) → Br⁻(g) − e⁻ — this shows formation of an anion by gaining an electron; that is electron affinity, not ionisation. The notation is also unusual.
5. Option C: ½Br₂(g) → Br⁺(g) + e⁻ — starts from a bromine molecule, so the energy change includes the atomisation energy of Br₂ as well as the ionisation energy. It is not simply the first ionisation energy.
6. Option D: ½Br₂(g) → Br⁻(g) − e⁻ — starts from a molecule and forms an anion; wrong on both counts.

Key Takeaways

• First ionisation energy ALWAYS starts from the gaseous atom and produces a gaseous 1+ ion plus a free electron.
• Do not confuse ionisation with atomisation (breaking a molecule into atoms) or with electron affinity (adding an electron to form a negative ion).

Common Mistakes

• Choosing C or D because they involve Br₂ — but the definition requires the starting material to be the gaseous atom, not the molecule.
• Choosing B because it involves removing an electron — but B actually forms an anion, which is the reverse of ionisation.

Things to Be Careful About

• The state symbol (g) is essential; the definition specifically requires gaseous atoms.
• The electron must appear on the product side for ionisation; if it appears on the reactant side, the process is electron gain, not removal.

Background Concept

This question combines redox chemistry with stoichiometry. A redox reaction can be split into oxidation and reduction half-equations. Electrons lost in oxidation must equal electrons gained in reduction. Once the half-equations are combined, the balanced overall equation gives the mole ratio between reactants and products.

Understanding the Question

We are told that hydrogen sulfide reacts with sulfur dioxide to produce sulfur. Two half-equations are given:

• H₂S is oxidised: H₂S → S + 2H⁺ + 2e⁻
• SO₂ is reduced: SO₂ + 4H⁺ + 4e⁻ → S + 2H₂O

We need the moles of H₂S required to produce 1 mol of sulfur overall.

Approach

1. Compare electrons: H₂S releases 2e⁻, SO₂ accepts 4e⁻.
2. Multiply the H₂S half-equation by 2 so both involve 4e⁻.
3. Add the half-equations and cancel H⁺ and e⁻ that appear on both sides.
4. Read the mole ratio H₂S : S from the overall equation.
5. Scale to 1 mol of S.

Step-by-Step Reasoning

1. Write the two half-equations:

   • Oxidation: H₂S → S + 2H⁺ + 2e⁻
   • Reduction: SO₂ + 4H⁺ + 4e⁻ → S + 2H₂O

2. Balance electrons: multiply the oxidation half-equation by 2:
   2H₂S → 2S + 4H⁺ + 4e⁻

3. Add the two half-equations:
   2H₂S + SO₂ + 4H⁺ + 4e⁻ → 2S + 4H⁺ + 4e⁻ + S + 2H₂O

4. Cancel the 4H⁺ and 4e⁻ that appear on both sides:
   2H₂S + SO₂ → 3S + 2H₂O

5. The overall equation shows 2 mol H₂S produce 3 mol S.
   So 1 mol S is produced by 2/3 mol H₂S.

Key Takeaways

• Always balance electrons before combining half-equations.
• The overall balanced equation is the source of mole ratios.
• Species appearing on both sides after addition, here H⁺ and e⁻, must be cancelled.

Common Mistakes

• Assuming a 1:1 ratio because both half-equations produce sulfur.
• Forgetting to multiply the H₂S half-equation by 2.
• Misreading the ratio: 2 mol H₂S : 3 mol S, not 3 mol H₂S : 2 mol S.

Things to Be Careful About

• Keep charges balanced in half-equations.
• Include state symbols if required.
• The question asks for moles needed to produce 1 mol of total sulfur, so use the overall ratio.

Background Concept

Relative atomic mass ($A_r$) is the weighted mean mass of an atom of an element compared with one-twelfth of the mass of one atom of carbon-12. It is a weighted mean because it takes into account the natural abundance of each isotope of the element. For example, naturally occurring chlorine is about 75% $^{35}\text{Cl}$ and 25% $^{37}\text{Cl}$, giving $A_r \approx 35.5$.

Relative isotopic mass is the mass of a specific isotope compared with one-twelfth of the mass of one atom of carbon-12. For a particular isotope, this is approximately equal to its mass number (nucleon number), e.g. $^{35}\text{Cl}$ has relative isotopic mass 35, and $^{24}\text{Mg}$ has relative isotopic mass 24.

Relative molecular mass ($M_r$) is the sum of the relative atomic masses of all the atoms in a molecule. Relative formula mass is the same concept applied to an ionic compound, where the formula unit is used instead of a molecule.

Understanding the Question

This is a one-mark multiple-choice question asking which of four statements about relative masses is correct. Each option refers to a different type of relative mass: relative atomic mass, relative molecular mass, relative formula mass, and relative isotopic mass. The key is to know the precise definition of each term and to apply it to the given species.

Approach

For each option, identify the type of relative mass being described and calculate or recall the correct value:

• For a specific isotope, the relative isotopic mass equals its mass number.
• For a molecule, sum the relative atomic masses of its constituent atoms.
• For an ionic compound, sum the relative atomic masses of the atoms in its formula unit.
• For a naturally occurring element, the relative atomic mass is a weighted mean of its isotopes.

Evaluate each statement in turn and select the one that is correct.

Step-by-Step Reasoning

Option A: "The relative atomic mass of a $^{35}\text{Cl}$ atom is 35.5."

This is incorrect. A $^{35}\text{Cl}$ atom is a specific isotope, so its relative isotopic mass is 35 (equal to its mass number). The value 35.5 is the relative atomic mass of naturally occurring chlorine, which is a weighted mean of $^{35}\text{Cl}$ and $^{37}\text{Cl}$. The statement confuses the relative isotopic mass of a single isotope with the relative atomic mass of the element.

Option B: "The relative molecular mass of $\text{O}_2$ is 16.0."

This is incorrect. The relative atomic mass of oxygen is 16.0. Since $\text{O}_2$ is a diatomic molecule containing two oxygen atoms, its relative molecular mass is $2 \times 16.0 = 32.0$. The statement gives the relative atomic mass of a single oxygen atom, not the relative molecular mass of the molecule.

Option C: "The relative formula mass of $\text{CaCO}_3$ is 100.1."

This is correct. $\text{CaCO}_3$ contains one calcium atom, one carbon atom, and three oxygen atoms. Using $A_r(\text{Ca}) = 40.1$, $A_r(\text{C}) = 12.0$, and $A_r(\text{O}) = 16.0$:

Option D: "The relative isotopic mass of a $^{24}\text{Mg}$ atom is 24.3."

This is incorrect. A $^{24}\text{Mg}$ atom is a specific isotope, so its relative isotopic mass is 24 (equal to its mass number). The value 24.3 is the relative atomic mass of naturally occurring magnesium, which is a weighted mean of its isotopes ($^{24}\text{Mg}$, $^{25}\text{Mg}$, and $^{26}\text{Mg}$). The statement confuses the relative isotopic mass with the relative atomic mass of the element.

Therefore, the only correct statement is C.

Key Takeaways

• Relative isotopic mass of a specific isotope is approximately its mass number.
• Relative atomic mass of an element is a weighted mean of its isotopes' masses, taking natural abundance into account.
• Relative molecular mass is the sum of relative atomic masses of atoms in a molecule.
• Relative formula mass is the same sum applied to an ionic compound's formula unit.
• Be careful not to confuse the relative atomic mass of an element with the relative isotopic mass of one of its isotopes.

Common Mistakes

• Confusing relative atomic mass with relative isotopic mass: e.g. saying the relative atomic mass of a $^{35}\text{Cl}$ atom is 35.5, when 35.5 is the weighted mean for the element, not the mass of one isotope.
• Forgetting to multiply by the number of atoms in a molecule: e.g. saying $M_r(\text{O}_2) = 16.0$ instead of $32.0$.
• Using the relative atomic mass of an element when a specific isotope is named: e.g. saying the relative isotopic mass of $^{24}\text{Mg}$ is 24.3 instead of 24.

Things to Be Careful About

• Read the question carefully to see whether it refers to a specific isotope or to a naturally occurring element.
• Always multiply the relative atomic mass of each element by its subscript in the formula.
• Use the correct terminology: "relative atomic mass" for an element, "relative isotopic mass" for a specific isotope, "relative molecular mass" for a covalent molecule, and "relative formula mass" for an ionic compound.

Background Concept

Hybridisation is the mixing of atomic orbitals of similar energy to form a set of equivalent hybrid orbitals that are used for bonding. In a linear, two-electron-pair arrangement the central atom is sp hybridised: one s orbital and one p orbital mix to give two sp orbitals pointing in opposite directions at 180° to each other, leaving the other two p orbitals unhybridised.

In $\text{N}_2$, each nitrogen atom (electron configuration $1\text{s}^2\,2\text{s}^2\,2\text{p}^3$) is sp hybridised. Each nitrogen therefore has:

• two sp hybrid orbitals — one used to form the $\sigma$ bond to the other nitrogen, and one holding a lone pair;
• two unhybridised p orbitals, which overlap sideways with the p orbitals of the other nitrogen to form two $\pi$ bonds.

The result is a triple bond (one $\sigma$ + two $\pi$) between the nitrogen atoms, with each nitrogen also carrying a lone pair. This is why $\text{N}_2$ is so unreactive: the very strong triple bond (N≡N) must be broken before nitrogen can react.

Understanding the Question

This multiple-choice question asks you to match two features of the bonding in $\text{N}_2$ with the correct description:

1. How the $\sigma$ bond between the nitrogen atoms is formed — specifically, which orbitals overlap.
2. Which type of orbital contains the lone pair of electrons on each nitrogen atom.

The stem already tells you that the bonding involves sp hybridisation, so the whole question turns on correctly assigning the sp orbitals and the p orbitals to their roles in the molecule.

Approach

Start by confirming the hybridisation of nitrogen in $\text{N}_2$ (sp, as stated in the stem). Then account for each orbital on one nitrogen atom:

• one sp orbital points toward the other nitrogen and forms the $\sigma$ bond by head-on overlap with the other nitrogen's sp orbital;
• the other sp orbital points away and holds the lone pair;
• the two unhybridised p orbitals form the two $\pi$ bonds.

Once this assignment is clear, the correct row follows directly: sp–sp overlap for the $\sigma$ bond, and an sp orbital for the lone pair.

Step-by-Step Reasoning

1. Confirm sp hybridisation. Each nitrogen in $\text{N}_2$ is sp hybridised because it has two regions of electron density around it (one $\sigma$ bond and one lone pair), which requires a linear, 180° arrangement of two sp orbitals.

2. Assign the sp orbitals. One sp orbital on each nitrogen points directly toward the other nitrogen. The $\sigma$ bond forms by head-on (end-on) overlap of these two sp orbitals — i.e. an sp orbital from one atom overlaps with an sp orbital of the other atom. This rules out rows C and D, which claim an s orbital overlaps a p orbital.

3. Locate the lone pair. The second sp orbital on each nitrogen points away from the other atom and contains the lone pair. So the lone pair is in an sp orbital, not a p orbital. This rules out row A, leaving row B as the only correct option.

4. Check the role of the p orbitals. The two unhybridised p orbitals on each nitrogen are perpendicular to the internuclear axis and overlap sideways to form the two $\pi$ bonds of the triple bond. They do not hold the lone pair.

So the correct row is B: the $\sigma$ bond forms by sp–sp overlap, and the lone pair is in an sp orbital.

Key Takeaways

• In $\text{N}_2$, each nitrogen is sp hybridised, giving a linear arrangement of two sp orbitals and leaving two p orbitals unhybridised.
• A $\sigma$ bond always forms by head-on overlap of orbitals pointing along the internuclear axis — here, sp–sp overlap.
• A lone pair occupies a hybrid orbital (here, sp), not a pure p orbital.
• The triple bond in $\text{N}_2$ consists of one $\sigma$ bond and two $\pi$ bonds.

Common Mistakes

• Placing the lone pair in a p orbital (row A): the lone pair occupies the sp orbital pointing away from the other nitrogen; the p orbitals are all used in the $\pi$ bonds.
• Describing the $\sigma$ bond as s–p overlap (rows C and D): in an sp-hybridised molecule the $\sigma$ bond forms between the two sp hybrid orbitals, not between an s and a p orbital.
• Forgetting that two p orbitals remain unhybridised: these are essential for the two $\pi$ bonds that complete the triple bond.

Things to Be Careful About

• The stem already states that sp hybridisation is involved — use that information rather than re-deriving the hybridisation from scratch.
• Distinguish clearly between the $\sigma$ bond (head-on, sp–sp) and the $\pi$ bonds (sideways, p–p); mixing these up is the most common error in this question.
• For an MCQ, test each row against both criteria: the correct row must be right for both the $\sigma$ bond formation and the lone-pair orbital.

Background Concept

VSEPR Theory (Valence Shell Electron Pair Repulsion) predicts the 3D shape of molecules based on the repulsion between electron pairs in the valence shell of the central atom. Key principles include:

• Electron pairs (both bonding and lone pairs) arrange themselves to minimize repulsion.
• Lone pairs occupy more space and repel more strongly than bonding pairs because they are closer to the central nucleus and not shared with another atom.
• The presence of lone pairs compresses bond angles relative to the ideal geometry defined by the number of electron domains.
• Ideal bond angles for 4 electron domains (tetrahedral electron geometry) is $109.5^\circ$.
  • 4 bonding pairs, 0 lone pairs (e.g., $\text{CH}_4$): $109.5^\circ$.
  • 3 bonding pairs, 1 lone pair (e.g., $\text{NH}_3$): $\sim 107^\circ$.
  • 2 bonding pairs, 2 lone pairs (e.g., $\text{H}_2\text{O}$): $\sim 104.5^\circ$.
• For heavier central atoms (period 3 and below, like S, P, Cl), bond angles are often smaller than expected because the central atom is larger, the bonding pairs are further from the nucleus, and repulsion between them is weaker, allowing lone pairs to compress the angle more (or the atom may use nearly pure p-orbitals for bonding, leading to angles near $90^\circ$, as in $\text{H}_2\text{S}$ at $\sim 92^\circ$).

Understanding the Question

The question asks to order three specific bond angles ($x$, $y$, $z$) in the molecule 2-aminoethanethiol ($\text{HS}-\text{CH}_2-\text{CH}_2-\text{NH}_2$) from largest to smallest.

• Angle $x$ is the H–S–C angle around the sulfur atom.
• Angle $y$ is the S–C–C angle around the first carbon atom.
• Angle $z$ is the C–N–H angle around the nitrogen atom.

We must determine the number of bonding pairs and lone pairs around each central atom to apply VSEPR theory and estimate the angles.

Approach

1. Identify the central atom for each labelled angle.
2. Count the number of bonding pairs (atoms bonded to the central atom) and lone pairs on that central atom.
3. Determine the electron domain geometry (tetrahedral for 4 domains) and the molecular shape.
4. Apply the rule: more lone pairs = smaller bond angle. Also consider the effect of the central atom's period (larger atoms like S have smaller bond angles).
5. Compare the estimated angles to find the decreasing order.

Step-by-Step Reasoning

• Angle $y$ (S–C–C): The central atom is Carbon. It is bonded to S, C, and two H atoms. Total = 4 bonding pairs, 0 lone pairs.

  • Geometry: Tetrahedral.
  • Angle: Ideal tetrahedral angle = $109.5^\circ$. This is the largest angle.

• Angle $z$ (C–N–H): The central atom is Nitrogen. It is bonded to C and two H atoms, and has 1 lone pair (shown as two dots). Total = 3 bonding pairs, 1 lone pair.

  • Geometry: Trigonal pyramidal (based on tetrahedral electron arrangement).
  • Angle: The single lone pair compresses the angle from $109.5^\circ$. Similar to ammonia ($\text{NH}_3$, $107^\circ$), this angle is approximately $107^\circ$. This is smaller than $y$ but larger than $x$.

• Angle $x$ (H–S–C): The central atom is Sulfur. It is bonded to H and C, and has 2 lone pairs (shown as four dots). Total = 2 bonding pairs, 2 lone pairs.

  • Geometry: Bent (based on tetrahedral electron arrangement).
  • Angle: Two lone pairs compress the angle significantly. Furthermore, sulfur is in Period 3. In $\text{H}_2\text{S}$, the bond angle is only $\sim 92^\circ$ because sulfur uses nearly unhybridized p-orbitals for bonding. In organic sulfides like this ($\text{R}-\text{S}-\text{H}$), the angle is slightly larger due to steric bulk of the carbon group, typically around $96^\circ$–$100^\circ$. Regardless of the exact value, it is the smallest angle due to having two lone pairs and being a period 3 element.

• Comparison: $y$ ($109.5^\circ$) > $z$ ($\sim 107^\circ$) > $x$ ($\sim 96^\circ$).

• Order: $y$, $z$, $x$. This corresponds to option C.

Key Takeaways

• Lone pairs repel more strongly than bonding pairs, reducing bond angles.
• The more lone pairs on a central atom (with the same number of electron domains), the smaller the bond angle (e.g., $\text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}$).
• For central atoms in Period 3 and below (like S, P), bond angles are often significantly smaller than their Period 2 counterparts (like O, N) due to larger atomic size and weaker repulsion between bonding pairs.

Common Mistakes

• Ignoring lone pairs: Failing to count lone pairs on S and N would lead to assuming all angles are $109.5^\circ$.
• Confusing electron geometry with molecular shape: While all three central atoms have 4 electron domains (tetrahedral electron geometry), their molecular shapes differ (tetrahedral, trigonal pyramidal, bent), which affects the bond angles.
• Overlooking the period of the central atom: Assuming S behaves exactly like O (in water, $104.5^\circ$). Sulfur compounds typically have much smaller bond angles (closer to $90^\circ$) because the larger central atom allows bonding pairs to be further apart with less repulsion, or the atom uses p-orbitals directly.

Things to Be Careful About

• State symbols and lone pairs: Always count lone pairs explicitly shown in the diagram (the dots). S has 4 dots (2 lone pairs), N has 2 dots (1 lone pair).
• Order of decreasing size: Ensure the answer is ordered from largest to smallest ($y \rightarrow z \rightarrow x$), not smallest to largest.
• Approximate values: You don't need to memorize exact angles for every molecule, but knowing the trend ($109.5^\circ \rightarrow <109.5^\circ \rightarrow <104.5^\circ$) and the effect of period 3 elements is sufficient to rank them correctly.

Background Concept

This question weaves together three ideas from AS chemistry:

1. Electron configuration and orbital filling. Electrons occupy orbitals in order of increasing energy ($1\text{s}$, $2\text{s}$, $2\text{p}$, $3\text{s}$, $3\text{p}$, ...). Within a sub-shell, Hund's rule says electrons fill each orbital singly before any pairing occurs. A $\text{p}$ sub-shell contains three orbitals ($\text{p}_x$, $\text{p}_y$, $\text{p}_z$), each holding up to two electrons. A "completely filled orbital" means an orbital holding its full two electrons.

2. First ionisation energy. The energy needed to remove one mole of electrons from one mole of gaseous atoms. Across Period 3 it generally increases as nuclear charge increases and atomic radius shrinks, but there are dips at Al (the $3\text{p}^1$ electron is easily removed) and at S (the fourth $3\text{p}$ electron experiences repulsion when pairing into an already-occupied orbital).

3. Structure and bonding of Period 3 elements. Na, Mg and Al are metals with giant metallic structures; Si is giant covalent; P, S and Cl are simple molecular; Ar is monatomic.

Understanding the Question

We are told X and Y are different Period 3 elements. Each has exactly one completely filled orbital in its highest occupied energy sub-shell. Y has a greater first ionisation energy than X. We must identify X and Y, then match their structure and bonding to the correct option.

The key clue is the electron configuration: only certain Period 3 elements have exactly one completely filled orbital in the highest occupied sub-shell. The ionisation energy clue then tells us which is X and which is Y.

Approach

1. Write out the electron configurations of all Period 3 elements (Na to Ar).
2. For each, look at the highest occupied sub-shell and count completely filled orbitals (orbitals holding 2 electrons). Keep only those with exactly one.
3. This narrows X and Y to two elements. Use the first ionisation energy comparison to assign which is which.
4. Recall the structure and bonding of each element and pick the matching option.

Step-by-Step Reasoning

Step 1 — Electron configurations of Period 3.

• Na: $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^1$ — the $3\text{s}$ orbital has only 1 electron, so it is not filled. ✗
• Mg: $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2$ — the $3\text{s}$ sub-shell is one orbital holding 2 electrons, so it is completely filled. ✓
• Al: $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^1$ — the highest occupied sub-shell is $3\text{p}$ (higher energy than $3\text{s}$); it holds just 1 electron, so no orbital is filled. ✗ (Even though $3\text{s}$ is filled, it is not the highest occupied sub-shell.)
• Si: $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^2$ — by Hund's rule the two $3\text{p}$ electrons occupy two different orbitals singly; none is filled. ✗
• P: $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^3$ — one electron in each $3\text{p}$ orbital; none is filled. ✗
• S: $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^4$ — four $3\text{p}$ electrons give $3\text{p}_x^2\,3\text{p}_y^1\,3\text{p}_z^1$; exactly one orbital is completely filled. ✓
• Cl: $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^5$ — five $3\text{p}$ electrons give $3\text{p}_x^2\,3\text{p}_y^2\,3\text{p}_z^1$; two orbitals are completely filled. ✗
• Ar: $1\text{s}^2\,2\text{s}^2\,2\text{p}^6\,3\text{s}^2\,3\text{p}^6$ — all three $3\text{p}$ orbitals are filled. ✗

So only Mg and S satisfy the condition; X and Y are Mg and S.

Step 2 — Assign X and Y by ionisation energy.

First ionisation energies: $\text{Mg} \approx 738\text{ kJ mol}^{-1}$ and $\text{S} \approx 1000\text{ kJ mol}^{-1}$, so S has the greater value. Since Y has the greater first ionisation energy, $Y = \text{S}$ and $X = \text{Mg}$.

Step 3 — Structure and bonding.

• Mg is a metal, so it has a giant metallic structure — positive ions in a sea of delocalised electrons, held by metallic bonding.
• S is a non-metal, so it has a simple molecular structure — $\text{S}_8$ molecules held together by weak van der Waals forces.

X (Mg) = giant metallic; Y (S) = simple molecular. This matches option B.

Why the distractors fail:

• A: both giant metallic — S is a non-metal, not metallic.
• C: X giant covalent — X is Mg, a metal, not covalent.
• D: both simple molecular — Mg is a metal, not molecular.

Key Takeaways

• Hund's rule: electrons fill orbitals singly before pairing; a $\text{p}$ sub-shell has three orbitals.
• "Completely filled orbital" means an orbital with its full two electrons — count carefully.
• First ionisation energy rises across Period 3 but dips at Al and S; Mg still has a lower IE than S.
• Period 3 structures: Na/Mg/Al giant metallic, Si giant covalent, P/S/Cl simple molecular, Ar monatomic.

Common Mistakes

• Treating Al as qualifying because $3\text{s}$ is filled — the highest occupied sub-shell is $3\text{p}$, which is not filled.
• Counting Cl as having one filled orbital — it has two ($3\text{p}_x^2$ and $3\text{p}_y^2$).
• Assuming the IE dip at S makes S's IE lower than Mg's — it does not; S is still well above Mg.
• Confusing the structure of S (simple molecular) with that of Si (giant covalent).

Things to Be Careful About

• "Highest occupied energy sub-shell": for Al onwards this is $3\text{p}$, not $3\text{s}$.
• Hund's rule matters — $3\text{p}^2$ (Si) has no filled orbital because the two electrons occupy separate orbitals.
• The IE trend has dips at Al and S; use the correct relative order ($\text{Na} < \text{Al} < \text{Mg} < \text{Si} < \text{S} < \text{P} < \text{Cl} < \text{Ar}$).
• Mg is metallic; never classify it as covalent or molecular.

Background Concept

The ideal gas equation $pV = nRT$ describes the behaviour of an ideal gas, where $p$ is pressure (Pa), $V$ is volume (m$^3$), $n$ is amount (mol), $R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$ and $T$ is temperature (K). Because $n = \frac{m}{M_r}$, the equation can be rearranged to find the molar mass of a gas from its density.

Understanding the Question

The question gives the density of a pure gas ($2.62\text{ g dm}^{-3}$), its pressure ($101\,000\text{ Pa}$) and its temperature ($25\,^\circ\text{C}$), and asks which of four expressions correctly gives $M_r$. The traps are: temperature must be in kelvin, and the density must be converted so that the units match those of $R$.

Approach

Start from $pV = nRT$, substitute $n = \frac{m}{M_r}$, rearrange to make $M_r$ the subject, then replace $\frac{m}{V}$ with the density $\rho$. Convert temperature to kelvin and density to $\text{g m}^{-3}$, then compare with the options.

Step-by-Step Reasoning

1. Write $pV = nRT$ and substitute $n = \frac{m}{M_r}$: $pV = \frac{m}{M_r}RT$.
2. Rearrange: $M_r = \frac{mRT}{pV}$.
3. Since density $\rho = \frac{m}{V}$, this becomes $M_r = \frac{\rho RT}{p}$.
4. Convert temperature: $T = 25 + 273 = 298\text{ K}$.
5. Convert density: $2.62\text{ g dm}^{-3} = 2.62 \times 10^3\text{ g m}^{-3}$ because $1\text{ dm}^3 = 10^{-3}\text{ m}^3$.
6. Substitute: $M_r = \frac{2.62 \times 10^3 \times 8.31 \times 298}{101\,000} = \frac{2.62 \times 8.31 \times 298}{101\,000 \times 0.001}$.
7. This matches option D.

Key Takeaways

• The ideal gas equation rearranged for molar mass is $M_r = \frac{\rho RT}{p}$.
• Temperature always goes into gas-law calculations in kelvin.
• Units of $R$ (Pa m$^3$) dictate that volume must be in m$^3$; a density in g dm$^{-3}$ needs the factor $10^{-3}$.

Common Mistakes

• Using 25 instead of 298 K — this is the most common error and produces options A and C.
• Forgetting to convert dm$^3$ to m$^3$, which changes the power of ten.
• Inverting the fraction so that pressure ends up in the numerator.

Things to Be Careful About

• $R = 8.31\text{ J K}^{-1}\text{ mol}^{-1} = 8.31\text{ Pa m}^3\text{ K}^{-1}\text{ mol}^{-1}$.
• $1\text{ dm}^3 = 10^{-3}\text{ m}^3$, so g dm$^{-3}$ = $10^3$ g m$^{-3}$.
• Check that the final value of $M_r$ is chemically sensible (here about 64, e.g. SO$_2$).

Background Concept

When a fuel such as a hydrocarbon burns, the chemical energy released is transferred to the surroundings. In a simple calorimetry experiment, this energy is used to heat a known mass of water, and the energy absorbed by the water is calculated using

where $m$ is the mass of water being heated (in grams), $c$ is the specific heat capacity of water ($4.18 \text{ J g}^{-1}\text{ K}^{-1}$), and $\Delta T$ is the temperature rise. The key idea is that the water is the substance that absorbs the heat, so its mass and temperature change are what matter — the mass of fuel burnt is not part of this calculation.

Understanding the Question

The question states that $0.47\text{ g}$ of a hydrocarbon is completely burnt in air, and the energy released heats $200\text{ g}$ of water from $23.7\,^\circ\text{C}$ to $41.0\,^\circ\text{C}$. We are asked to find the energy absorbed by the water, in Joules, and to select the correct expression from four options. The options are deliberately constructed to trap two common errors: using the mass of the hydrocarbon instead of the water, and adding 273 to the temperature change.

Approach

The correct approach is to apply $q = mc\Delta T$ directly:

• Use the mass of the water, $m = 200\text{ g}$.
• Use the specific heat capacity of water, $c = 4.18\text{ J g}^{-1}\text{ K}^{-1}$.
• Calculate the temperature change: $\Delta T = 41.0 - 23.7 = 17.3\,^\circ\text{C}$.
• Substitute into the equation and match the resulting expression to the options.

The mass of the hydrocarbon ($0.47\text{ g}$) is a distractor — it tells us about the fuel, not about the water that absorbs the energy.

Step-by-Step Reasoning

1. Identify the substance that absorbs the energy: the water, with mass $200\text{ g}$.
2. Determine the temperature change: $$\Delta T = 41.0 - 23.7 = 17.3\,^\circ\text{C}$$ Because a temperature difference is the same in Celsius and Kelvin, we do not add 273. Adding 273 would only be correct for an absolute temperature, not for a change in temperature.
3. Substitute into the calorimetry equation: $$q = mc\Delta T = 200 \times 4.18 \times 17.3$$ This matches option C exactly.
4. Check the distractors:
   • Options A and B use $0.47$ (the mass of the hydrocarbon) instead of $200$ (the mass of water). The hydrocarbon is the source of energy, not the substance absorbing it, so its mass is irrelevant here.
   • Options B and D add 273 to the temperature change. This is wrong because $\Delta T$ is a difference, and $1\,^\circ\text{C}$ equals $1\text{ K}$; no conversion is needed.

Key Takeaways

• The calorimetry equation is $q = mc\Delta T$, where $m$ is the mass of the substance being heated.
• The temperature change in Celsius is numerically identical to the temperature change in Kelvin, so 273 is never added to a $\Delta T$.
• Always ask which substance actually absorbs the energy — here it is the water, not the fuel.

Common Mistakes

• Using the mass of the hydrocarbon ($0.47\text{ g}$) instead of the mass of water ($200\text{ g}$). The fuel releases energy; the water absorbs it.
• Adding 273 to the temperature change. This is a common error when students confuse absolute temperature with a temperature change. Since $\Delta T$ is a difference, $17.3\,^\circ\text{C} = 17.3\text{ K}$.

Things to Be Careful About

• The temperature change is a difference, so Celsius and Kelvin are interchangeable; never add 273 to a $\Delta T$.
• Units: $\text{g} \times \text{J g}^{-1}\text{K}^{-1} \times \text{K} = \text{J}$, which is consistent with the question asking for energy in Joules.
• The specific heat capacity of water is given as $4.18\text{ J g}^{-1}\text{ K}^{-1}$; use it as provided without modification.

Background Concept

This question tests two linked ideas: the electron-transfer definition of redox, and the classification of reactions as exothermic or endothermic.

Redox (reduction–oxidation) can be defined in terms of electron transfer. Oxidation is loss of electrons; reduction is gain of electrons. A useful mnemonic is OIL RIG (Oxidation Is Loss, Reduction Is Gain). When a metal such as iron reacts with oxygen, it is oxidised: each iron atom loses electrons and its oxidation state rises from 0 to +3.

An exothermic reaction releases heat energy to the surroundings, so the surroundings (and the reaction mixture) get hotter. An endothermic reaction absorbs heat energy from the surroundings, so the surroundings get colder. The sign of $\Delta H$ is negative for exothermic and positive for endothermic.

Understanding the Question

The heat pad contains iron, activated carbon and water. When it is activated by air (oxygen), it gets hotter. The question asks us to choose the statement that correctly describes the chemical reaction: whether it is exothermic or endothermic, and whether iron gains or loses electrons.

The key observation is that the pad gets hotter — this tells us heat is being released, so the reaction must be exothermic. The chemistry is the familiar oxidation of iron by oxygen, i.e. rusting. In rusting, iron is oxidised and therefore loses electrons.

Approach

1. Identify the reaction: iron reacting with oxygen in the air (rusting/oxidation).
2. Decide whether iron gains or loses electrons: oxidation is loss of electrons, so iron loses electrons.
3. Decide whether the reaction is exothermic or endothermic: the pad gets hotter, meaning heat is released, so the reaction is exothermic.
4. Match these two conclusions to the options.

Step-by-Step Reasoning

• The reaction in the heat pad is the oxidation of iron by oxygen:

• In this reaction, each iron atom goes from oxidation state 0 to +3, losing three electrons:

So iron loses electrons — it is oxidised. Oxygen is the oxidising agent (it accepts electrons) and iron is the reducing agent.

• The pad gets hotter when activated, which means heat energy is released to the surroundings. A reaction that releases heat is exothermic.
• Combining these two facts, the correct statement is: the reaction is exothermic and iron loses electrons. This is option D.

Why the other options are wrong:

• A and B say the reaction is endothermic. This is incorrect because the pad gets hotter — heat is given out, not absorbed.
• C says iron gains electrons. This is incorrect because iron is oxidised (it loses electrons). Gaining electrons would be reduction, which is what happens to oxygen, not iron.

Key Takeaways

• Oxidation is loss of electrons; reduction is gain of electrons (OIL RIG).
• A metal reacting with oxygen is oxidised and loses electrons.
• A reaction that makes its surroundings hotter is exothermic (releases heat).
• Rusting/oxidation of iron is an exothermic redox reaction.

Common Mistakes

• Confusing oxidation and reduction: thinking iron gains electrons because it 'combines' with oxygen. Iron actually loses electrons (it is oxidised), while oxygen gains them.
• Associating 'getting hot' with endothermic: heat released to the surroundings means exothermic. Endothermic reactions make the surroundings colder.
• Forgetting that oxidation number increases on oxidation: iron goes from 0 to +3, which is oxidation (loss of electrons).

Things to Be Careful About

• Always connect the temperature change of the surroundings to the enthalpy classification: hotter surroundings = exothermic, colder surroundings = endothermic.
• In redox questions, identify the oxidation state change of the element in question before deciding gain/loss of electrons.
• The mark scheme for this type of question requires both halves of the statement to be correct — the reaction type (exo/endothermic) and the electron transfer (gain/loss) — so check both before selecting an option.

Background Concept

The oxidation number of an element in a species is a bookkeeping number that represents the charge an atom would have if all bonds were ionic. The key rules are:

• The oxidation number of an element in its free state is 0.
• For a neutral compound, the sum of all oxidation numbers is 0.
• Oxygen usually has oxidation number $-2$.
• Hydrogen usually has oxidation number $+1$.
• Group 1 metals such as sodium have oxidation number $+1$.
• Halogens usually have oxidation number $-1$ when bonded to less electronegative elements.

When a compound contains more than one atom of the same element, the oxidation number may be different for each atom. The average oxidation number is found by treating all atoms of that element as having the same value, $x$, and solving the equation that makes the total charge equal to the charge on the species.

Understanding the Question

This question asks which substance contains sulfur with the highest average oxidation number. Four substances are given: $\text{S}_8$, $\text{Na}_2\text{S}_4\text{O}_6$, $\text{Na}_2\text{S}_2\text{O}_3$, and $\text{SO}_2\text{Cl}_2$. The word "average" is important because three of the compounds contain more than one sulfur atom, and in some of them the individual sulfur atoms have different oxidation numbers. We are not asked for individual oxidation numbers, only the average value for each compound.

Approach

For each neutral compound, set the sum of all oxidation numbers equal to zero. Let the oxidation number of sulfur be $x$ when there is one sulfur atom, or the average oxidation number of sulfur be $x$ when there are several. Then solve the resulting linear equation for $x$ and compare the values.

Step-by-Step Reasoning

A $\text{S}_8$ is the element sulfur in its standard state. Any element in its free state has oxidation number 0, so the average oxidation number of sulfur here is 0.

B $\text{Na}_2\text{S}_4\text{O}_6$ contains two sodium atoms, four sulfur atoms, and six oxygen atoms. Sodium is $+1$, oxygen is $-2$. Let the average oxidation number of sulfur be $x$.

So the average oxidation number of sulfur in tetrathionate is $+2.5$.

C $\text{Na}_2\text{S}_2\text{O}_3$ contains two sodium atoms, two sulfur atoms, and three oxygen atoms. Again sodium is $+1$ and oxygen is $-2$. Let the average oxidation number of sulfur be $x$.

So the average oxidation number of sulfur in thiosulfate is $+2$.

D $\text{SO}_2\text{Cl}_2$ contains one sulfur atom, two oxygen atoms, and two chlorine atoms. Oxygen is $-2$ and chlorine is $-1$. Let the oxidation number of sulfur be $x$.

Comparing the four values:

• $\text{S}_8$: 0
• $\text{Na}_2\text{S}_4\text{O}_6$: $+2.5$
• $\text{Na}_2\text{S}_2\text{O}_3$: $+2$
• $\text{SO}_2\text{Cl}_2$: $+6$

The highest average oxidation number is $+6$, so the correct answer is D.

Key Takeaways

• The oxidation number of a free element is always 0.
• In a neutral compound, the sum of all oxidation numbers is 0.
• When an element appears more than once in a formula, its average oxidation number is found by solving for $x$ after summing all contributions.
• Oxygen is usually $-2$ and chlorine is usually $-1$, so in $\text{SO}_2\text{Cl}_2$ sulfur must be $+6$.

Common Mistakes

• Forgetting that oxygen is $-2$ in $\text{SO}_2\text{Cl}_2$ and treating it as $+2$.
• Forgetting that chlorine is $-1$ in $\text{SO}_2\text{Cl}_2$.
• Not multiplying the oxidation number by the number of atoms present, e.g. writing $+1 + x - 2 = 0$ for $\text{Na}_2\text{S}_2\text{O}_3$ instead of $2(+1) + 2x + 3(-2) = 0$.
• Confusing individual oxidation numbers with the average. In thiosulfate, the two sulfur atoms have different oxidation numbers, but the question asks for the average.
• Assuming $\text{S}_8$ has a positive oxidation number because sulfur can show positive oxidation numbers; in its elemental form it is 0.

Things to Be Careful About

• Always check the charge of the species. All four substances here are neutral, so the sum of oxidation numbers must be 0.
• Include every atom in the formula when writing the equation.
• An average oxidation number can be fractional, as in $+2.5$ for $\text{Na}_2\text{S}_4\text{O}_6$; this is acceptable because it is an average.
• In $\text{SO}_2\text{Cl}_2$, both oxygen and chlorine are more electronegative than sulfur, so they take negative oxidation numbers and sulfur takes a high positive value.

Background Concept

A system at dynamic equilibrium has equal forward and backward rates, so concentrations remain constant. If a change is imposed, the position of equilibrium shifts to oppose that change (Le Chatelier's principle). For a reaction quotient comparison:

At equilibrium $Q_c = K_c$. Increasing a reactant concentration makes the denominator larger, so immediately after the addition $Q_c < K_c$; the reaction therefore proceeds in the forward direction, consuming reactant and producing products, until $Q_c$ rises back to $K_c$.

Understanding the Question

This is a qualitative equilibrium question. The mixture has volume $1.0\text{ dm}^3$ and is initially at equilibrium with $[\text{Q}] = 10\text{ mol dm}^{-3}$. Then $5.0\text{ mol}$ of pure Q is added and completely dissolves. Because the volume does not change, the immediate concentration of Q becomes $15\text{ mol dm}^{-3}$. The question asks for the new equilibrium concentration, not the concentration immediately after addition. The options test whether you know both the direction and the extent of the shift.

Approach

Use Le Chatelier's principle: adding a reactant shifts the equilibrium to the product side, which consumes some Q. That tells us the new [Q] must be less than 15. To see why it must still be greater than 10, consider that consuming all 5 mol of added Q would return [Q] to 10, but would also change P, R and S in a way that makes the reaction quotient larger than $K_c$; equilibrium is reached before that point. So the answer is the range between 10 and 15.

Step-by-Step Reasoning

1. The volume is $1.0\text{ dm}^3$, so adding $5.0\text{ mol}$ of Q raises its concentration by $5.0\text{ mol dm}^{-3}$.
2. Immediately after addition, $[\text{Q}] = 15\text{ mol dm}^{-3}$; the concentrations of P, R and S are unchanged.
3. The reaction quotient is now $Q_c = [\text{R}]^2[\text{S}]/([\text{P}][\text{Q}])$ with a larger denominator, so $Q_c < K_c$.
4. To restore equilibrium, the forward reaction is favoured: P and Q are consumed, R and S are formed, and [Q] falls.
5. The forward reaction stops when $Q_c = K_c$. It cannot consume all 5 mol of added Q: if it did, [Q] would be back to 10, but [P] would be lower and [R] and [S] higher than initially, so $Q_c$ would already be greater than $K_c$. Hence the new [Q] lies between 10 and 15.

Distractors:

• A ($15$) ignores the equilibrium shift.
• C ($10$) assumes the shift exactly cancels the addition, which would overshoot.
• D (between 5 and 10) is the wrong direction; adding a reactant shifts the equilibrium to the right, not the left.

Key Takeaways

• Le Chatelier's principle predicts the direction of shift when a concentration is changed.
• The shift opposes the change but does not completely reverse it; the new equilibrium concentration lies between the original and the disturbed value.
• The reaction quotient $Q_c$ compared with $K_c$ explains why the shift stops before all added reactant is consumed.
• For a fixed volume, moles added can be converted directly into a concentration change.

Common Mistakes

• Choosing A: forgetting that the system responds by shifting, so [Q] does not stay at 15.
• Choosing C: thinking equilibrium 'cancels' the disturbance. Le Chatelier only opposes it; returning exactly to 10 would require consuming all added Q, which is not the equilibrium position.
• Choosing D: reversing the direction of the shift. Adding a reactant favours the forward reaction, so [Q] must fall from 15, not from 10.
• Treating the question as a calculation requiring $K_c$; no $K_c$ value is needed.

Things to Be Careful About

• The volume is fixed at $1.0\text{ dm}^3$; 'completely dissolved' means the added Q is fully available and does not change the volume.
• Keep units as $\text{mol dm}^{-3}$.
• The question asks for the new equilibrium concentration, not the concentration immediately after addition.
• Le Chatelier's principle gives a qualitative range, not an exact value, unless $K_c$ and other concentrations are known.

Background Concept

For a gaseous equilibrium, the equilibrium constant $K_p$ is written in terms of the partial pressures of the gases. For a general reaction

The units of $K_p$ depend on the change in the number of moles of gas, $\Delta n = (c+d) - (a+b)$. If $\Delta n = 0$, $K_p$ has no units; otherwise it has units of pressure raised to $\Delta n$.

$K_p$ is constant for a given reaction at a given temperature. Changing pressure, adding a catalyst, or changing concentration does not change $K_p$; it may change the position of equilibrium, but not the value of the equilibrium constant. A catalyst lowers the activation energy for both the forward and reverse reactions equally, so it increases the rate at which equilibrium is reached but does not shift the equilibrium position.

Understanding the Question

This question asks which of four statements about the Contact process and the Haber process is correct. The statements test three ideas: the units of $K_p$, the effect of pressure on $K_p$, and the nature and effect of catalysts. You need to recall the balanced equations for both industrial processes and apply the definitions of $K_p$ and catalysts.

Approach

Write the balanced equations for both processes, derive the $K_p$ expressions, and check each statement in turn.

• For statement A, determine the units of $K_p$ from the stoichiometry of each reaction.
• For statement B, recall that $K_p$ changes only with temperature, not with pressure.
• For statement C, identify the phase of the catalyst and of the reactants/products in the Haber process.
• For statement D, recall that a catalyst does not change the position of equilibrium.

Step-by-Step Reasoning

Statement A:
The Contact process is