Chemistry 9701/11 — October/November 2024

Background Concept

Two periodic trends are needed here.

First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions. Down a group, the atomic radius increases and there is more shielding by inner electron shells. The outer electron is therefore less strongly attracted to the nucleus, so less energy is needed to remove it. First ionisation energy decreases down Group 2: Mg has a higher first ionisation energy than Ca.

Covalent bond energy is the energy required to break one mole of a covalent bond in the gaseous state. For the halogen molecules, the X–X bond energy decreases down Group 17. As atomic radius increases down the group, the X–X bond length increases and the overlap between the two atomic orbitals becomes weaker, so the bond is weaker. Thus Cl–Cl has a higher bond energy than Br–Br.

Understanding the Question

The question asks you to choose one compound from four possible combinations of a Group 2 element (Mg or Ca) and a halogen (Cl or Br). The compound must contain:

• the Group 2 element with the higher first ionisation energy;
• the halogen with the higher Q–Q bond energy.

No calculation is needed. You simply need to compare the two elements within each group and then combine the winners.

Approach

Handle the two trends separately:

1. Compare Mg and Ca using first ionisation energy.
2. Compare Cl and Br using Q–Q bond energy.
3. Combine the two chosen elements and write the correct formula.

The formula is determined by the charges of the ions: magnesium and calcium both form 2+ ions, while chlorine and bromine both form 1- ions. So the compound will have the general formula $\text{MX}_2$.

Step-by-Step Reasoning

Step 1: Choose the Group 2 element.

Mg is above Ca in Group 2. Down the group, the atomic radius increases and shielding increases, so the outer electron is held less tightly. Therefore the first ionisation energy of Mg is higher than that of Ca. The required Group 2 element is Mg.

Step 2: Choose the halogen.

Cl is above Br in Group 17. The Cl–Cl bond is shorter and stronger than the Br–Br bond because the atoms are smaller and the orbital overlap is better. Therefore the Q–Q bond energy of Cl is higher. The required halogen is Cl.

Step 3: Write the formula.

Magnesium forms $\text{Mg}^{2+}$ and chlorine forms $\text{Cl}^-$. The neutral compound is therefore $\text{MgCl}_2$.

This corresponds to option A.

The other options are incorrect because:

• $\text{MgBr}_2$ contains the wrong halogen (Br has lower Q–Q bond energy).
• $\text{CaCl}_2$ contains the wrong Group 2 element (Ca has lower first ionisation energy).
• $\text{CaBr}_2$ contains neither of the required elements.

Key Takeaways

• First ionisation energy decreases down a group because atomic radius and shielding increase.
• Halogen X–X bond energy decreases down Group 17 because bond length increases and orbital overlap weakens.
• When a question asks you to combine two separate trends, deal with each trend independently before writing the final formula.
• Ionic formula writing depends on the charges of the ions, not on the periodic trends.

Common Mistakes

• Saying that Ca has the higher first ionisation energy because it is heavier. Mass is not the deciding factor; atomic radius and shielding are.
• Saying that Br–Br has the higher bond energy because Br has more electrons. More electrons do not make a covalent bond stronger; shorter bonds with better orbital overlap are stronger.
• Choosing CaCl2 or CaBr2 because calcium is more reactive. Reactivity is not the property asked about.
• Writing MgCl instead of MgCl2. Magnesium is in Group 2 and forms a 2+ ion, so two chloride ions are needed.

Things to Be Careful About

• Read “higher first ionisation energy” carefully; it is not the same as electronegativity or reactivity.
• “Q–Q bond energy” refers to the bond in the halogen molecule, e.g. Cl–Cl, not to a bond in a compound.
• Compare elements only within their own group; do not compare Mg with Cl directly.
• Make sure the final formula is electrically neutral.

Background Concept

Electrons occupy sub-shells made of orbitals. A p sub-shell contains three orbitals, each able to hold up to two electrons. Hund's rule says that electrons first enter each orbital singly before any orbital receives a second electron. So the way orbitals are occupied tells us exactly how many p electrons an atom has:

• one occupied orbital: $p^1$
• two occupied orbitals: $p^2$
• three occupied orbitals, all singly: $p^3$
• three occupied orbitals, one doubly occupied: $p^4$
• three occupied orbitals, two doubly occupied: $p^5$
• three occupied orbitals, all doubly occupied: $p^6$

Combining this p-electron count with the period (the shell number) identifies the element.

Understanding the Question

This is a deduction question: from orbital-occupancy clues, identify two elements Y and Z, then choose the formula of the compound X they form. The stem gives two clues: Y is in Period 2 and its p sub-shell has all three orbitals occupied but only one fully occupied; Z is in Period 3 and only two of its p orbitals are occupied. We need to translate orbital occupancy into electron configuration and then into element identity.

Approach

Count p electrons from the orbital description. For Y, all three orbitals are occupied and only one is full: one orbital with two electrons and two orbitals with one electron each, so four p electrons. Since Y is in Period 2, its outer shell is $n = 2$, so it has $2p^4$; the element is oxygen. For Z, only two orbitals are occupied, meaning two p electrons; Period 3 gives $3p^2$; the element is silicon. Then identify the compound of oxygen and silicon among the options.

Step-by-Step Reasoning

1. Identify Y. A p sub-shell always has three orbitals. 'All three orbitals are occupied' means each has at least one electron. 'Only one of these three orbitals is fully occupied' means exactly one orbital has two electrons and the other two have one each. Total p electrons = $2 + 1 + 1 = 4$, so Y has configuration ending $2p^4$. In Period 2, the element with $2p^4$ is oxygen (O).

2. Identify Z. 'Only two orbitals are occupied' means the p sub-shell has two electrons, each in its own orbital: $3p^2$. In Period 3, this is silicon (Si).

3. Choose the formula. Oxygen and silicon combine as silicon dioxide, $\text{SiO}_2$. This matches option C.

4. Check the other options. $\text{CCl}_4$ would need Y = C (Period 2, but $2p^2$, not $2p^4$) and Z = Cl (Period 3, but $3p^5$, not $3p^2$). $\text{SiCl}_4$ has Y = Si, which is Period 3, not Period 2. $\text{SO}_2$ has Y = S (Period 3) and Z = O (Period 2), with the period clues reversed and the wrong occupancies. Only $\text{SiO}_2$ fits both clues.

Key Takeaways

• The number of occupied p orbitals and whether any are doubly occupied fixes the number of p electrons.
• The period number gives the principal quantum shell, so period + p-electron count identifies the element.
• Once the two elements are known, the formula of their compound follows from normal valency: silicon forms four bonds and oxygen forms two, giving $\text{SiO}_2$.

Common Mistakes

• Counting only the 'fully occupied' orbital and forgetting the two singly occupied orbitals: Y would be wrongly identified as $2p^2$ (carbon) instead of $2p^4$ (oxygen).
• Thinking 'only two orbitals are occupied' means three p electrons. Hund's rule prevents pairing before all three orbitals are singly occupied, so two occupied orbitals means $p^2$, not $p^3$.
• Mixing up Period 2 and Period 3 elements, for example choosing S and O for $\text{SO}_2$; the period clues must be matched to the correct element.
• Writing the wrong ratio in the formula: silicon dioxide is $\text{SiO}_2$, not $\text{SiO}$ or $\text{Si}_2\text{O}$.

Things to Be Careful About

• A 'fully occupied' orbital contains two electrons; an 'occupied' orbital contains at least one.
• Apply Hund's rule consistently when converting orbital occupancy to electron count.
• Use the period to fix the shell number: Period 2 means $n = 2$, Period 3 means $n = 3$.
• In the final formula, silicon's valency is 4 and oxygen's is 2, so the simplest ratio is 1:2, giving $\text{SiO}_2$.

Background Concept

Glauber's salt is a hydrated salt, meaning each formula unit contains a fixed number of water molecules trapped in the crystal lattice. These water molecules are called water of crystallisation. When the salt is heated, this water is driven off, leaving the anhydrous salt. The ratio of water molecules to salt formula units is given by $x$ in $\text{Na}_2\text{SO}_4 \cdot x\text{H}_2\text{O}$.

Understanding the Question

The question gives the mass of water removed ($1.91\text{ g}$) and the mass of anhydrous salt left behind ($1.51\text{ g}$). To find $x$, we need the mole ratio of water to anhydrous salt — this ratio IS the value of $x$.

Approach

1. Calculate the molar mass of anhydrous $\text{Na}_2\text{SO}_4$.
2. Convert the mass of anhydrous salt to moles.
3. Convert the mass of water to moles (molar mass $18.0\text{ g mol}^{-1}$).
4. Divide moles of water by moles of salt to get $x$.

Step-by-Step Reasoning

Step 1 — Molar mass of $\text{Na}_2\text{SO}_4$:
$M_r = 2(23.0) + 32.1 + 4(16.0) = 46.0 + 32.1 + 64.0 = 142.1\text{ g mol}^{-1}$

Step 2 — Moles of anhydrous salt:
$n = \frac{m}{M} = \frac{1.51}{142.1} = 0.01063\text{ mol}$

Step 3 — Moles of water:
$n = \frac{1.91}{18.0} = 0.1061\text{ mol}$

Step 4 — Mole ratio:
$x = \frac{0.1061}{0.01063} = 9.98 \approx 10$

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

Key Takeaways

• Water of crystallisation is found by comparing moles of water to moles of anhydrous salt.
• Always use the molar mass of the anhydrous salt, not the hydrated one, when converting the remaining mass.
• The mole ratio is a pure number — it has no units.

Common Mistakes

• Using the mass of the hydrated salt instead of the anhydrous salt in the mole calculation.
• Forgetting to divide by the molar mass of water (using $18$ incorrectly, or using $M_r$ of the whole hydrated salt).
• Rounding too early: the ratio $9.98$ rounds to $10$, and a student might incorrectly pick $8.85$ if they divide masses directly instead of moles.

Things to Be Careful About

• The masses given are for two different substances (water vs salt); never combine them directly.
• Check that the final $x$ is a sensible whole number — hydrated salts typically have small integer values of $x$.
• Option B ($8.85$) and D ($11.25$) are traps derived from incorrect mass-to-mole handling; option A ($1$) is far too small and would imply almost no water.

Background Concept

The number of particles in a sample depends on the amount of substance in moles, not directly on its mass or volume. One mole of any substance contains Avogadro's constant, $6.02 \times 10^{23}$ particles. At room temperature and pressure (r.t.p.), one mole of any gas occupies $24\text{ dm}^3$.

Understanding the Question

We are asked to compare the number of named particles in four different samples:

• $6.0\text{ dm}^3$ of argon atoms
• $6.0\text{ g}$ of carbon dioxide molecules
• $6.0\text{ g}$ of magnesium atoms
• $6.0\text{ g}$ of water molecules

Because the units are different, we cannot compare the numbers directly. We must convert each quantity to moles.

Approach

Convert every option to moles:

• For a gas volume: divide by the molar volume at r.t.p., $24\text{ dm}^3\text{ mol}^{-1}$.
• For a mass: divide by the molar mass ($M_r$ or $A_r$).

Then compare the mole values. The larger the number of moles, the larger the number of particles.

Step-by-Step Reasoning

1. Option A: $6.0\text{ dm}^3$ of argon at r.t.p.
   $n = \frac{6.0\text{ dm}^3}{24\text{ dm}^3\text{ mol}^{-1}} = 0.25\text{ mol}$
   Argon is monatomic, so this is $0.25$ mol of Ar atoms.

2. Option B: $6.0\text{ g}$ of carbon dioxide.
   $M(\text{CO}_2) = 12 + 2(16) = 44\text{ g mol}^{-1}$
   $n = \frac{6.0\text{ g}}{44\text{ g mol}^{-1}} = 0.136\text{ mol}$
   This is $0.136$ mol of CO$_2$ molecules.

3. Option C: $6.0\text{ g}$ of magnesium.
   $A_r(\text{Mg}) = 24\text{ g mol}^{-1}$
   $n = \frac{6.0\text{ g}}{24\text{ g mol}^{-1}} = 0.25\text{ mol}$
   This is $0.25$ mol of Mg atoms.

4. Option D: $6.0\text{ g}$ of water.
   $M(\text{H}_2\text{O}) = 2(1) + 16 = 18\text{ g mol}^{-1}$
   $n = \frac{6.0\text{ g}}{18\text{ g mol}^{-1}} = 0.333\text{ mol}$
   This is $0.333$ mol of H$_2$O molecules.

Comparing: $0.333 > 0.25 = 0.25 > 0.136$. Therefore, option D contains the greatest number of particles.

Key Takeaways

• The number of particles is proportional to the amount in moles, not to mass or volume.
• For gases at r.t.p., use $24\text{ dm}^3\text{ mol}^{-1}$.
• For solids, liquids, or gases given in grams, use the molar mass to find moles.
• Always compare moles when asked to compare numbers of particles.

Common Mistakes

• Using $22.4\text{ dm}^3\text{ mol}^{-1}$ instead of $24\text{ dm}^3\text{ mol}^{-1}$ for room conditions. The question specifies room conditions, so $24$ is correct.
• Thinking that a larger mass always means more particles. Mass must be converted to moles first.
• Using the wrong molar mass, e.g. treating Mg as $24.3$ instead of $24$; this would still not change the final answer because option D is clearly larger.
• Forgetting that argon exists as atoms, not molecules, but this does not affect the comparison here.

Things to Be Careful About

• Use the correct molar masses: CO$_2$ = 44, H$_2$O = 18, Mg = 24.
• Recognise that the named particle is different in each option: atoms for Ar and Mg, molecules for CO$_2$ and H$_2$O.
• At r.t.p. the molar volume is $24\text{ dm}^3\text{ mol}^{-1}$; at s.t.p. it would be $22.4\text{ dm}^3\text{ mol}^{-1}$, but the question says room conditions.

Background Concept

Phosphorus has five valence electrons. In $\text{PH}_3$, phosphorus forms three covalent bonds to hydrogen atoms, using three of these electrons. The remaining two valence electrons stay on phosphorus as a lone pair.

A dative covalent bond (also called a coordinate bond) is a covalent bond in which both of the shared electrons are supplied by one atom. The atom that supplies the electron pair is the donor; the atom that accepts the pair is the acceptor. Once formed, a dative covalent bond is no different in strength or nature from an ordinary covalent bond.

A hydrogen ion, $\text{H}^+$, has no electrons of its own. It is therefore an electron-pair acceptor. When it meets a molecule with a lone pair, such as $\text{PH}_3$, it can accept that lone pair to form a new covalent bond.

Understanding the Question

The question asks which type of interaction occurs between $\text{PH}_3$ and $\text{H}^+$. This is a classification question: you must identify the bonding interaction, not just describe it. The options are dative covalent bond, dipole–dipole forces, hydrogen bond, and ionic bond.

The key observation is that $\text{PH}_3$ has a lone pair on phosphorus, and $\text{H}^+$ is a bare proton with no electrons. This is exactly the situation that produces a coordinate bond, just as $\text{NH}_3$ forms $\text{NH}_4^+$ with $\text{H}^+$.

Approach

1. Determine the electron structure of $\text{PH}_3$: three bonding pairs and one lone pair on P.
2. Recognise that $\text{H}^+$ is an electron-pair acceptor.
3. Identify the bond formed as one where the lone pair on P is shared with $\text{H}^+$: a dative covalent bond.
4. Eliminate the other options by recognising that they describe different situations.

Step-by-Step Reasoning

• Phosphorus is in Group 15 and has five valence electrons. In $\text{PH}_3$, three electrons form P–H bonds, leaving one lone pair on P.
• $\text{H}^+$ is a hydrogen atom that has lost its electron, so it is just a proton with no valence electrons.
• The lone pair on phosphorus can be donated to $\text{H}^+$, forming $\text{PH}_4^+$. Both electrons in the new P–H bond come from the phosphorus lone pair, so the bond is dative covalent.
• Why not the other options?
  • Dipole–dipole forces are intermolecular attractions between neutral molecules. Here $\text{H}^+$ is an ion, and a covalent bond is being formed, not an intermolecular attraction.
  • A hydrogen bond requires a hydrogen atom covalently bonded to a very electronegative atom such as N, O, or F. Phosphorus is not sufficiently electronegative, and the interaction here is a bond formation, not a hydrogen bond.
  • An ionic bond involves transfer of electrons and attraction between opposit charged ions. $\text{H}^+$ is already an ion, but it gains a shared electron pair rather than a complete transfer, so the product is a covalently bonded ion, $\text{PH}_4^+$.

Key Takeaways

• A lone pair can be donated to an electron-pair acceptor to form a dative covalent bond.
• $\text{PH}_3$ behaves like $\text{NH}_3$ in this respect, although phosphorus is less electronegative than nitrogen.
• Dative covalent bonds are common whenever a molecule with a lone pair reacts with a species that needs an electron pair, such as $\text{H}^+$.

Common Mistakes

• Choosing hydrogen bond: this is tempting because $\text{PH}_3$ contains hydrogen, but hydrogen bonding requires H bonded to N, O, or F, and the interaction here is a bond-forming reaction, not an intermolecular force.
• Choosing ionic bond: $\text{H}^+$ is an ion, but the product is formed by sharing an electron pair, not by complete electron transfer.
• Thinking that dative bonds are weaker or different from ordinary covalent bonds after formation; they are identical once formed.

Things to Be Careful About

• Recognise that "dative covalent" and "coordinate" mean the same thing.
• Remember that $\text{PH}_3$ has a lone pair even though phosphorus is not very electronegative.
• Do not confuse an intermolecular force, such as dipole–dipole or hydrogen bonding, with an intramolecular covalent bond formed between $\text{PH}_3$ and $\text{H}^+$.

Background Concept

Across Period 3 (Na, Mg, Al, Si, P, S, Cl), four key physical properties exhibit distinct trends due to changes in atomic structure, nuclear charge, and bonding type:

• Atomic radius decreases steadily because the increasing nuclear charge pulls the electron shell closer, with electrons added to the same principal quantum level.
• First ionisation energy generally increases for the same reason, but with two notable dips: one at Al (electron removed from a higher-energy 3p orbital rather than a 3s orbital) and one at S (electron removed from a paired 3p orbital, increasing electron-electron repulsion).
• Melting point rises from Na to Si due to the transition from metallic bonding to a giant covalent structure (Si), then falls sharply for P, S, and Cl because these exist as simple molecular substances with weak van der Waals forces.
• Electronegativity (the ability of an atom to attract bonding electrons in a covalent bond) increases steadily across the period. This is because nuclear charge increases while atomic radius decreases, resulting in a stronger electrostatic attraction for shared electrons.

Understanding the Question

The question provides four line graphs (A, B, C, D) plotting different physical properties against the Period 3 elements Na through Cl. The task is to identify which graph represents the trend for electronegativity.

Approach

The strategy is to match the distinctive shape of each graph to the known trends of Period 3 properties. By eliminating the graphs that correspond to atomic radius, ionisation energy, and melting point, the remaining graph must be electronegativity.

Step-by-Step Reasoning

• Graph A shows a value that rises, plateaus slightly, peaks at Si, and then drops sharply. This matches the melting point trend: metallic bonding in Na, Mg, Al increases in strength, Si has a giant covalent lattice with a very high melting point, and P, S, Cl are simple molecules with low melting points.
• Graph B shows a continuous, smooth decrease from Na to Cl. This matches the atomic radius trend, as the increasing nuclear charge without additional electron shells pulls the outermost electrons closer to the nucleus.
• Graph C shows a general upward trend but with distinct dips at Al and S. This matches the first ionisation energy trend. The dip at Al occurs because the outermost electron is in a 3p orbital (higher energy, further from nucleus) rather than a 3s orbital. The dip at S occurs because the outermost electron is removed from a paired 3p orbital, where electron-electron repulsion makes it easier to remove.
• Graph D shows a steady, continuous increase from Na to Cl with no dips or drops. This matches the electronegativity trend. Because nuclear charge increases and atomic radius decreases steadily across the period, the attraction for bonding electrons increases smoothly without the quantum-mechanical anomalies that affect ionisation energy.

Key Takeaways

Memorise the characteristic shapes of the four main Period 3 property trends. Electronegativity is unique in that it increases steadily across the period without the dips seen in first ionisation energy.

Common Mistakes

Confusing the dip at Al in ionisation energy with the trend for electronegativity. Electronegativity does not have dips; it increases steadily because it is a continuous measure of nuclear attraction for shared electrons, not a discrete energy required to remove an electron.

Things to Be Careful About

Ensure you do not confuse atomic radius (which decreases) with electronegativity (which increases). Both are driven by the same underlying factors (increasing nuclear charge, constant shell number), but they measure opposite aspects of the atom's pull on electrons.

Background Concept

Allotropes are different structural forms of the same element in the same physical state. Tin exists as grey tin and white tin, and the two forms have different arrangements of atoms, so they have different physical properties.

The key idea is that physical properties can be used to deduce the type of giant structure:

• Giant covalent structures, such as diamond, consist of atoms joined by a network of strong covalent bonds. There are no free electrons or ions, so they do not conduct electricity in the solid or liquid state. They are often hard but brittle.
• Giant metallic structures consist of positive ions in a regular lattice surrounded by a sea of delocalised electrons. The delocalised electrons carry charge, so metals conduct electricity in both the solid and liquid states. Layers of ions can slide over each other, making metals malleable and ductile.
• Giant ionic structures consist of oppositely charged ions in a lattice. In the solid state the ions are fixed, so the solid does not conduct electricity; when molten or dissolved in water, the ions are free to move and the substance conducts.

The very high boiling points of both grey tin and white tin show that both are giant structures with strong bonding, but the difference in conductivity and malleability reveals that the type of giant structure is different.

Understanding the Question

The question provides a table of properties for grey tin and white tin: boiling point, electrical conductivity, and malleability. It asks which structural change might take place when grey tin changes to white tin.

This is not a question about a chemical reaction or about the elements involved. It is a question about identifying the type of giant structure from physical properties, then selecting the correct change from the four options.

Approach

The best strategy is to treat the properties as fingerprints for the type of structure:

1. Look at electrical conductivity in both the solid and liquid states.
2. Look at malleability.
3. Deduce the structure of grey tin.
4. Deduce the structure of white tin.
5. Match the change to the options.

Conductivity is especially useful because it distinguishes metallic, ionic, and covalent structures. Malleability then confirms whether a structure is metallic.

Step-by-Step Reasoning

Grey tin:

• It has no electrical conductivity in the solid state or the liquid state.
• A metallic structure would conduct in the solid state because delocalised electrons are free to move.
• A giant ionic structure would not conduct as a solid, but it would conduct when molten because the ions become mobile. Grey tin does not conduct when liquid, so it cannot be ionic.
• A giant covalent structure has no free electrons or mobile ions, so it does not conduct in either state. This matches grey tin.
• Grey tin is also brittle, which is consistent with a giant covalent network rather than a metallic structure.

So grey tin is a giant covalent structure.

White tin:

• It has good electrical conductivity in both the solid and liquid states.
• This is the classic behaviour of a metallic structure because the sea of delocalised electrons remains available to carry charge in both states.
• White tin is malleable, which is also characteristic of a metal: layers of positive ions can slide over one another without breaking the metallic bonding.

So white tin is a giant metallic structure.

Choosing the option:

The change is therefore from a giant covalent structure to a giant metallic structure.

• A is wrong because white tin is not giant ionic.
• B is correct.
• C is wrong because grey tin is not giant ionic.
• D is wrong because grey tin is not giant ionic.

Key Takeaways

• Physical properties such as electrical conductivity and malleability can be used to deduce the type of giant structure.
• Conductivity in the solid state distinguishes metallic structures from ionic and covalent structures.
• Conductivity in the liquid state distinguishes ionic structures from covalent network structures.
• Malleability is a strong indicator of a metallic structure.
• Allotropes of the same element can have very different structures and therefore very different physical properties.

Common Mistakes

• Assuming that a substance that does not conduct as a solid must be ionic. In fact, ionic solids do not conduct either; they only conduct when molten or in aqueous solution.
• Assuming that all non-metals form simple molecular structures. Grey tin is a non-metal allotrope but forms a giant covalent structure.
• Forgetting to consider conductivity in the liquid state. This is the key piece of evidence that rules out a giant ionic structure for grey tin.
• Confusing malleability with hardness. Malleability means the substance can be hammered or bent without breaking, which is a metallic property.

Things to Be Careful About

• Read the table carefully: grey tin has no conductivity in either state, while white tin conducts well in both states. The phrase "in solid or liquid" is crucial.
• The high boiling points of both forms are consistent with giant structures, but they do not by themselves distinguish covalent from metallic or ionic.
• An allotrope change is a physical change in structure, not a chemical change producing a new element.
• When matching to options, make sure the direction of the change is correct: grey tin to white tin, not the reverse.

Background Concept

Solids can be classified by the type of particles they contain and how those particles are held together.

• Ionic lattice: oppositely charged ions held by strong electrostatic forces, e.g. calcium fluoride.
• Metallic lattice: positive metal ions in a sea of delocalised electrons, e.g. nickel.
• Giant covalent lattice: a network of atoms joined by covalent bonds, e.g. silicon(IV) oxide.
• Simple molecular lattice: discrete molecules held together by weak intermolecular forces, e.g. sulfur.

The key distinction is whether the solid is made of separate molecules or of one continuous network of atoms/ions.

Understanding the Question

This is a one-mark recall question. It asks you to choose, from four common solids, the one that has a simple molecular lattice. The command is essentially "identify" — you need to know the structure of each solid, not calculate anything.

Approach

Go through each option and recall its structure and bonding. Eliminate any solid that is ionic, metallic, or giant covalent. The remaining one must be simple molecular.

Step-by-Step Reasoning

1. Calcium fluoride, CaF$_2$ — a compound of a metal and a non-metal. It forms an ionic lattice of Ca$^{2+}$ and F$^-$ ions. Not molecular.
2. Nickel, Ni — a metal. Its atoms are held in a metallic lattice with delocalised electrons. Not molecular.
3. Silicon(IV) oxide, SiO$_2$ — each silicon atom is covalently bonded to four oxygen atoms and each oxygen to two silicon atoms, forming a giant covalent network. It has no discrete molecules, so it is not simple molecular.
4. Sulfur, S$_8$ — sulfur exists as discrete S$_8$ rings. Within each ring the S atoms are joined by covalent bonds, but between rings only weak van der Waals forces act. This is a simple molecular lattice.

Therefore the correct option is D.

Key Takeaways

• Simple molecular solids are made of discrete molecules; the weak forces between molecules explain their low melting and boiling points.
• Giant covalent solids such as SiO$_2$ are not molecular even though they contain covalent bonds.
• Recognising the type of particle (molecule, ion, atom, metal ion) is the fastest way to classify a solid's lattice.

Common Mistakes

• Choosing silicon(IV) oxide because it contains covalent bonds. Covalent bonds alone do not make a solid molecular — the structure must consist of discrete molecules.
• Thinking all non-metal solids are molecular. Silicon(IV) oxide is a counterexample because it is a giant covalent network.
• Confusing a metallic lattice with a molecular one because both may be described as "regular arrangements".

Things to Be Careful About

• Sulfur is usually written as S$_8$, not S, when referring to its molecular structure.
• Simple molecular solids are soft, have low melting points, and do not conduct electricity, unlike ionic, metallic, or giant covalent solids.
• The term "lattice" applies to all four structures; the key word is "simple molecular", which specifically means discrete molecules.

Background Concept

This question tests Hess's law: the enthalpy change of a reaction is independent of the route taken, depending only on the initial and final states. Two useful consequences are:

• The enthalpy of combustion of a compound can be found from the enthalpies of combustion of its constituent elements:
  $\Delta H_c^\ominus(\text{compound}) = \sum \Delta H_c^\ominus(\text{products}) - \sum \Delta H_c^\ominus(\text{reactants})$
• The enthalpy of formation is the enthalpy change when one mole of a compound forms from its elements in their standard states.

Understanding the Question

We are given:

• $\Delta H_c^\ominus(\text{C}) = -394\text{ kJ mol}^{-1}$ (combustion of carbon)
• $\Delta H_c^\ominus(\text{H}_2) = -286\text{ kJ mol}^{-1}$ (combustion of hydrogen)
• $\Delta H_f^\ominus(\text{C}_4\text{H}_{10}) = -129\text{ kJ mol}^{-1}$ (formation of butane)

We must find the standard enthalpy change of combustion of butane, $\text{C}_4\text{H}_{10}$.

Approach

The combustion of butane produces $4\text{CO}_2$ and $5\text{H}_2\text{O}$. The same products can be reached by first forming butane from its elements (formation), then burning those elements. Equating these two routes via Hess's law gives a direct formula.

Step-by-Step Reasoning

1. Write the combustion equation for butane:
   $\text{C}_4\text{H}_{10} + 6\tfrac{1}{2}\text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}$

2. Identify the elements and their combustion enthalpies:

   • Carbon: 4 moles, each $-394$ → $4 \times (-394) = -1576$
   • Hydrogen: 5 moles of $\text{H}_2$, each $-286$ → $5 \times (-286) = -1430$

3. Apply Hess's law:
   The combustion enthalpy of butane equals the sum of the combustion enthalpies of its elements minus the enthalpy of formation of butane (because formation is the reverse of "decomposing" the compound into elements):
   $\Delta H_c^\ominus(\text{C}_4\text{H}_{10}) = 4(-394) + 5(-286) - (-129)$

4. Calculate:
   $-1576 - 1430 + 129 = -2877\text{ kJ mol}^{-1}$

5. Match to the options: Option B is $-2877\text{ kJ mol}^{-1}$.

Key Takeaways

• The general formula for combustion of a hydrocarbon $\text{C}_x\text{H}_y$ is:
  $\Delta H_c^\ominus = x\Delta H_c^\ominus(\text{C}) + \tfrac{y}{2}\Delta H_c^\ominus(\text{H}_2) - \Delta H_f^\ominus(\text{compound})$
• Formation enthalpy is subtracted because forming the compound from elements is the reverse of the "decomposition" step in the cycle.

Common Mistakes

• Forgetting to subtract the formation enthalpy — this gives $-3006$ (not an option, but a common slip).
• Adding the formation enthalpy instead of subtracting — this gives $-3135$ (option C), a classic distractor.
• Using the wrong number of moles of $\text{H}_2$ — butane has 10 H atoms, so 5 moles of $\text{H}_2$ are needed, not 10.

Things to Be Careful About

• All values are negative (exothermic), so watch the signs when combining.
• The formation enthalpy of butane is negative, so subtracting it means adding its magnitude.
• Double-check the stoichiometry of the combustion equation before multiplying.

Things to Be Careful About

• All values are negative (exothermic), so watch the signs when combining.
• The formation enthalpy of butane is negative, so subtracting it means adding its magnitude.
• Double-check the stoichiometry of the combustion equation before multiplying.

Background Concept

The enthalpy change of a process, $\Delta H$, is the heat energy transferred at constant pressure. Its sign tells us the direction of energy flow between the system and the surroundings:

• Exothermic ($\Delta H$ negative): the system releases energy to the surroundings, usually as heat. The products are more stable (lower in energy) than the reactants.
• Endothermic ($\Delta H$ positive): the system absorbs energy from the surroundings. The products are higher in energy than the reactants.

Certain classes of reaction are always (or almost always) exothermic because bond-forming or particle-attraction steps release more energy than is consumed:

• Neutralisation: acid + base → salt + water. The formation of the strong O–H bonds in water releases energy.
• Combustion: a fuel reacts with oxygen. The C=O and O–H bonds formed in $\text{CO}_2$ and $\text{H}_2\text{O}$ are very strong, releasing far more energy than is needed to break the fuel's bonds.
• Condensation (gas → liquid) and freezing (liquid → solid): as particles come closer together, intermolecular forces form, releasing energy. The reverse processes (vaporisation, melting) are endothermic.

Understanding the Question

This question presents three chemical or physical processes and asks which statement about their enthalpy changes is correct. The options all concern whether "none", or "only one" of the processes has a positive enthalpy change. The key task is to assign the correct sign of $\Delta H$ to each process by recognising the type of process it is. No numerical calculation is needed — this is purely qualitative recognition.

Approach

For each process, identify what kind of transformation it represents:

1. $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})$ — this is the ionic equation for a neutralisation reaction.
2. $\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})$ — this is the combustion of methane.
3. $\text{NH}_3(\text{g}) \rightarrow \text{NH}_3(\text{l})$ — this is condensation of ammonia from gas to liquid.

Then recall the sign of $\Delta H$ for each class and decide whether any is positive.

Step-by-Step Reasoning

Process 1 — neutralisation.
$\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})$

This is the net ionic equation for any acid–base neutralisation. The standard enthalpy of neutralisation for a strong acid and strong base is about $-57 \text{ kJ mol}^{-1}$, which is exothermic. The formation of the covalent O–H bonds in water releases more energy than is required to separate the ions. So $\Delta H < 0$.

Process 2 — combustion.
$\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})$

This is the complete combustion of methane, the standard enthalpy of combustion of methane being about $-890 \text{ kJ mol}^{-1}$. Breaking the bonds in $\text{CH}_4$ and $\text{O}_2$ requires energy, but forming the very strong C=O bonds in $\text{CO}_2$ and O–H bonds in water releases much more. The net effect is a large release of energy, so $\Delta H < 0$.

Process 3 — condensation.
$\text{NH}_3(\text{g}) \rightarrow \text{NH}_3(\text{l})$

When ammonia gas condenses, its molecules come closer together and hydrogen bonds (and weaker van der Waals forces) form between them. Forming these intermolecular forces releases energy. Condensation is therefore exothermic, so $\Delta H < 0$. (The reverse process, vaporisation, is endothermic.)

Conclusion. All three processes are exothermic. None has a positive enthalpy change, so the correct option is A.

Key Takeaways

• The sign of $\Delta H$ is determined by the type of process: neutralisation, combustion, and condensation are all exothermic (negative $\Delta H$); vaporisation, melting, and thermal decomposition are typically endothermic (positive $\Delta H$).
• Recognising the class of a reaction from its equation is often all that is needed to assign the sign of $\Delta H$.
• Condensation and freezing release energy because forming intermolecular forces is exothermic; the reverse changes (vaporisation, melting) absorb energy.

Common Mistakes

• Thinking condensation is endothermic because "going from gas to liquid" sounds like gaining order. In fact, forming bonds/forces always releases energy; it is vaporisation that absorbs energy.
• Confusing the sign convention: a negative $\Delta H$ means exothermic (energy released), and a positive $\Delta H$ means endothermic (energy absorbed).
• Assuming all reactions with gases are endothermic — combustion involves gases but is strongly exothermic because of the strong bonds formed in the products.

Things to Be Careful About

• The question asks which statement is correct about a positive enthalpy change. Since all three processes are exothermic, the answer is the option that says "none" have a positive enthalpy change.
• Note that process 3 is a physical change (condensation), not a chemical reaction; the same sign logic applies because it involves forming intermolecular forces.
• No calculation is required here, but in a calculation question you would need to be careful with the sign of $\Delta H$ and with state symbols, since the state of water (liquid vs gas) affects the value of $\Delta H$ for combustion.

Background Concept

This question tests redox balancing. In a redox reaction, oxidation and reduction happen simultaneously, and the total number of electrons lost by the reducing agent must exactly equal the total number of electrons gained by the oxidising agent. The quickest way to find the stoichiometric ratio is to compare the change in oxidation number of each element.

Understanding the Question

The permanganate ion $\text{MnO}_4^-$ is reduced to $\text{MnO}_2$ in alkaline solution, while the sulfite ion $\text{SO}_3^{2-}$ is oxidised to sulfate $\text{SO}_4^{2-}$. We are asked for the ratio of $\text{MnO}_4^-$ to $\text{SO}_3^{2-}$ in the balanced equation — i.e. how many of each ion must react so that electrons balance.

Approach

1. Find the oxidation number of Mn in both $\text{MnO}_4^-$ and $\text{MnO}_2$.
2. Find the oxidation number of S in both $\text{SO}_3^{2-}$ and $\text{SO}_4^{2-}$.
3. Work out electrons gained per $\text{MnO}_4^-$ and lost per $\text{SO}_3^{2-}$.
4. Scale both so the electrons gained equal the electrons lost.

Step-by-Step Reasoning

Step 1 — Oxidation number of Mn.
In $\text{MnO}_4^-$, O is $-2$ each, so Mn is $+7$. In $\text{MnO}_2$, O is $-2$ each, so Mn is $+4$. Change: $+7 \rightarrow +4$, a gain of $3$ electrons per $\text{MnO}_4^-$.

Step 2 — Oxidation number of S.
In $\text{SO}_3^{2-}$, O is $-2$ each, so S is $+4$. In $\text{SO}_4^{2-}$, O is $-2$ each, so S is $+6$. Change: $+4 \rightarrow +6$, a loss of $2$ electrons per $\text{SO}_3^{2-}$.

Step 3 — Balance electrons.
The lowest common multiple of $3$ and $2$ is $6$. Therefore $2$ permanganate ions gain $6$ electrons and $3$ sulfite ions lose $6$ electrons.

Step 4 — Ratio.
$\text{MnO}_4^- : \text{SO}_3^{2-} = 2 : 3$, which is option A.

As a check, the balanced equation is:
$2\text{MnO}_4^- + 3\text{SO}_3^{2-} + \text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + 3\text{SO}_4^{2-} + 2\text{OH}^-$
Atoms and charge balance correctly.

Key Takeaways

• Oxidation number changes directly tell you how many electrons each species transfers.
• The stoichiometric ratio is found by making electrons gained equal electrons lost.
• Always check the final equation balances for both atoms and charge.

Common Mistakes

• Using the wrong oxidation number for Mn in $\text{MnO}_4^-$ ($+7$, not $+6$).
• Reversing the ratio — the question asks for $\text{MnO}_4^-$ first, then $\text{SO}_3^{2-}$.
• Forgetting that $\text{SO}_3^{2-}$ is oxidised (loses electrons), not reduced.

Things to Be Careful About

• In alkaline solution the reduction product of permanganate is $\text{MnO}_2$ (Mn $+4$), not $\text{Mn}^{2+}$ (which forms in acid).
• Oxygen is always $-2$ in these ions (except in peroxides), so oxidation numbers are easy to assign.
• Read the ratio order in the question carefully before selecting an option.

Background Concept

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and re-establish equilibrium.

For the Haber process equilibrium:

• Pressure effect: The left side has 4 moles of gas; the right side has 2 moles. Increasing pressure shifts equilibrium to the right (fewer moles of gas). Decreasing pressure shifts it to the left (more moles of gas).
• Concentration effect: Removing a reactant shifts equilibrium to the left to replace it. Removing a product shifts equilibrium to the right to replace it.

Lithium is a reactive metal that reacts readily with nitrogen at room temperature to form lithium nitride:

Understanding the Question

We have three connected vessels:

• Left: evacuated (vacuum, zero gas)
• Middle: equilibrium mixture of $\text{N}_2$, $\text{H}_2$, and $\text{NH}_3$
• Right: solid lithium

We want to find which action maximises the amount of $\text{NH}_3$ in the middle vessel at the new equilibrium.

Approach

Evaluate each option by asking: does the disturbance shift the equilibrium to the right (more $\text{NH}_3$) or to the left (less $\text{NH}_3$)?

• Opening tap 1 → volume change → pressure change → Le Chatelier shift
• Opening tap 2 → chemical reaction removing a reactant → Le Chatelier shift
• Both closed → no disturbance → original equilibrium maintained

Step-by-Step Reasoning

Option A — Keep both taps closed:
No disturbance is applied. The system remains at its original equilibrium. The amount of $\text{NH}_3$ is whatever it was initially. This is our baseline.

Option C — Open tap 1 only:
The evacuated vessel is connected to the middle vessel. The gas now occupies twice the volume (two equal vessels instead of one). Total pressure drops. Le Chatelier's principle: the system shifts to counteract the pressure drop by producing more gas molecules. The left side has 4 moles of gas vs 2 on the right, so equilibrium shifts left. Amount of $\text{NH}_3$ decreases.

Option D — Open tap 2 only:
Lithium reacts with nitrogen gas: $6\text{Li}(\text{s}) + \text{N}_2(\text{g}) \rightarrow 2\text{Li}_3\text{N}(\text{s})$. Nitrogen is removed from the equilibrium mixture. Le Chatelier's principle: the system shifts to replace the lost $\text{N}_2$, so equilibrium shifts left. Amount of $\text{NH}_3$ decreases.

Option B — Open both taps:
Both disturbances act simultaneously. Pressure drops (tap 1) AND nitrogen is removed (tap 2). Both shifts are to the left. Amount of $\text{NH}_3$ decreases the most.

Therefore, keeping both taps closed (Option A) preserves the original equilibrium and gives the most ammonia.

Key Takeaways

• Le Chatelier's principle applies to any disturbance: pressure changes, concentration changes, temperature changes.
• Removing a reactant always shifts equilibrium toward the reactants (left), reducing product yield.
• Increasing volume (decreasing pressure) shifts equilibrium toward the side with more moles of gas.
• The option that causes no disturbance preserves the maximum product amount.

Common Mistakes

• Assuming opening tap 2 would increase $\text{NH}_3$: Students sometimes think lithium "helps" the reaction. In reality, lithium consumes $\text{N}_2$, a reactant, which starves the equilibrium and shifts it left.
• Confusing the direction of the pressure shift: Remember: fewer moles of gas on the right (2) vs more on the left (4). Pressure decrease → shift to MORE moles → left → less $\text{NH}_3$.
• Thinking "evacuated" means it absorbs gas: The evacuated vessel simply provides extra volume, reducing pressure. It does not chemically interact.

Things to Be Careful About

• Count moles of gas on each side of the equilibrium carefully: $1 + 3 = 4$ on the left, $2$ on the right.
• State symbols matter: $\text{Li}_3\text{N}$ is solid, so it doesn't contribute to gas moles.
• The question asks for the "most ammonia" — this means the option that causes the least reduction (or no reduction) from the original equilibrium amount.
• Keep taps closed = no change = original equilibrium = maximum $\text{NH}_3$ under these conditions. Any action taken makes things worse.

Background Concept

For a reversible reaction at equilibrium, the equilibrium constant $K_c$ is the ratio of the equilibrium concentrations of products to reactants, each raised to the power of its stoichiometric coefficient.

For the reaction

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

Although $K_c$ is defined using concentrations in mol dm$^{-3}$, when all species are in the same container of volume $V$, each concentration is $n/V$. Substituting these into the expression gives:

So the volume cancels and equilibrium mole amounts can be used directly.

Understanding the Question

The question gives the initial amounts of H$_2$ and I$_2$, and the equilibrium amount of HI. We are asked to choose the correct expression for $K_c$ from four options. The key is that $K_c$ must use equilibrium amounts, not initial amounts, so we first need to work out how much H$_2$ and I$_2$ remain at equilibrium.

The container volume is not given, but that does not matter because it cancels in the $K_c$ expression for this reaction.

Approach

1. Use the stoichiometry of the equation to find how much H$_2$ and I$_2$ have reacted.
2. Subtract the reacted amounts from the initial amounts to find the equilibrium amounts.
3. Write the $K_c$ expression and substitute the equilibrium amounts.

Step-by-Step Reasoning

• Initial amounts: H$_2$ = 0.20 mol, I$_2$ = 0.15 mol, HI = 0 mol.
• At equilibrium, HI = 0.26 mol, so the change in HI is +0.26 mol.
• From the equation, 2 mol HI are produced from 1 mol H$_2$ and 1 mol I$_2$. Therefore, the amount of H$_2$ consumed is:

$\frac{0.26}{2} = 0.13 \text{ mol}$

and the amount of I$_2$ consumed is also 0.13 mol.

• Equilibrium amount of H$_2$:

$0.20 - 0.13 = 0.07 \text{ mol}$

• Equilibrium amount of I$_2$:

$0.15 - 0.13 = 0.02 \text{ mol}$

• Substitute into the $K_c$ expression:

This matches option C.

Why the other options are wrong:

• Option A uses $2 \times 0.26$ in the numerator, which incorrectly treats the coefficient 2 as a multiplier of the amount rather than as a power.
• Option B squares $2 \times 0.26$, making the same error and also using initial amounts in the denominator.
• Option D uses 0.13 and 0.13 in the denominator, which are the amounts consumed, not the equilibrium amounts remaining.

Key Takeaways

• Always use equilibrium amounts, not initial amounts, in $K_c$.
• Use the stoichiometry of the balanced equation to find how much of each reactant is consumed.
• The stoichiometric coefficients become powers in the $K_c$ expression.
• When all species are in the same container, the volume cancels and mole amounts can be used directly.

Common Mistakes

• Using initial moles (0.20 and 0.15) instead of equilibrium moles (0.07 and 0.02).
• Forgetting that 0.26 mol of HI is produced from 0.13 mol of each reactant, not 0.26 mol.
• Writing the numerator as $2 \times 0.26$ instead of $(0.26)^2$.
• Inverting the expression, putting reactants over products.

Things to Be Careful About

• The coefficient 2 in front of HI means the concentration of HI is squared in the numerator.
• The volume is not needed because it cancels, but if the volume were different for different species, concentrations would have to be used.
• For this reaction, the units of $K_c$ cancel because there are two moles of gas on each side of the equation, so $K_c$ is dimensionless.

Background Concept

The rate of a reaction at a specific instant (the instantaneous rate) is given by the gradient of the tangent to the concentration-time curve at that time. Since concentration is directly proportional to absorbance (according to Beer-Lambert law), the gradient of an absorbance-time curve is also proportional to the rate of reaction. A calibration curve is used to establish the exact proportionality constant between absorbance and concentration, allowing the conversion of an absorbance gradient into a concentration gradient.

Understanding the Question

We are given two graphs: a calibration curve plotting iodine concentration against absorbance, and an absorbance-time graph for the reaction. We need to find the rate of reaction at $t = 20 \text{ s}$. This requires finding the rate of change of iodine concentration at that specific time, which means finding the gradient of the tangent to the absorbance-time graph at $t = 20 \text{ s}$ and then converting that gradient into concentration units using the calibration curve.

Approach

1. Use the calibration curve to find the conversion factor between absorbance and concentration of $\text{I}_2$.
2. Draw (or estimate) a tangent to the absorbance-time curve at $t = 20 \text{ s}$ and calculate its gradient ($\Delta A / \Delta t$).
3. Multiply the absorbance gradient by the conversion factor to get the rate of change of concentration. The rate of reaction is the positive value of this gradient (since $\text{I}_2$ is a reactant, its concentration decreases).

Step-by-Step Reasoning

Step 1: Calibration curve analysis
Looking at the left graph, the y-axis is concentration of iodine $\times 10^{-3} \text{ mol dm}^{-3}$ and the x-axis is absorbance. The line passes through $(0.5, 0.25)$. This means an absorbance of $0.5$ corresponds to a concentration of $0.25 \times 10^{-3} \text{ mol dm}^{-3}$. Since the line passes through the origin, the relationship is:

Step 2: Tangent gradient at $t = 20 \text{ s}$
On the right graph, at $t = 20 \text{ s}$, the absorbance is $0.3$. The curve is decreasing. To find the instantaneous rate, we need the gradient of the tangent at this point. Observing the points on the curve: at $t = 10 \text{ s}$, $A = 0.4$; at $t = 20 \text{ s}$, $A = 0.3$; at $t = 30 \text{ s}$, $A = 0.2$. These three points are nearly collinear, so the tangent at $t = 20 \text{ s}$ can be approximated by the line connecting $(10, 0.4)$ and $(30, 0.2)$.

This means the absorbance is decreasing at a rate of $0.01$ per second.

Step 3: Calculate the rate
The rate of reaction with respect to $\text{I}_2$ is: