Chemistry 9701/12 — May/June 2024

Background Concept

Ionic compounds are electrically neutral overall. The formula of an ionic compound is written so that the total positive charge equals the total negative charge. This means you must know the charge on each ion:

• Ag⁺, K⁺, NH₄⁺ are +1
• Zn²⁺ is +2
• CO₃²⁻, SO₄²⁻, PO₄³⁻ are polyatomic anions
• HCO₃⁻ (hydrogencarbonate) is −1
• NO₃⁻ (nitrate) is −1
• PO₄³⁻ (phosphate) is −3

When combining ions, use subscripts to balance charges. For example, Ca²⁺ and PO₄³⁻ combine as Ca₃(PO₄)₂ because 3 × (+2) = +6 balances 2 × (−3) = −6.

Understanding the Question

The question gives four pairs of formulae and asks which pair is entirely correct. You must check every formula in each pair — a pair is only correct if both formulae are correct. This is a common exam style: one wrong formula in a pair makes the whole option wrong.

Approach

For each option, write the ions, note their charges, and check whether the subscripts balance the total charge. Move quickly through the options, eliminating any that contain a single incorrect formula.

Step-by-Step Reasoning

Option A:

• Ag₂CO₃: Ag⁺ and CO₃²⁻ → 2 × (+1) + (−2) = 0, so Ag₂CO₃ is correct.
• (NH₄)₃NO₃: NH₄⁺ and NO₃⁻ are both 1:1 ions, so the formula should be NH₄NO₃. The subscript 3 on NH₄ is wrong. This option is incorrect.

Option B:

• K₂HCO₃: K⁺ and HCO₃⁻ are both 1:1 ions, so the formula should be KHCO₃. The subscript 2 on K is wrong. This option is incorrect.
• Zn₃(PO₄)₂: Zn²⁺ and PO₄³⁻ → 3 × (+2) = +6 and 2 × (−3) = −6, so Zn₃(PO₄)₂ is correct.

Option C:

• AgHCO₃: Ag⁺ and HCO₃⁻ are both 1:1 ions, so AgHCO₃ is correct.
• K₃PO₄: K⁺ and PO₄³⁻ → 3 × (+1) = +3 and 1 × (−3) = −3, so K₃PO₄ is correct.
• Both are correct, so C is the answer.

Option D:

• ZnCO₃: Zn²⁺ and CO₃²⁻ are both 2:2, so ZnCO₃ is correct.
• (NH₄)₂PO₄: NH₄⁺ is +1 and PO₄³⁻ is −3, so three NH₄⁺ are needed: (NH₄)₃PO₄. The subscript 2 is wrong. This option is incorrect.

Key Takeaways

• Always balance total positive and total negative charges when writing ionic formulae.
• Common 1:1 pairs like K⁺/HCO₃⁻ or Ag⁺/HCO₃⁻ need no subscripts.
• Polyatomic ions with charges of different magnitude (e.g. +1 vs −3) require brackets and subscripts: (NH₄)₃PO₄.
• In a "which pair is correct" question, one wrong formula disqualifies the entire option — check every formula.

Common Mistakes

• Assuming that because one formula in a pair is correct, the whole pair is correct.
• Forgetting the charge on a polyatomic ion (e.g. phosphate is −3, not −2).
• Writing (NH₄)₃NO₃ instead of NH₄NO₃ — the ammonium and nitrate charges are both 1, so no subscript is needed on either.

Things to Be Careful About

• Memorise the charges of common polyatomic ions: nitrate NO₃⁻, hydrogencarbonate HCO₃⁻, carbonate CO₃²⁻, sulfate SO₄²⁻, phosphate PO₄³⁻.
• Brackets are used only when a polyatomic ion appears more than once in the formula (e.g. Zn₃(PO₄)₂, (NH₄)₃PO₄).

Background Concept

The mole is the amount of substance that contains the Avogadro constant, $L$, of particles ($L = 6.02 \times 10^{23} \text{ mol}^{-1}$). The number of particles in a sample is found by multiplying the amount in moles by $L$:

$\text{number of particles} = n \times L$

To find $n$ from a given mass, divide the mass by the molar mass of the substance:

$n = \frac{m}{M_r}$

A key trap in this question is identifying the particle. White phosphorus exists as $\text{P}_4$ molecules, not individual P atoms. The molar mass must therefore be calculated for the whole $\text{P}_4$ molecule, not for a single phosphorus atom.

Understanding the Question

The question gives a mass of solid white phosphorus, 62 g, and asks how many molecules are present. The options are all expressed in terms of $L$, the Avogadro constant. So the answer must be a simple multiple or fraction of $L$. The command word is implicit — a calculation. The crucial detail is that the species is $\text{P}_4$, so each molecule contains four phosphorus atoms and has a molar mass of 124 g mol⁻¹.

Approach

1. Find the molar mass of $\text{P}_4$ by multiplying the relative atomic mass of phosphorus (31) by 4.
2. Convert the given mass (62 g) into moles using $n = m/M_r$.
3. Multiply the number of moles by $L$ to get the number of molecules, and match the result to the options.

Step-by-Step Reasoning

Step 1 — Molar mass of P₄.
Phosphorus has a relative atomic mass of 31. Since one $\text{P}_4$ molecule contains four phosphorus atoms:

$M_r(\text{P}_4) = 4 \times 31 = 124 \text{ g mol}^{-1}$

Step 2 — Amount in moles.
Using $n = m/M_r$:

$n = \frac{62}{124} = 0.5 \text{ mol}$

Step 3 — Number of molecules.
Each mole contains $L$ molecules, so 0.5 mol contains:

$0.5 \times L = \frac{L}{2}$

This matches option C. Option A ($L$) would be the number of molecules in 124 g; option B ($2L$) would correspond to 248 g; option D ($L/4$) would correspond to 31 g, i.e. the number of P atoms in 62 g of $\text{P}_4$ if one mistakenly treated the mass as one mole of atoms.

Key Takeaways

• Always identify the actual particle in the formula: $\text{P}_4$ molecules, not P atoms.
• The number of particles = (mass ÷ molar mass) × $L$.
• When the answer options are expressed in terms of $L$, the working naturally produces a multiple or fraction of $L$.

Common Mistakes

• Using the atomic mass of P (31) instead of the molecular mass of $\text{P}_4$ (124). This would give 2 mol and the wrong answer B ($2L$).
• Confusing atoms with molecules: 62 g of $\text{P}_4$ contains 2 mol of P atoms but only 0.5 mol of $\text{P}_4$ molecules.
• Forgetting to divide by the molar mass before multiplying by $L$.

Things to Be Careful About

• State the particle clearly — the question asks for molecules, so the $\text{P}_4$ unit is the counting particle.
• Keep units consistent: mass in grams with molar mass in g mol⁻¹ gives moles directly.
• The relative atomic mass of phosphorus is 31 (to the nearest whole number), which is the standard value used in this syllabus.

Background Concept

Successive ionisation energies increase gradually as electrons are removed from the same principal energy level (shell). However, a very large jump in ionisation energy occurs when an electron is removed from a shell closer to the nucleus, because inner-shell electrons experience a much greater effective nuclear charge and are held more tightly. The position of this large jump indicates the number of valence (outer-shell) electrons, which directly corresponds to the element's group number in the Periodic Table.

Understanding the Question

We are given two graphs showing the first eight successive ionisation energies for two unknown elements, both confirmed to be in Period 3. We must use the position of the large jumps in each graph to determine the number of outer-shell electrons for each element, identify their groups, deduce their identities, and finally write the formula of the ionic compound they form.

Approach

1. Identify the number of valence electrons for the element in Graph 1 by finding where the large jump occurs.
2. Identify the number of valence electrons for the element in Graph 2 using the same method.
3. Map these group numbers to Period 3 elements to find their chemical symbols and typical ionic charges.
4. Combine the ions in a ratio that balances the overall charge to write the empirical formula.

Step-by-Step Reasoning

• Graph 1: The ionisation energy increases steadily from the 1st to the 7th electron removed. Between the 7th and 8th electrons, there is a very large jump. This means the first 7 electrons are removed from the outer shell, and the 8th electron is being removed from a full inner shell (the noble gas core). The element has 7 valence electrons, placing it in Group 17. The Period 3 Group 17 element is chlorine (Cl). Chlorine gains one electron to achieve a stable octet, forming a Cl⁻ ion.
• Graph 2: The ionisation energy increases steadily from the 1st to the 2nd electron removed. Between the 2nd and 3rd electrons, there is a large jump. This indicates the first 2 electrons are removed from the outer shell, and the 3rd is removed from a full inner shell. The element has 2 valence electrons, placing it in Group 2. The Period 3 Group 2 element is magnesium (Mg). Magnesium loses two electrons to achieve a stable octet, forming a Mg²⁺ ion.
• Formula construction: To form a neutral ionic compound, the total positive charge must equal the total negative charge. One Mg²⁺ ion (charge +2) requires two Cl⁻ ions (charge -1 each) to balance. The resulting empirical formula is MgCl₂.
• Checking options: Option A is MgCl₂, which matches our derivation. Options B, C, and D include elements like Ca, K, Br, and Se, which are not in Period 3 (Ca and K are Period 4; Br and Se are Period 4), so they can be immediately discarded.

Key Takeaways

• A large jump in successive ionisation energies indicates the removal of an electron from a new, inner principal energy level closer to the nucleus.
• The number of electrons removed before the large jump equals the number of valence electrons, which determines the group number.
• Combining group numbers with the constraint that the specified elements must belong to the stated period allows accurate prediction of ionic compound formulas.

Common Mistakes

• Misreading the graph and counting the jump incorrectly (e.g., thinking the jump is after the 6th electron instead of the 7th).
• Forgetting to check the period constraint, leading to incorrect element identification (e.g., choosing fluorine or oxygen instead of chlorine, ignoring that they are Period 2).
• Confusing the total number of electrons removed with the number of valence electrons.

Things to Be Careful About

• Always count the number of electrons removed before the large jump to find the number of valence electrons. The jump itself is between electron $n$ and electron $n+1$.
• Ensure the identified elements actually belong to the period specified in the question (Period 3 here). This is a critical filter that eliminates distractor options like CaBr₂ or K₂Se.
• When writing the ionic formula, ensure the charges are balanced to give a neutral compound; do not simply concatenate the symbols without considering the stoichiometry (e.g., writing MgCl instead of MgCl₂).

Background Concept

A free radical is any species that has one or more unpaired electrons. Free radicals are highly reactive because an unpaired electron strongly tends to pair up with another electron. The definition does not require the species to be neutral: charged species can also be radicals if they contain unpaired electrons.

To decide whether a species is a radical, count its total electrons and write the outer electron configuration. Then apply Hund's rule: electrons occupy orbitals singly before pairing.

Understanding the Question

The question lists four pairs of species and asks in which pairs both species are free radicals. It is not enough for one member of a pair to be a radical; both must have unpaired electrons.

Approach

For each species, determine the number of electrons after applying the charge, then look at the outermost p sub-shell. If the p sub-shell is not completely filled (not p⁶), there will usually be unpaired electrons.

Step-by-Step Reasoning

1. Cl

   • Neutral chlorine has 17 electrons: [Ne] 3s² 3p⁵.
   • 3p⁵ has one unpaired electron.
   • Cl is a free radical.

2. O

   • Neutral oxygen has 8 electrons: 1s² 2s² 2p⁴.
   • 2p⁴ has two unpaired electrons.
   • O is a free radical.

3. Cl⁻

   • Chloride has 18 electrons: [Ne] 3s² 3p⁶.
   • 3p⁶ is completely filled; all electrons are paired.
   • Cl⁻ is not a free radical.

4. O²⁻

   • Oxide ion has 10 electrons: 1s² 2s² 2p⁶.
   • 2p⁶ is completely filled; all electrons are paired.
   • O²⁻ is not a free radical.

5. Cl⁺

   • Cl⁺ has 16 electrons: [Ne] 3s² 3p⁴.
   • 3p⁴ has two unpaired electrons.
   • Cl⁺ is a free radical, even though it is positively charged.

6. O²⁺

   • O²⁺ has 6 electrons: 1s² 2s² 2p².
   • 2p² has two unpaired electrons.
   • O²⁺ is a free radical, even though it is positively charged.

Now check each pair:

• Pair 1: Cl and O → both radicals ✔
• Pair 2: Cl⁻ and O²⁻ → neither is a radical ✘
• Pair 3: Cl and O⁻ → both radicals ✔
• Pair 4: Cl⁺ and O²⁺ → both radicals ✔

Therefore, the correct answer is A.

Key Takeaways

• A free radical is defined by the presence of unpaired electrons, not by whether the species is neutral.
• Noble-gas configurations such as 3p⁶ or 2p⁶ have all electrons paired and are not radicals.
• p², p³, p⁴ and p⁵ configurations all contain unpaired electrons.

Common Mistakes

• Assuming that ions cannot be radicals. Cl⁺ and O²⁺ are charged but still have unpaired electrons.
• Thinking that a positive charge means all electrons are removed from the outer shell. Only the required number of electrons is removed, leaving unpaired electrons in the p sub-shell.
• Forgetting that O²⁻ has a filled 2p⁶ configuration and therefore no unpaired electrons.

Things to Be Careful About

• Always apply the charge when counting electrons.
• Remember Hund's rule: electrons do not pair until each orbital in a sub-shell has one electron.
• Check both species in each pair, not just one.

Background Concept

VSEPR (Valence Shell Electron Pair Repulsion) theory predicts molecular shape from the number of electron pairs (both bonding and lone) surrounding a central atom. Electron pairs repel each other and arrange themselves as far apart as possible. The total number of electron pairs determines the electron-pair geometry, while lone pairs, which repel more strongly than bonding pairs, compress bond angles and alter the observed molecular shape.

Key electron-pair geometries:

• 2 electron pairs → linear (e.g. BeCl2)
• 3 electron pairs → trigonal planar (e.g. BF3)
• 4 electron pairs → tetrahedral (e.g. CH4)

When lone pairs are present, the molecular shape differs from the electron-pair geometry:

• 4 pairs with 1 lone pair → pyramidal (e.g. NH3)
• 4 pairs with 2 lone pairs → bent/angular (e.g. H2O)

Understanding the Question

The question lists four combinations of bonded electron pairs and lone pairs, each paired with a shape, and asks which combination is correctly predicted by VSEPR theory. The trap is that the total number of electron pairs (bonding + lone) sets the electron-pair geometry, and the lone pairs then replace atom positions, changing the molecular shape.

Approach

1. For each option, add the bonded pairs and lone pairs to get the total number of electron pairs.
2. Determine the electron-pair geometry from the total.
3. Account for the lone pairs: each lone pair occupies a position but holds no atom, altering the molecular shape.
4. Compare the resulting molecular shape with the shape given in the option.

Step-by-Step Reasoning

• A (2 bonded, 2 lone): Total = 4 pairs → tetrahedral electron-pair arrangement. With 2 lone pairs, the shape is bent (angular), e.g. H2O. Not linear. ✗
• B (2 bonded, 2 lone): Same as A — bent, not tetrahedral. ✗
• C (3 bonded, 1 lone): Total = 4 pairs → tetrahedral electron-pair arrangement. With 1 lone pair, the shape is pyramidal, e.g. NH3. ✓
• D (3 bonded, 1 lone): Same electron-pair arrangement as C — pyramidal, not trigonal planar. Trigonal planar requires 3 bonded pairs and 0 lone pairs. ✗

Therefore, C is the correct answer.

Key Takeaways

• The total number of electron pairs (bonding + lone) determines the electron-pair geometry.
• Lone pairs reduce the number of atoms but keep the electron-pair geometry, changing the molecular shape.
• 4 electron pairs with 1 lone pair → pyramidal; with 2 lone pairs → bent.

Common Mistakes

• Confusing electron-pair geometry with molecular shape. Both NH3 and H2O have tetrahedral electron-pair geometry, but their molecular shapes are pyramidal and bent respectively.
• Thinking 2 bonded + 2 lone gives a linear shape — it actually gives a bent shape.
• Forgetting that trigonal planar requires exactly 3 bonded pairs and no lone pairs.

Things to Be Careful About

• Always count lone pairs as electron pairs — they are not invisible; they affect the shape.
• Remember that lone pairs repel more strongly than bonding pairs, so bond angles are compressed (e.g. NH3 is about 107°, H2O about 104.5°).
• Read the options carefully — both C and D have the same electron-pair counts but different shapes; only C is correct.

Background Concept

Atoms in covalent molecules are usually described as having a complete outer shell when they are surrounded by eight valence electrons, known as the octet rule. For second-period elements such as boron, carbon, nitrogen, oxygen and fluorine, the valence shell is the second shell, which can hold a maximum of eight electrons. Some atoms, especially boron, can be stable with only six valence electrons around them; such species are described as electron-deficient or having an incomplete octet.

Understanding the Question

The question asks which underlined atom has an incomplete outer shell. We must count the electrons that actually surround the underlined atom in its molecule or ion, not just the electrons it contributes as a free atom. For each option, add the electrons in bonds (two per bond) and any lone pairs on the atom, and adjust for any charge on the species.

Approach

For each species, identify the central/underlined atom, count its valence electrons in the free atom, then add shared electrons from bonds and lone pairs, and apply the charge. If the total is 8, the octet is complete; if it is 6, the outer shell is incomplete.

Step-by-Step Reasoning

• A: BF3 — Boron has 3 valence electrons. It forms three single B-F bonds, so there are 3 bonding pairs = 6 electrons around B. There is no lone pair. Total = 6, so the outer shell is incomplete.
• B: CH3- — Carbon has 4 valence electrons. It forms three C-H bonds and carries one lone pair. Total around C = 3 bonding pairs (6) + 1 lone pair (2) = 8. The octet is complete.
• C: F2O — Oxygen has 6 valence electrons. It forms two O-F bonds and has two lone pairs. Total around O = 2 bonding pairs (4) + 2 lone pairs (4) = 8. The octet is complete.
• D: H3O+ — Oxygen has 6 valence electrons. It forms three O-H bonds and has one lone pair. Total around O = 3 bonding pairs (6) + 1 lone pair (2) = 8. The octet is complete.
  Only BF3 has an incomplete outer shell, so the answer is A.

Key Takeaways

• The octet rule counts all electrons around an atom: bonding pairs and lone pairs.
• Boron compounds such as BF3 are classic electron-deficient species with only six valence electrons.
• A negative or positive charge changes the electron count and must be included.

Common Mistakes

• Counting only the atom's own valence electrons and forgetting the electrons contributed by bonds.
• Thinking that three bonds means a full octet; three bonds plus no lone pair gives only six electrons.
• Forgetting to include the lone pair on carbon in CH3- or on oxygen in F2O and H3O+.

Things to Be Careful About

• In H3O+, the positive charge means one electron has been removed, but oxygen still has three bonding pairs and one lone pair, giving eight electrons.
• The underlined atom is the one to focus on; the other atoms in the formula are not being tested.

Background Concept

The ideal gas equation, $pV = nRT$, relates the pressure, volume, temperature and amount of a gas. The gas constant $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$ fixes the units that must be used: pressure in pascals (Pa), volume in cubic metres (m³), temperature in kelvin (K) and amount in moles (mol). Because the question gives the mass of gas rather than the amount, we first convert mass to moles using $n = m/M_r$. Nitrogen is a diatomic molecule, $\text{N}_2$, so its molar mass is $2 \times 14.0 = 28.0 \text{ g mol}^{-1}$.

Understanding the Question

We are told to assume nitrogen behaves as an ideal gas. We are given 1.00 g of $\text{N}_2$ at 50.0 °C and 120 kPa, and asked to select the volume occupied from four options. This is a direct application of $pV = nRT$; the challenge is handling the units correctly — temperature must be in kelvin, pressure in pascals, and the final volume converted from m³ to dm³.

Approach

The strategy is to convert every quantity into the units required by $pV = nRT$, then rearrange for $V$.

1. Find the amount of nitrogen in moles: $n = m/M_r$.
2. Convert the temperature from °C to K by adding 273.15.
3. Convert the pressure from kPa to Pa (multiply by 1000).
4. Rearrange $pV = nRT$ to give $V = nRT/p$ and substitute.
5. Convert the volume from m³ to dm³ (multiply by 1000) to match the answer options.

Step-by-Step Reasoning

Step 1 — amount of nitrogen.

$M_r(\text{N}_2) = 2 \times 14.0 = 28.0$

$n = \frac{1.00}{28.0} = 0.0357 \text{ mol}$

Step 2 — temperature in kelvin.

$T = 50.0 + 273.15 = 323.15 \text{ K}$

Step 3 — pressure in pascals.

$p = 120 \text{ kPa} = 1.20 \times 10^5 \text{ Pa}$

Step 4 — apply the ideal gas equation.

Step 5 — convert to dm³.

Since $1 \text{ m}^3 = 1000 \text{ dm}^3$:

$V = 7.99 \times 10^{-4} \times 1000 = 0.799 \text{ dm}^3$

This matches option B.

Why the distractors are wrong:

• Option C (1.60 dm³) arises from using $M_r = 14.0$ (for a single nitrogen atom) instead of 28.0 for $\text{N}_2$. That doubles the number of moles and hence doubles the volume.
• Option A (0.124 dm³) arises from using $T = 50 \text{ K}$ — forgetting to add 273.15 — which makes the volume too small by roughly a factor of 6.5.
• Option D (22.4 dm³) is the molar volume of an ideal gas at STP (0 °C, 1 atm). It ignores both the mass (1 g, not 1 mol) and the given conditions.

Key Takeaways

• Always express temperature in kelvin when using $pV = nRT$.
• Always use SI units: pascals for pressure, cubic metres for volume.
• Remember that nitrogen is diatomic: $M_r(\text{N}_2) = 28.0$, not 14.0.
• Convert the final volume to the unit requested by the question (dm³).
• The molar volume $22.4 \text{ dm}^3$ applies only at STP and for one mole — it is not a general constant.

Common Mistakes

• Using $M_r = 14.0$ for nitrogen instead of 28.0 — this doubles the moles and gives option C.
• Forgetting to convert °C to K, giving option A.
• Mixing units, e.g. using kPa with m³, which produces a volume about 1000 times too large.
• Assuming the volume is $22.4 \text{ dm}^3$ regardless of conditions (option D).

Things to Be Careful About

• The gas constant $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$ requires p in Pa and V in m³. If you use kPa, you must convert to Pa first.
• $1 \text{ dm}^3 = 10^{-3} \text{ m}^3$, so converting from m³ to dm³ means multiplying by 1000.
• Report the final answer to an appropriate number of significant figures — here the data are given to three significant figures, so 0.799 dm³ is appropriate.

Background Concept

The boiling point of a molecular substance is the temperature at which its vapour pressure equals atmospheric pressure. It reflects the total strength of the intermolecular forces that must be overcome. The main types, in order of increasing strength, are induced dipole–induced dipole (London/van der Waals) forces, permanent dipole–dipole forces, and hydrogen bonding.

For non-polar molecules, van der Waals forces depend on surface area and number of electrons. Branched isomers have a more compact shape, less contact surface, and therefore weaker van der Waals forces than their straight-chain isomers. For polar molecules, the functional group controls the dominant interaction: an O–H group can form strong hydrogen bonds, whereas an S–H group is much less polar and does not hydrogen bond nearly as effectively.

Understanding the Question

We are given four small organic compounds: two C4 alkanes (one branched, one straight-chain), a C3 alcohol, and a C3 thiol. The question asks for the order of increasing boiling point, lowest first. This is a conceptual ranking, not a calculation. The options differ mainly in two places: whether the branched alkane or the straight-chain alkane is lower, and whether the thiol or the alcohol is higher.

Approach

First separate the non-polar alkanes from the polar functionalised compounds. For the two alkanes, compare branching and surface area. For the two functionalised compounds, compare the strength of the intermolecular forces available: O–H hydrogen bonding versus the weaker S–H interactions. Then combine the two groups into a single order.

Step-by-Step Reasoning

1. Compounds 1 and 4 are both C4H10 alkanes: 2-methylpropane and butane. They are non-polar, so only van der Waals forces act. Compound 1 is branched, giving a more compact shape and less surface contact, so its van der Waals forces are weaker and it boils lower than compound 4. Thus 1 < 4.

2. Compound 2 is propan-1-ol. The O–H group can form hydrogen bonds, which are much stronger than van der Waals forces. It therefore has the highest boiling point of the four compounds.

3. Compound 3 is propan-1-thiol. The S–H bond is less polar than O–H, and sulfur does not form hydrogen bonds as effectively as oxygen. The main interactions in the thiol are dipole–dipole and van der Waals forces, which are stronger than those in alkanes but weaker than the hydrogen bonding in the alcohol. Hence 3 < 2.

4. Combining these comparisons gives 1 < 4 < 3 < 2, which is option B.

For reference, approximate boiling points are: 2-methylpropane about -12 °C, butane about -0.5 °C, propan-1-thiol about 68 °C, and propan-1-ol about 97 °C. These values confirm the ranking.

Key Takeaways

• Boiling point generally follows: branched non-polar < straight-chain non-polar < polar (dipole–dipole) < hydrogen-bonded.
• Branching lowers the boiling point of an alkane by reducing surface area and weakening van der Waals forces.
• O–H hydrogen bonding is much stronger than the interactions available in a thiol, so an alcohol boils higher than the corresponding thiol.

Common Mistakes

• Assuming all C4 alkanes have the same boiling point; branching matters.
• Thinking that a thiol hydrogen-bonds as strongly as an alcohol; S–H is much less polar than O–H.
• Confusing “increasing” with “decreasing” and choosing the highest-boiling compound first.
• Using molar mass alone; compounds 2 and 3 have lower molar masses than the alkanes but still boil higher because of their polar interactions.

Things to Be Careful About

• Read “increasing” as lowest first.
• Recognise that compounds 1 and 4 are structural isomers, so the branching effect is the key difference between them.
• Strong hydrogen bonding requires H attached to N, O, or F; S–H does not give comparable hydrogen bonding.
• In an MCQ, eliminate options by testing the two key comparisons: 1 vs 4 and 2 vs 3.

Background Concept

Bond energy (bond dissociation enthalpy) is the energy required to break one mole of a specific covalent bond in the gaseous state. Breaking bonds is endothermic (energy must be put in), while forming bonds is exothermic (energy is released).

For any reaction, the enthalpy change can be approximated using:

$\Delta H = \sum(\text{bond energies of bonds broken}) - \sum(\text{bond energies of bonds formed})$

This is an approximation because tabulated bond energies are average values taken over many different compounds.

Understanding the Question

We are given the reaction:

$\text{C}_2\text{H}_6 + 2\text{F}_2 \rightarrow \text{C}_2\text{H}_4\text{F}_2 + 2\text{HF} \qquad \Delta H = -960 \text{ kJ mol}^{-1}$

Bond energies are provided for C–H, C–C, F–F and H–F. The C–F bond energy is unknown and must be deduced from the overall enthalpy change.

Approach

1. Write out the structural formulae so every bond is visible.
2. Count every bond broken in the reactants and sum their energies.
3. Count every bond formed in the products, treating the C–F bond energy as an unknown $x$.
4. Substitute into $\Delta H = \sum\text{broken} - \sum\text{formed}$ and solve for $x$.

Step-by-Step Reasoning

Step 1 — Bonds in the reactants.

• Ethane, $\text{C}_2\text{H}_6$ (structure $\text{H}_3\text{C}{-}\text{CH}_3$), contains 1 C–C bond and 6 C–H bonds.
  • Energy = $350 + 6(410) = 350 + 2460 = 2810 \text{ kJ}$.
• Two molecules of $\text{F}_2$ contain 2 F–F bonds.
  • Energy = $2(158) = 316 \text{ kJ}$.
• Total energy to break all reactant bonds = $2810 + 316 = 3126 \text{ kJ}$.

Step 2 — Bonds in the products.

• 1,2-Difluoroethane, $\text{FCH}_2{-}\text{CH}_2\text{F}$, contains 1 C–C, 4 C–H and 2 C–F bonds.
  • Energy = $350 + 4(410) + 2x = 350 + 1640 + 2x = 1990 + 2x$.
• Two molecules of HF contain 2 H–F bonds.
  • Energy = $2(562) = 1124 \text{ kJ}$.
• Total energy released on forming all product bonds = $1990 + 2x + 1124 = 3114 + 2x$.

Step 3 — Apply the bond-energy equation.
$\Delta H = \sum\text{broken} - \sum\text{formed}$
$-960 = 3126 - (3114 + 2x)$
$-960 = 3126 - 3114 - 2x$
$-960 = 12 - 2x$
$2x = 972$
$x = 486 \text{ kJ mol}^{-1}$

So the C–F bond energy is 486 kJ mol⁻¹, which is option C.

Key Takeaways

• Always convert molecular formulae into structural formulae before counting bonds.
• The bond-energy method uses $\Delta H = \sum\text{broken} - \sum\text{formed}$; a negative $\Delta H$ means more energy is released in bond formation than is absorbed in bond breaking.
• When one bond energy is unknown, set it as a variable and solve.

Common Mistakes

• Counting 6 C–H bonds in the product instead of 4 — the two hydrogens replaced by fluorine are gone.
• Forgetting that $2\text{F}_2$ means two F–F bonds, not one.
• Forgetting the two C–F bonds in the product — the molecule is 1,2-difluoroethane, so there are two fluorine atoms.
• Mixing up which side is "broken" and which is "formed", producing the wrong sign in the equation.

Things to Be Careful About

• The reaction is exothermic ($\Delta H$ is negative), so the total bond energy of products is greater than that of reactants.
• Bond energies are average values, so the result is an approximation.
• Keep units consistent — all energies here are in kJ mol⁻¹.
• Check the arithmetic at the end: $2x = 972$ gives $x = 486$, not 972 (which is the energy for two C–F bonds).

Background Concept

The standard enthalpy change of formation, $\Delta H_f^\ominus$, is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (usually 298 K and 1 atm pressure). For sodium oxide, the elements are sodium metal, $\text{Na(s)}$, and oxygen gas, $\text{O}_2\text{(g)}$. The standard state of an element is its most stable physical form at those conditions: sodium is a solid, oxygen is a gas.

Because enthalpy is an extensive property, the enthalpy change for a reaction is proportional to the amount of substance. Forming two moles of $\text{Na}_2\text{O}$ releases twice the enthalpy change of forming one mole. This is why the definition insists on exactly one mole of product.

Understanding the Question

This is a definition-based multiple-choice question. It asks which equation, among four, has an enthalpy change equal to the standard enthalpy of formation of sodium oxide. You are not given any numerical data; you only need to apply the definition precisely. The command word is implicit: identify the equation that matches the definition.

Approach

Apply the definition step by step:

1. The product must be exactly one mole of $\text{Na}_2\text{O(s)}$.
2. The reactants must be the elements in their standard states: $\text{Na(s)}$ and $\text{O}_2\text{(g)}$.
3. The equation must be balanced.

Then check each option against these three requirements.

Step-by-Step Reasoning

• Option A: $\text{Na(s)} + \frac{1}{4}\text{O}_2\text{(g)} \rightarrow \frac{1}{2}\text{Na}_2\text{O(s)}$ forms only half a mole of sodium oxide. The enthalpy change would be half of the standard enthalpy of formation, so A is incorrect.
• Option B: $\text{Na(s)} + \text{O}_2\text{(g)} \rightarrow \text{Na}_2\text{O(s)}$ is not balanced: there is one Na atom on the left but two on the right, and two O atoms on the left but one on the right. It cannot represent the formation of sodium oxide, so B is incorrect.
• Option C: $2\text{Na(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{Na}_2\text{O(s)}$ is balanced, has one mole of $\text{Na}_2\text{O}$ as product, and uses the elements in their standard states. This is exactly the definition, so C is correct.
• Option D: $4\text{Na(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{Na}_2\text{O(s)}$ is balanced, but it forms two moles of sodium oxide. The enthalpy change would be twice the standard enthalpy of formation, so D is incorrect.

Therefore, the correct answer is C.

Key Takeaways

• The standard enthalpy of formation always refers to forming one mole of a compound from its elements in their standard states.
• Fractional coefficients are allowed for reactants in a formation equation, because the product must have coefficient 1.
• Always check state symbols: the elements must be in their standard states.
• Enthalpy is extensive: doubling the amount of substance doubles the enthalpy change.

Common Mistakes

• Choosing D because it is a balanced equation with whole-number coefficients. It forms two moles of product, so it represents twice the standard enthalpy of formation.
• Choosing B without checking that it is balanced. A formation equation must be balanced.
• Thinking that fractional coefficients are not allowed. In enthalpy of formation equations, fractional coefficients for reactants are standard and necessary.
• Forgetting that sodium's standard state is solid and oxygen's is gas; state symbols matter.

Things to Be Careful About

• The definition requires exactly one mole of product; the coefficient of $\text{Na}_2\text{O}$ must be 1.
• The reactants must be elements in their standard states: $\text{Na(s)}$ and $\text{O}_2\text{(g)}$, not $\text{Na(l)}$ or $\text{O(g)}$.
• Check balancing of both atoms and charge (though charge is not relevant here).
• In an exam, if you are unsure, write the formation equation yourself and compare it with the options.

Background Concept

Disproportionation is a redox reaction in which the same element is simultaneously oxidised and reduced. One part of the element increases in oxidation number (loses electrons) while another part decreases in oxidation number (gains electrons).

To detect this, assign oxidation numbers using the standard rules:

• The oxidation number of oxygen is usually -2.
• The oxidation number of hydrogen is usually +1.
• In a neutral molecule, the sum of all oxidation numbers is zero.

An oxidising agent is the species that is reduced (gains electrons); a reducing agent is the species that is oxidised (loses electrons).

Understanding the Question

This question gives the reaction between nitrogen dioxide and water:

$2\text{NO}_2 + \text{H}_2\text{O} \rightarrow \text{HNO}_2 + \text{HNO}_3$

It asks which statement about the reaction is correct. The options test whether you can follow the oxidation numbers of nitrogen, oxygen, and hydrogen, and whether you can identify disproportionation and the role of water.

Approach

Assign oxidation numbers to every element in the reactants and products. Then compare the oxidation numbers of each element before and after the reaction. Look for one element that appears with a higher oxidation number in one product and a lower oxidation number in another product. That is the signature of disproportionation.

Step-by-Step Reasoning

1. Oxidation number of N in NO$_2$

   Let the oxidation number of N be $x$.

   $x + 2(-2) = 0$
   $x - 4 = 0$
   $x = +4$

   So nitrogen is +4 in $\text{NO}_2$.

2. Oxidation number of N in HNO$_2$

   $+1 + x + 2(-2) = 0$
   $x - 3 = 0$
   $x = +3$

   So nitrogen is +3 in $\text{HNO}_2$.

3. Oxidation number of N in HNO$_3$

   $+1 + x + 3(-2) = 0$
   $x - 5 = 0$
   $x = +5$

   So nitrogen is +5 in $\text{HNO}_3$.

4. Compare oxidation numbers

   Nitrogen changes from +4 in $\text{NO}_2$ to +3 in $\text{HNO}_2$ and to +5 in $\text{HNO}_3$. The same element is both reduced (+4 to +3) and oxidised (+4 to +5). This is disproportionation.

5. Check the other options

   • A is incorrect. Oxygen remains -2 in water and in both products; oxygen atoms do not gain electrons.
   • C is incorrect. Hydrogen remains +1 in water and in both acids; its oxidation number is unchanged.
   • D is incorrect. Water is not reduced, so it does not act as an oxidising agent. The oxidising and reducing behaviour comes from $\text{NO}_2$ itself.

Therefore, the correct statement is B.

Key Takeaways

• Oxidation numbers are a bookkeeping tool for tracking electron transfer.
• Disproportionation occurs when one element is both oxidised and reduced in the same reaction.
• To identify disproportionation, assign oxidation numbers to all atoms and look for one element appearing at two different oxidation numbers in the products.
• An oxidising agent is reduced; a reducing agent is oxidised.

Common Mistakes

• Incorrectly assigning the oxidation number of nitrogen in $\text{HNO}_2$ or $\text{HNO}_3$. Remember that the sum of oxidation numbers in a neutral molecule must be zero.
• Thinking oxygen gains electrons because oxygen is part of the product acids. Oxygen remains -2 throughout.
• Confusing the oxidising agent with the species that is oxidised. Water is not oxidised or reduced here.
• Forgetting that disproportionation requires the same element to appear at both higher and lower oxidation numbers in the products.

Things to Be Careful About

• Use the oxidation number of oxygen as -2 and hydrogen as +1 unless a special case such as a peroxide is involved.
• Check every element in the equation, not just the one mentioned in the options.
• In a neutral molecule, the oxidation numbers must sum to zero; in an ion, they must sum to the ionic charge.
• The reaction is already balanced, so no balancing is needed, but the oxidation-number analysis must still be done carefully.

Background Concept

This question combines redox chemistry with balancing equations. In redox reactions, oxidation numbers change because electrons are transferred. Phosphorus is oxidised from 0 in elemental P to +5 in phosphoric acid, while sulfur is reduced from +6 in sulfuric acid to +4 in sulfur dioxide. Balancing a redox equation requires equal total electrons lost and gained.

Understanding the Question

We are given an unbalanced equation with coefficients a–e and asked to find the sum of those coefficients. The clue "balanced by using oxidation numbers" tells us to use electron transfer rather than trial-and-error atom balancing.

Approach

1. Assign oxidation numbers to each element.
2. Determine the electron change per atom for the oxidised and reduced species.
3. Find the smallest whole-number ratio that makes electrons lost equal electrons gained.
4. Balance atoms other than H and O.
5. Balance H and O using water.
6. Sum all coefficients.

Step-by-Step Reasoning

1. Oxidation numbers:

   • P in elemental P = 0.
   • P in H3PO4: H is +1, O is −2, so P must be +5.
   • S in H2SO4: H is +1, O is −2, so S must be +6.
   • S in SO2: O is −2, so S must be +4.

2. Each P atom loses 5 electrons: 0 → +5.
   Each S atom gains 2 electrons: +6 → +4.

3. The lowest common multiple of 5 and 2 is 10.

   • 2 P atoms lose 10 electrons.
   • 5 S atoms gain 10 electrons.
     Therefore b = 2 and d = 5.

4. Phosphorus atoms: c = b = 2.
   Sulfur atoms: a = d = 5.

5. Hydrogen atoms:

   • Left: 5 H2SO4 gives 10 H.
   • Right: 2 H3PO4 gives 6 H, so water must provide 4 H.
   • Therefore e = 2.

6. Oxygen check:

   • Left: 5 × 4 = 20 O.
   • Right: 2 × 4 + 5 × 2 + 2 × 1 = 8 + 10 + 2 = 20 O.

7. Sum of coefficients:
   a + b + c + d + e = 5 + 2 + 2 + 5 + 2 = 16.

Key Takeaways

• The oxidation-number method balances electron transfer first, then atoms.
• Use the oxidised/reduced species ratio to fix two coefficients immediately.
• Always finish by checking both atoms and, where relevant, charge.

Common Mistakes

• Forgetting that elemental phosphorus has oxidation number 0.
• Using the wrong oxidation number for P in H3PO4, such as +3.
• Balancing atoms before balancing electrons, which often gives incorrect whole-number coefficients.
• Forgetting to include water when balancing hydrogen and oxygen.

Things to Be Careful About

• Sulfur in H2SO4 is +6, not +4.
• Phosphorus in H3PO4 is +5.
• Use the smallest whole-number coefficients; multiplying the equation would change the sum.
• Verify every element at the end, especially oxygen.

Background Concept

The Haber process involves the reversible, exothermic reaction: $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$, $\Delta H = -92 \text{ kJ mol}^{-1}$.

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. For an exothermic reaction, increasing the temperature favors the reverse (endothermic) reaction, shifting the equilibrium to the left and decreasing the yield of products.

Reaction rate is affected by temperature, pressure, concentration, and catalysts. Increasing temperature increases the kinetic energy of particles, leading to more frequent and more energetic collisions, thus increasing the rate of reaction. A catalyst increases the rate by providing an alternative pathway with a lower activation energy, but it does not affect the position of equilibrium.

Understanding the Question

We are given a graph of the volume of ammonia produced against time for two experiments. Experiment 1 is the baseline (45 °C, 200 atm). Experiment 2 differs in only one condition. The graph shows that experiment 2 has a steeper initial gradient (faster initial rate) but a lower final plateau (less ammonia at equilibrium). We need to identify which single change from the options causes both of these effects.

Approach

We analyze the two features of the graph separately:

1. Rate of reaction: The steeper initial slope of experiment 2 means the reaction is faster. This points to a higher temperature, higher pressure, or the presence of a catalyst.
2. Equilibrium position: The lower final volume of NH₃ in experiment 2 means the equilibrium has shifted to the left (towards N₂ and H₂). We apply Le Chatelier's principle to see which of the rate-increasing changes (higher T, higher P, catalyst) also shifts the equilibrium to the left for this specific exothermic reaction.

Step-by-Step Reasoning

• Option A (Iron catalyst): A catalyst increases the rate of both forward and reverse reactions equally. It would give a steeper initial slope, but the final volume of NH₃ (equilibrium position) would remain the same. This contradicts the graph.
• Option B (2 mol helium gas): Helium is an inert gas. Adding it at constant volume does not change the partial pressures of the reacting gases, so it has no effect on the rate or equilibrium position. If added at constant pressure, the volume increases, partial pressures decrease, rate decreases, and equilibrium shifts to the side with more moles (left), but the rate would be slower, not faster. This contradicts the graph.
• Option C (Pressure of 250 atm): Increasing pressure from 200 atm to 250 atm increases the concentration of gases, increasing the rate (steeper slope). However, Le Chatelier's principle predicts that increasing pressure shifts the equilibrium to the side with fewer moles of gas. The reactants have 4 moles (1 N₂ + 3 H₂) and the products have 2 moles (2 NH₃). Thus, increasing pressure shifts the equilibrium to the right, producing more NH₃. The plateau would be higher, not lower. This contradicts the graph.
• Option D (Temperature of 600 °C): Increasing the temperature from 45 °C to 600 °C increases the kinetic energy of the particles, leading to a faster reaction rate (steeper initial slope). The forward reaction is exothermic ($\Delta H = -92 \text{ kJ mol}^{-1}$). Increasing the temperature shifts the equilibrium to the left (the endothermic direction) to absorb the excess heat. This results in a lower yield of NH₃ at equilibrium (lower plateau). This matches both features of the graph perfectly.

Key Takeaways

When interpreting equilibrium graphs, always consider both the rate (initial slope) and the equilibrium position (final plateau) separately. A change like temperature affects both, but in opposite directions for exothermic reactions (higher T = faster rate but lower yield). Catalysts only affect the rate, not the yield. Pressure changes affect both, but shift equilibrium towards the side with fewer gas moles.

Common Mistakes

• Assuming a catalyst changes the equilibrium position: A catalyst speeds up the attainment of equilibrium but does not change the final yield.
• Forgetting the stoichiometry of the gases when applying Le Chatelier's principle to pressure changes: 4 moles of reactant gas vs 2 moles of product gas means increased pressure favors the product.
• Confusing the effect of temperature on exothermic vs endothermic reactions: For exothermic reactions, higher temperature shifts equilibrium to the left (reactants).

Things to Be Careful About

• Read the graph axes carefully: volume of NH₃ vs time. The plateau represents the equilibrium amount, not the rate.
• Ensure you identify the correct feature (slope vs plateau) for rate vs equilibrium position.
• Check the sign of $\Delta H$ to determine if the forward reaction is exothermic or endothermic before applying Le Chatelier's principle for temperature changes.

Background Concept

For a gaseous equilibrium, the equilibrium constant $K_p$ is written using partial pressures. Its units are determined by the stoichiometry of the balanced equation. If the total number of moles of gaseous products equals the total number of moles of gaseous reactants, the pressure terms in the expression cancel completely, leaving a dimensionless constant. If the numbers are different, the expression retains pressure units such as atm, Pa, or kPa raised to a power.

Understanding the Question

The question asks which reaction has a $K_p$ with no units. This is purely a stoichiometric test: count the moles of gas on each side of each balanced equation. The reaction where the total gaseous coefficients are equal on both sides will have a unitless $K_p$.

Approach

For each option, calculate $\Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)}$. If $\Delta n = 0$, the pressure units cancel and $K_p$ has no units. If $\Delta n \neq 0$, the units of $K_p$ are $(\text{pressure})^{\Delta n}$.

Step-by-Step Reasoning

• Option A: $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$

  • Reactants: 1 + 1 = 2 moles of gas
  • Products: 2 moles of gas
  • $\Delta n = 2 - 2 = 0$, so $K_p$ has no units.

• Option B: $3\text{H}_2\text{(g)} + \text{N}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$

  • Reactants: 3 + 1 = 4 moles of gas
  • Products: 2 moles of gas
  • $\Delta n = 2 - 4 = -2$, so $K_p$ would have units of $\text{pressure}^{-2}$, e.g. $\text{atm}^{-2}$.

• Option C: $2\text{NO}_2\text{(g)} \rightleftharpoons \text{N}_2\text{O}_4\text{(g)}$

  • Reactants: 2 moles of gas
  • Products: 1 mole of gas
  • $\Delta n = 1 - 2 = -1$, so $K_p$ would have units of $\text{pressure}^{-1}$, e.g. $\text{atm}^{-1}$.

• Option D: $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$

  • Reactants: 2 + 1 = 3 moles of gas
  • Products: 2 moles of gas
  • $\Delta n = 2 - 3 = -1$, so $K_p$ would have units of $\text{pressure}^{-1}$, e.g. $\text{atm}^{-1}$.

Only Option A has $\Delta n = 0$, so only its $K_p$ has no units.

Key Takeaways

• The units of $K_p$ depend on the change in the number of moles of gas, $\Delta n$.
• When $\Delta n = 0$, $K_p$ is dimensionless.
• Always count only gaseous species when determining the units of $K_p$; pure solids and liquids do not appear in the expression.

Common Mistakes

• Counting all species, including solids and liquids, when determining $\Delta n$. Only gases contribute to the pressure terms.
• Confusing $K_p$ with $K_c$: $K_c$ units depend on concentration, but the same $\Delta n$ logic applies using mol dm$^{-3}$.
• Forgetting that coefficients in the balanced equation are the powers in the equilibrium expression, so they must be used when counting moles of gas.

Things to Be Careful About

• Ensure the equation is balanced before counting coefficients.
• If $\Delta n = 0$, the numerical value of $K_p$ is independent of the pressure units used, because the units cancel exactly.
• For reactions with $\Delta n \neq 0$, the numerical value of $K_p$ changes if the pressure unit changes, because the units do not cancel.

Background Concept

A Boltzmann distribution curve shows the distribution of kinetic energies of particles in a gas or liquid at a specific temperature. The x-axis represents kinetic energy (increasing to the right) and the y-axis represents the number of particles. The curve starts at the origin, rises to a peak (the most probable energy), and then tails off to the right. The total area under the curve represents the total number of particles in the sample.

The area under the curve to the right of a vertical line drawn at the activation energy ($E_a$) represents the number of particles with kinetic energy greater than or equal to $E_a$. These are the particles that have sufficient energy to undergo successful collisions and react.

A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy. It does this without being consumed in the overall reaction. Crucially, a catalyst does not change the temperature of the system, nor does it change the total number of particles or their overall energy distribution.

Understanding the Question

The question describes the decomposition of gas Q at room temperature and shows its Boltzmann distribution curve with the activation energy $E_a$ marked. We are asked to identify what happens to the diagram and the particles when a catalyst is added. The temperature is held constant at room temperature.

Approach

We need to evaluate each option by considering two key facts about adding a catalyst at constant temperature:

1. The temperature is constant, so the average kinetic energy and the shape/position of the Boltzmann distribution curve remain unchanged.
2. The activation energy $E_a$ is lowered. On an energy axis increasing to the right, a lower $E_a$ means the vertical line moves to the left.

We then check how these changes affect the area under the curve to the right of the $E_a$ line.

Step-by-Step Reasoning

• Option A: "The peak of the curve moves to the right." The position of the peak (most probable energy) and the overall shape of the curve depend only on temperature. Since the temperature is constant (room temperature), the curve does not shift. This option is incorrect.
• Option B: "The number of particles with enough energy to decompose increases." A catalyst lowers $E_a$, moving the vertical dotted line to the left. The area under the curve to the right of this new, lower $E_a$ line is larger than the original area. This larger area represents a greater number of particles with energy $\ge E_a$. Therefore, more particles have enough energy to decompose. This option is correct.
• Option C: "The kinetic energy of the unreacted particles increases." The average kinetic energy of the particles is directly proportional to the absolute temperature. Since the temperature is constant, the average kinetic energy of all particles (reacted or unreacted) remains unchanged. This option is incorrect.
• Option D: "The value of $E_a$ decreases, moving the vertical dotted line to the right on the diagram." While it is true that $E_a$ decreases, the x-axis represents energy increasing to the right. A decrease in the $E_a$ value means the line must move to the left (towards lower energy values), not the right. This option is incorrect.

Key Takeaways

• A catalyst lowers the activation energy ($E_a$) by providing an alternative pathway.
• On a Boltzmann distribution graph, this is represented by the vertical $E_a$ line moving to the left.
• The curve itself does not change shape or position unless the temperature changes.
• The area under the curve to the right of $E_a$ increases, meaning a greater proportion of particles can react.

Common Mistakes

• Confusing the direction of the $E_a$ line movement: Students often think "lower activation energy" means moving right because they associate catalysts with "increasing" the rate. Remember, the x-axis is energy increasing to the right, so a lower energy value is to the left.
• Thinking the curve shifts: Some students believe the curve shifts right or changes shape when a catalyst is added. The curve only changes if the temperature changes (shifting right and flattening for higher temperatures).
• Confusing catalyst effects with temperature effects: A catalyst increases the number of successful collisions by lowering the energy threshold ($E_a$), whereas increasing temperature increases the number of successful collisions by giving more particles higher kinetic energy (shifting the curve).

Things to Be Careful About

• Always check the direction of the axes. Energy increases to the right, so a decrease in $E_a$ is a leftward shift.
• Ensure you distinguish between the number of particles with enough energy (area to the right of $E_a$) and the average kinetic energy (related to the peak/temperature). A catalyst affects the former without affecting the latter.

Background Concept

Chlorides of elements show a trend in bonding across the Periodic Table. Electropositive metals such as sodium form ionic chlorides, $\text{Na}^+\text{Cl}^-$, with a giant ionic lattice. As elements become more electronegative and the cation becomes smaller and more highly charged, the chloride becomes increasingly covalent: $\text{AlCl}_3$ is covalent and exists as the dimer $\text{Al}_2\text{Cl}_6$, not as a giant ionic lattice. Non-metal chlorides such as $\text{SiCl}_4$ and $\text{PCl}_5$ are covalent molecular compounds. Many covalent chlorides react with water by hydrolysis, because water can attack the electron-deficient central atom and the chloride ligands leave as $\text{HCl}$. The resulting solution conducts electricity only if it contains mobile ions.

Understanding the Question

This multiple-choice question asks which of four statements about chlorides is correct. It tests two linked ideas: the type of bonding/lattice in a chloride, and what happens when that chloride is added to water. Statement D is the one to verify: phosphorus(V) chloride reacts with water to give acidic products that ionise, producing a conducting solution.

Approach

Check each statement against known chemistry. For A, recall the structure of $\text{AlCl}_3$. For B, recall what happens when an ionic lattice dissolves. For C, recall whether $\text{SiCl}_4$ reacts with water. For D, write the hydrolysis of $\text{PCl}_5$ and decide whether the products give ions in water.

Step-by-Step Reasoning

• A is incorrect. $\text{AlCl}_3$ is covalent, not a giant ionic lattice. In the solid and in non-aqueous conditions it forms the dimer $\text{Al}_2\text{Cl}_6$ with bridging chlorine atoms. The $\text{Al}^{3+}$ ion is small and highly charged, so it strongly polarises the chloride ions, making the bonding predominantly covalent.
• B is incorrect. When $\text{NaCl}$ dissolves in water, the lattice breaks up and the ions are hydrated: $\text{NaCl}(\text{s}) \rightarrow \text{Na}^+(\text{aq}) + \text{Cl}^-(\text{aq})$. It does not form $\text{HCl}$ and $\text{NaOH}$; those would be the products of electrolysis of brine, not simple dissolution.
• C is incorrect. $\text{SiCl}_4$ is a covalent molecular liquid that hydrolyses readily: $\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{aq})$. The fact that its covalent bonds are strong does not prevent reaction; water attacks silicon and forms strong Si-O bonds, while HCl is released.
• D is correct. $\text{PCl}_5$ hydrolyses vigorously: $\text{PCl}_5(\text{s}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_3\text{PO}_4(\text{aq}) + 5\text{HCl}(\text{aq})$. Both $\text{H}_3\text{PO}_4$ and $\text{HCl}$ are acids that ionise in water, giving $\text{H}^+$, $\text{Cl}^-$, $\text{H}_2\text{PO}_4^-$ and related ions. Because the solution contains mobile ions, it conducts electricity.

Key Takeaways

• Bonding in chlorides changes from ionic to covalent across a period as cation charge density increases.
• Ionic chlorides dissolve by dissociation into hydrated ions.
• Covalent chlorides often hydrolyse with water, releasing $\text{HCl}$ and forming an oxide or oxoacid.
• Electrical conductivity of an aqueous solution requires the presence of mobile ions.

Common Mistakes

• Choosing A because $\text{AlCl}_3$ contains aluminium and chlorine; it is covalent, not a giant ionic lattice.
• Choosing B because the formula suggests $\text{NaCl}$ and water could give $\text{HCl}$ and $\text{NaOH}$; dissolution is simply ion separation.
• Choosing C because "strong covalent bonds" sounds like a reason for unreactivity; hydrolysis can still occur and break those bonds.
• Writing an unbalanced hydrolysis equation for $\text{PCl}_5$ or $\text{SiCl}_4$; check atoms on both sides.

Things to Be Careful About

• Include state symbols in hydrolysis equations: $\text{PCl}_5(\text{s})$, $\text{H}_2\text{O}(\text{l})$, $\text{H}_3\text{PO}_4(\text{aq})$, $\text{HCl}(\text{aq})$.
• Remember that "conducts electricity" refers to mobile ions in solution, not to the covalent compound itself.
• Do not confuse $\text{AlCl}_3$ with ionic chlorides such as $\text{NaCl}$; the small, highly charged $\text{Al}^{3+}$ ion causes covalent character.
• In hydrolysis of $\text{SiCl}_4$, the solid product is $\text{SiO}_2$, not silicic acid unless excess water is considered; either is acceptable at this level, but the key point is that $\text{HCl}$ is produced.

Background Concept

Group 2 carbonates and nitrates both undergo thermal decomposition, but they give different sets of products.

• A Group 2 carbonate decomposes to the metal oxide and carbon dioxide:

• An anhydrous Group 2 nitrate decomposes to the metal oxide, nitrogen dioxide and oxygen:

This is different from Group 1 nitrates, which usually give the nitrite and oxygen. Because calcium and strontium are both Group 2 metals, their carbonates and nitrates follow the same patterns above. The phrase "anhydrous" tells us there is no water of crystallisation, so no water is produced.

Understanding the Question

The mixture contains four solids: CaCO3, Ca(NO3)2, SrCO3 and Sr(NO3)2. All four decompose completely. The question asks for the number of different products formed. This is a counting question about distinct chemical substances, not about the total number of product molecules or the total number of decomposition equations.

Approach

Write (or recall) the decomposition of each compound. Then list the products and remove duplicates. Carbon dioxide appears from both carbonates; nitrogen dioxide and oxygen appear from both nitrates; the metal oxides are different for calcium and strontium. Count the unique species.

Step-by-Step Reasoning

1. Calcium carbonate:

1. Strontium carbonate:

1. Calcium nitrate:

1. Strontium nitrate:

Now collect the distinct products:

• CaO (from CaCO3 and Ca(NO3)2)
• SrO (from SrCO3 and Sr(NO3)2)
• CO2 (from both carbonates)
• NO2 (from both nitrates)
• O2 (from both nitrates)

That gives five different products: CaO, SrO, CO2, NO2 and O2. The correct option is B.

Key Takeaways

• Group 2 carbonates: oxide + CO2.
• Anhydrous Group 2 nitrates: oxide + NO2 + O2.
• "Different products" means distinct chemical species; common products are counted once.

Common Mistakes

• Counting total product molecules across all four equations (2 + 2 + 3 + 3 = 10) instead of distinct products. This would give 10, which is not even an option, but it shows the wrong approach.
• Forgetting that CO2 is produced by both carbonates, or that NO2 and O2 are produced by both nitrates, leading to overcounting.
• Using the Group 1 nitrate decomposition (nitrite + O2) for calcium and strontium nitrates. Group 2 nitrates give the oxide, NO2 and O2.
• Forgetting the oxygen gas produced by nitrate decomposition, which would give only 4 products and option A.
• Thinking CaO and SrO are the same because both are Group 2 oxides; they are different compounds and both must be counted.

Things to Be Careful About

• The question says the substances are anhydrous. If water of crystallisation were present, water would be an additional product.
• "Different products" refers to different chemical substances, not different numbers of moles or different product molecules.
• Balance the nitrate equations to see that O2 is a product; an unbalanced or incorrect nitrate decomposition is the most likely source of error.
• State symbols are not needed for the count, but they help confirm that CO2, NO2 and O2 are gases and therefore separate products.

Background Concept

Group 2 metals form ionic oxides such as $\text{MgO}$ and $\text{BaO}$. These are basic oxides: they react with water to form the metal hydroxide, giving an alkaline solution. By contrast, non-metal oxides such as phosphorus(V) oxide are acidic and produce acidic solutions with water. Another key trend is that the solubility of Group 2 sulfates decreases down the group: $\text{MgSO}_4$ is soluble, while $\text{BaSO}_4$ is very insoluble and appears as a white precipitate.

Understanding the Question

The question gives two observations about solid W: it reacts with water to give an alkaline solution, and addition of two drops of dilute $\text{H}_2\text{SO}_4$ to that solution produces a white precipitate. We need the solid whose chemistry matches both observations. The first observation rules out acidic oxides; the second distinguishes between possible metal hydroxides by their sulfate solubility.

Approach

Start with observation 1: identify which solids can react with water to give an alkaline solution. This eliminates phosphorus oxide. Then use observation 2: add sulfate ions (from dilute sulfuric acid) to the alkaline solution and ask which cation forms an insoluble white sulfate. Only barium does so among the options.

Step-by-Step Reasoning

1. Phosphorus oxide, $\text{P}_4\text{O}_{10}$, reacts with water to form phosphoric acid, so the solution would be acidic, not alkaline. Eliminate D.
2. Magnesium hydroxide is already a hydroxide and does not react with water to form a new alkaline product; it is only sparingly soluble. Even if its suspension is alkaline, adding sulfate gives soluble $\text{MgSO}_4$, so no white precipitate forms. Eliminate A.
3. Magnesium oxide does react with water: $\text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2$, giving an alkaline solution. However, $\text{MgSO}_4$ is soluble, so dilute sulfuric acid would not give a white precipitate. Eliminate B.
4. Barium oxide reacts with water: $\text{BaO} + \text{H}_2\text{O} \rightarrow \text{Ba(OH)}_2$, giving an alkaline solution. On adding dilute sulfuric acid, barium ions react with sulfate ions: $\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})$. Barium sulfate is white and insoluble, so a white precipitate is observed. Therefore W is barium oxide, option C.

The phrase 'two drops' simply means a small amount of sulfuric acid is added; it supplies sulfate ions without dissolving the precipitate, because $\text{BaSO}_4$ is insoluble even in dilute acid.

Key Takeaways

• Metal oxides are generally basic; non-metal oxides are generally acidic.
• Group 2 oxides react with water to form alkaline hydroxide solutions.
• The solubility of Group 2 sulfates decreases down the group; $\text{BaSO}_4$ is a classic white precipitate.
• Qualitative observations such as 'white precipitate with sulfate' can identify a cation.

Common Mistakes

• Choosing magnesium oxide because it gives an alkaline solution, while forgetting that its sulfate is soluble and would not precipitate.
• Choosing phosphorus oxide because it is an oxide, while ignoring that it gives an acidic, not alkaline, solution.
• Assuming that any alkaline solution will give a white precipitate with sulfuric acid; only cations forming insoluble sulfates do.
• Writing the precipitate as barium sulfite or forgetting the (s) state symbol.

Things to Be Careful About

• Use the correct formula $\text{Ba(OH)}_2$ and balance the equation for oxide with water.
• The precipitate is $\text{BaSO}_4$, white and insoluble; do not confuse it with soluble $\text{MgSO}_4$.
• State symbols matter: $\text{BaO}(\text{s})$, $\text{Ba(OH)}_2(\text{aq})$, $\text{BaSO}_4(\text{s})$.
• Remember the acid-base nature of oxides: metal oxides are basic, non-metal oxides are acidic.

Background Concept

This question tests disproportionation, a special type of redox reaction in which the same element is both oxidised and reduced. To recognise it, you assign oxidation numbers to each atom before and after the reaction.

Key oxidation-number rules used here:

• An element in its free state has oxidation number 0, so each chlorine atom in $\text{Cl}_2$ has oxidation number 0.
• In a monatomic ion, the oxidation number equals the ionic charge, so chlorine in $\text{Cl}^-$ has oxidation number $-1$.
• In a polyatomic ion such as $\text{ClO}^-$, the sum of oxidation numbers equals the charge of the ion. Oxygen is normally $-2$, so for $\text{ClO}^-$: $x + (-2) = -1$, giving $x = +1$.

Reduction is a decrease in oxidation number (gain of electrons); oxidation is an increase in oxidation number (loss of electrons). When one element simultaneously increases and decreases in oxidation number, it has been disproportionated.

Understanding the Question

The question asks which statement correctly describes what happens to chlorine when $\text{Cl}_2$ reacts with cold aqueous sodium hydroxide. The four options describe whether chlorine is oxidised, reduced, both, or neither. You are not asked to write the equation, but the equation is the key to deciding the answer.

The reaction with cold NaOH is:

This is a standard Group 17 reaction and is a classic example of disproportionation.

Approach

1. Write the balanced equation for chlorine with cold aqueous sodium hydroxide.
2. Assign oxidation numbers to chlorine in the reactant and in each chlorine-containing product.
3. Compare the oxidation numbers: if chlorine goes from 0 to a lower value, it is reduced; if it goes to a higher value, it is oxidised.
4. If both changes occur, chlorine is both oxidised and reduced, which is disproportionation.

Step-by-Step Reasoning

In $\text{Cl}_2$, each chlorine atom has oxidation number 0 because it is an element in its standard state.

In the product $\text{NaCl}$, chlorine is present as $\text{Cl}^-$, so its oxidation number is $-1$. This is a decrease from 0, so chlorine has been reduced.

In the product $\text{NaClO}$, the chlorine is part of the hypochlorite ion, $\text{ClO}^-$. Since oxygen is $-2$ and the overall charge is $-1$, chlorine must have oxidation number $+1$. This is an increase from 0, so chlorine has been oxidised.

Therefore, in the same reaction, some chlorine atoms are reduced and some are oxidised. This is exactly what is meant by disproportionation. The correct option is therefore A.

Why the other options are wrong:

• B says chlorine is neither oxidised nor reduced. This is false because the oxidation number clearly changes from 0 to $-1$ and to $+1$.
• C says chlorine is oxidised but not reduced. This ignores the formation of $\text{Cl}^-$ in $\text{NaCl}$.
• D says chlorine is reduced but not oxidised. This ignores the formation of $\text{ClO}^-$ in $\text{NaClO}$.

Key Takeaways

• Disproportionation occurs when the same element is both oxidised and reduced in one reaction.
• Chlorine with cold aqueous NaOH gives chloride, $\text{Cl}^-$, and hypochlorite, $\text{ClO}^-$, showing oxidation states $-1$ and $+1$.
• Always assign oxidation numbers systematically before deciding whether a species is oxidised or reduced.
• The product depends on conditions: cold NaOH gives hypochlorite, while hot concentrated NaOH gives chlorate, $\text{ClO}_3^-$, where chlorine reaches $+5$.

Common Mistakes

• Forgetting that $\text{Cl}_2$ is the free element and therefore has oxidation number 0.
• Assigning chlorine in $\text{ClO}^-$ an oxidation number of $-1$ by treating it as a simple halide ion. The correct value is $+1$.
• Thinking that because chlorine forms $\text{Cl}^-$, it is only reduced. The hypochlorite product shows simultaneous oxidation.
• Confusing cold and hot alkali conditions. Cold NaOH gives $\text{ClO}^-$; hot concentrated NaOH gives $\text{ClO}_3^-$.

Things to Be Careful About

• Oxidation numbers are bookkeeping charges, not actual ionic charges in covalent species such as $\text{ClO}^-$.
• In $\text{ClO}^-$, oxygen is $-2$ and the sum of oxidation numbers must equal the ion charge, $-1$.
• The question is multiple choice, so the final answer is simply the correct option letter, but you should still be able to justify it with the balanced equation and oxidation-number changes.