Chemistry 9701/13 — May/June 2024

Background Concept

Group 2 metals form +2 ions, so their carbonate and chloride formulae are $\text{MCO}_3$ and $\text{MCl}_2$. When a carbonate reacts with an acid, carbon dioxide is released:

$\text{MCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})$

The key stoichiometric fact is that one formula unit of carbonate produces one formula unit of chloride, because each contains one metal ion. The sample is impure: 57% by mass is $\text{MCO}_3$ and the rest is inert. Only the carbonate reacts with the acid, so the amount of chloride formed tells us the amount of pure carbonate that was present.

Understanding the Question

We are given 7.4 g of impure X, told it is 57% $\text{MCO}_3$ by mass, and told that the impurities do not react with HCl. Reaction with excess HCl produces 0.050 mol of the Group 2 metal chloride. The question asks us to identify M. Because the acid is in excess, all of the carbonate reacts; the only limiting factor is the amount of pure $\text{MCO}_3$ in the sample. The identity of M is found from its relative atomic mass, which we obtain by working out the molar mass of the carbonate.

Approach

1. Calculate the mass of pure $\text{MCO}_3$: multiply the sample mass by 57%.
2. Use the balanced equation to convert moles of $\text{MCl}_2$ into moles of $\text{MCO}_3$ (1:1).
3. Divide the pure mass by the number of moles to get the molar mass of $\text{MCO}_3$.
4. Subtract the molar mass of the carbonate ion, 60 g mol$^{-1}$, to get $A_r$ of the metal.
5. Match this value to the periodic table.

Step-by-Step Reasoning

First, find the mass of pure carbonate:

$m(\text{MCO}_3) = 0.57 \times 7.4 = 4.218\ \text{g}$

The balanced equation shows one mole of $\text{MCO}_3$ gives one mole of $\text{MCl}_2$:

$\text{MCO}_3 + 2\text{HCl} \rightarrow \text{MCl}_2 + \text{CO}_2 + \text{H}_2\text{O}$

Therefore:

$n(\text{MCO}_3) = n(\text{MCl}_2) = 0.050\ \text{mol}$

Now calculate the molar mass of the carbonate:

$M(\text{MCO}_3) = \frac{4.218}{0.050} = 84.36\ \text{g mol}^{-1} \approx 84.4\ \text{g mol}^{-1}$

The carbonate ion has:

$M(\text{CO}_3^{2-}) = 12 + 3(16) = 60\ \text{g mol}^{-1}$

So the metal has:

$A_r(\text{M}) = 84.4 - 60 = 24.4 \approx 24$

Magnesium has $A_r \approx 24.3$, so M is Mg. The correct option is A.

Key Takeaways

• Percentage purity must be applied before using mass in a stoichiometric calculation.
• A balanced equation gives the mole ratio; here it is 1:1 between carbonate and chloride.
• Molar mass can be used to identify an unknown element by subtracting the known part of a formula.
• Excess reactant ensures that the amount of product is controlled by the limiting reactant, which is the pure carbonate.

Common Mistakes

• Using 7.4 g as the mass of pure $\text{MCO}_3$. This gives a molar mass of 148 g mol$^{-1}$ and would incorrectly suggest Sr. Always multiply by 0.57 first.
• Forgetting that the impurities do not react, so they contribute mass but no moles of chloride.
• Using the wrong mole ratio, e.g. assuming two moles of carbonate per mole of chloride because there are two chlorides in $\text{MCl}_2$. The ratio is 1:1 because both compounds contain one metal ion.
• Subtracting 44 (the molar mass of $\text{CO}_2$) instead of 60 (the molar mass of $\text{CO}_3^{2-}$).
• Rounding 84.36 to 84 before subtracting 60, which still gives 24, but rounding too early can be risky in other problems.

Things to Be Careful About

• 57% by mass means $0.57 \times 7.4$ g, not 57 g and not 43% of the sample.
• Include units: molar mass is in g mol$^{-1}$.
• The acid is in excess, so the amount of chloride formed is determined by the carbonate, not by the acid.
• The carbonate ion is $\text{CO}_3^{2-}$, molar mass 60 g mol$^{-1}$; do not confuse it with carbon dioxide, $\text{CO}_2$, molar mass 44 g mol$^{-1}$.
• When comparing with the periodic table, use the approximate relative atomic mass: Mg = 24.3, Ca = 40.1, Sr = 87.6, Ba = 137.3.

Background Concept

The mole is the amount of substance that contains $6.02 \times 10^{23}$ particles (Avogadro's constant). The molar mass of a substance in $\text{g mol}^{-1}$ is numerically equal to its relative molecular mass, $M_r$. To compare numbers of atoms in different samples, convert each mass into moles of the substance and then multiply by the number of atoms in one molecule or atom of that substance. This gives the total amount (in mol) of atoms, which is proportional to the actual number of atoms because the same Avogadro constant applies to every mole.

Understanding the Question

This question asks which gas sample contains the same number of atoms as 1 g of hydrogen gas. It is important to notice that the comparison is between numbers of atoms, not numbers of molecules and not masses. Hydrogen exists as diatomic $\text{H}_2$ molecules, so its molar mass is 2 $\text{g mol}^{-1}$, and 1 g is only 0.5 mol of molecules. Each option gives a mass and an $M_r$; you must work out the moles of atoms in each.

Approach

1. Find the amount of $\text{H}_2$ in 1 g: $n = \frac{m}{M_r}$.
2. Multiply by 2 because each $\text{H}_2$ molecule contains two atoms. This gives the target number of moles of atoms.
3. For each option, calculate $n = \frac{m}{M_r}$ for the gas, then multiply by the number of atoms per molecule (3 for $\text{CO}_2$, 5 for $\text{CH}_4$, 1 for Ne, 3 for $\text{O}_3$).
4. Select the option whose total moles of atoms equals the target.

Step-by-Step Reasoning

Target sample:

Since each $\text{H}_2$ molecule has two H atoms:

Option A, 22 g $\text{CO}_2$ ($M_r = 44$):

Each $\text{CO}_2$ molecule contains 3 atoms, so total atoms $= 0.5 \times 3 = 1.5$ mol. This is too many.

Option B, 8 g $\text{CH}_4$ ($M_r = 16$):

Each $\text{CH}_4$ molecule contains 5 atoms, so total atoms $= 0.5 \times 5 = 2.5$ mol. This is too many.

Option C, 20 g Ne ($M_r = 20$):

Neon is monatomic, so 1.0 mol of Ne atoms is present. This matches the target of 1.0 mol atoms.

Option D, 8 g $\text{O}_3$ ($M_r = 48$):

Each $\text{O}_3$ molecule contains 3 atoms, so total atoms $= \frac{1}{6} \times 3 = 0.5$ mol. This is too few.

Therefore the correct option is C.

Key Takeaways

• Always compare amounts in moles, not masses, when asked about numbers of particles.
• Distinguish between moles of molecules and moles of atoms; multiply by the number of atoms per molecule.
• Remember that hydrogen, oxygen, nitrogen, and the halogens are diatomic, while noble gases are monatomic.

Common Mistakes

• Comparing 20 g with 1 g directly: masses are not proportional to atom numbers because molar masses differ.
• Using 1 g mol^{-1} for hydrogen instead of 2 g mol^{-1}; hydrogen gas is $\text{H}_2$.
• Forgetting to multiply by atoms per molecule, so choosing A (0.5 mol molecules) instead of seeing it has 1.5 mol atoms.
• Treating neon as diatomic; Ne is monatomic.

Things to Be Careful About

• Use the $M_r$ of the molecule as given, not the atomic mass where they differ.
• Count atoms correctly: $\text{CO}_2$ has 3 atoms, $\text{CH}_4$ has 5, $\text{O}_3$ has 3, Ne has 1.
• Keep units consistent: mass in g divided by molar mass in $\text{g mol}^{-1}$ gives moles.
• The phrase 'same number of atoms' means equal moles of atoms, since 1 mol always contains the same number of particles.

Background Concept

The ammonium ion, $\mathrm{NH_4^+}$, is a polyatomic ion. Its subatomic particle count is found by adding the contributions of one nitrogen atom and four hydrogen atoms, then adjusting the electron count for the +1 charge.

Understanding the Question

The question gives a relative formula mass of 21 for the ammonium ion and asks for the total number of protons, neutrons and electrons. The relative formula mass equals the total number of protons plus neutrons (the mass number).

Approach

1. Count protons: nitrogen has proton number 7; each hydrogen has proton number 1. Total = 11.
2. Count electrons: a neutral $\mathrm{NH_4}$ molecule would have 11 electrons; the +1 charge means one electron is lost, giving 10.
3. Count neutrons: mass number 21 − number of protons 11 = 10.

Step-by-Step Reasoning

• Protons: $7 + 4(1) = 11$
• Electrons: $11 - 1 = 10$
• Neutrons: $21 - 11 = 10$

These match option A.

Key Takeaways

• For an ion, the electron count differs from the proton count by the magnitude of the charge.
• The mass number (relative formula mass) is the sum of protons and neutrons.

Common Mistakes

• Forgetting to subtract an electron for the +1 charge.
• Adding the charge to the neutron count instead of the electron count.

Things to Be Careful About

• The relative formula mass given (21) is unusual for a standard $\mathrm{NH_4^+}$ (which would be 18); this implies an isotopic form of nitrogen or hydrogen, but the proton and electron counts remain fixed by the element identities and charge.

Background Concept

First ionisation energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions. It depends on nuclear charge, atomic radius, shielding by inner electrons, and the stability of the electron configuration.

Understanding the Question

The question asks two things: which element has the higher first ionisation energy, and which type of orbital (s or p) the removed electron comes from for Mg and Ne. The table rows combine these two pieces of information.

Approach

Write the full electron configurations of Mg and Ne. Identify the highest-energy occupied orbital in each. Compare the ionisation energies using the periodic trend and the stability of the noble gas configuration.

Step-by-Step Reasoning

1. Mg (Z = 12): 1s² 2s² 2p⁶ 3s². The outermost electron is in a 3s orbital, so the first ionisation energy removes an s electron.
2. Ne (Z = 10): 1s² 2s² 2p⁶. The outermost electrons are in 2p orbitals, so the first ionisation energy removes a p electron.
3. Ne has a much higher first ionisation energy than Mg. Its outer electrons are in the n = 2 shell, closer to the nucleus, and it has a stable filled octet. Mg's outer electron is in the n = 3 shell, further from the nucleus and more shielded, so it is easier to remove.
4. Therefore the correct row is: Ne > Mg; Mg removes s; Ne removes p.

Key Takeaways

• First ionisation energy generally increases across a period, but the jump from Mg to Ne is large because Ne has a stable noble gas configuration.
• The electron removed is always the highest-energy electron: for Mg this is 3s, for Ne this is 2p.
• Always connect the ionisation energy trend to the actual orbital from which the electron is removed.

Common Mistakes

• Choosing Mg > Ne because Mg has more protons. Nuclear charge is not the only factor; the principal shell and stability of the noble gas configuration matter more here.
• Saying Mg removes a p electron. Mg's valence electron is in the 3s orbital.
• Saying Ne removes an s electron. Ne's outer electrons are in the 2p subshell.

Things to Be Careful About

• Read the table carefully: the row asks for both the comparison and the orbital types.
• Remember that ionisation energy is always positive for removing an electron from a neutral gaseous atom.
• Use the full electron configuration, not just the outer shell, when deciding between s and p.

Background Concept

VSEPR (valence shell electron pair repulsion) theory states that the electron pairs around a central atom repel one another and arrange themselves as far apart as possible. The shape of a molecule is determined by the number of electron pairs around the central atom and whether any of them are lone pairs.

When the central atom is from Period 3 or below, it can have an expanded octet, meaning it can accommodate more than eight electrons in its valence shell. This allows elements such as arsenic and selenium to form more than the usual number of covalent bonds.

For a molecule with no lone pairs on the central atom:

• 5 bonding pairs give a trigonal bipyramidal shape with bond angles of 90°, 120° and 180°.
• 6 bonding pairs give an octahedral shape with bond angles of 90° and 180°.

The question tells us to ignore 180° bond angles, so we focus only on the other angles.

Understanding the Question

We are told that arsenic and selenium each form a fluoride in which the central atom has no lone pair of electrons and there are no dative bonds. This means every valence electron of the central atom is used in ordinary covalent bonds to fluorine atoms. We must decide in which of these two fluorides there are two different bond angles, ignoring 180°.

The key is to work out the formula of each fluoride from the number of valence electrons of the central atom, then use VSEPR to predict the shape and the bond angles.

Approach

1. Determine the number of outer electrons for arsenic and selenium.
2. Since there are no lone pairs, the central atom forms one bond to fluorine for each valence electron.
3. Predict the shape using VSEPR.
4. List the bond angles present, ignoring 180°.
5. Compare the two compounds.

Step-by-Step Reasoning

Arsenic is in Group 15, so it has five outer electrons: $4\text{s}^2 4\text{p}^3$. With no lone pairs, all five electrons are used to form five covalent bonds to fluorine atoms, giving $\text{AsF}_5$. The arsenic atom is surrounded by five bonding pairs, so VSEPR predicts a trigonal bipyramidal shape. In this shape there are three different bond angles: 90° between axial and equatorial bonds, 120° between equatorial bonds, and 180° between the two axial bonds. Ignoring 180°, there are still two different bond angles: 90° and 120°. So arsenic fluoride has two different bond angles.

Selenium is in Group 16, so it has six outer electrons: $4\text{s}^2 4\text{p}^4$. With no lone pairs, all six electrons are used to form six covalent bonds to fluorine atoms, giving $\text{SeF}_6$. The selenium atom is surrounded by six bonding pairs, so VSEPR predicts an octahedral shape. In an octahedron all adjacent bond angles are 90°, and opposite bonds are 180°. Ignoring 180°, there is only one bond angle: 90°. So selenium fluoride has only one bond angle.

Therefore arsenic fluoride only has two different bond angles, and the correct option is B.

Key Takeaways

• The number of valence electrons of the central atom, combined with the condition of no lone pairs, determines the number of bonds it forms.
• Five bonding pairs give a trigonal bipyramidal shape (90° and 120°).
• Six bonding pairs give an octahedral shape (90° only, after ignoring 180°).
• Always read special instructions such as "ignore 180° bond angles" before deciding how many different angles are present.

Common Mistakes

• Assuming arsenic forms $\text{AsF}_3$ or selenium forms $\text{SeF}_4$. These would have lone pairs on the central atom, which the question explicitly rules out.
• Counting 180° as a bond angle when the question tells you to ignore it.
• Confusing the trigonal bipyramidal shape with the octahedral shape, or mixing up the bond angles of the two shapes.
• Forgetting that Period 3 and below elements can have an expanded octet, allowing five or six bonds.

Things to Be Careful About

• Use the group number to find the number of outer electrons: Group 15 gives five, Group 16 gives six.
• "No dative bonds" means all bonds are ordinary covalent bonds, so the number of bonds equals the number of valence electrons used.
• The phrase "two different bond angles" requires you to compare the non-180° angles only, so state 90° and 120° for the trigonal bipyramid and 90° only for the octahedron.

Background Concept

Benzene ($\text{C}_6\text{H}_6$) is a planar, six-membered ring where each carbon atom is $sp^2$ hybridised. Each carbon contributes one electron from its unhybridised $p$-orbital to a delocalised $\pi$ system containing 6 electrons. Borazole ($\text{B}_3\text{N}_3\text{H}_6$), often called "inorganic benzene", is a six-membered ring of alternating boron and nitrogen atoms. To form a delocalised $\pi$ system analogous to benzene, the ring atoms must be $sp^2$ hybridised and planar. Nitrogen has a lone pair of electrons, while boron has an empty $p$-orbital. The lone pairs on the nitrogen atoms can delocalise into the empty $p$-orbitals of the boron atoms.

Understanding the Question

The question asks for the molecular shape of borazole and the number of $\pi$ electrons in its delocalised system. The structure shows a six-membered ring with alternating B and N atoms, each bonded to one H atom, with formal charges shown (B$^-$, N$^+$). We must determine the hybridisation of the ring atoms, count the electrons in the $\pi$ system, and deduce the overall shape.

Approach

1. Determine the hybridisation of the ring atoms (B and N) based on their bonding (3 $\sigma$ bonds each).
2. Identify the source of $\pi$ electrons: N atoms have lone pairs in $p$-orbitals, B atoms have empty $p$-orbitals.
3. Calculate the total number of $\pi$ electrons by summing the electrons contributed by each atom to the $\pi$ system.
4. Conclude the shape based on the requirement for $p$-orbital overlap in a delocalised system.

Step-by-Step Reasoning

• Hybridisation and Shape: Each boron and nitrogen atom in the ring is bonded to three other atoms (two in the ring, one hydrogen). This requires three $\sigma$ bonds, meaning the atoms are $sp^2$ hybridised. $sp^2$ hybridisation gives a trigonal planar geometry around each atom. For the $p$-orbitals to overlap and form a delocalised $\pi$ system across the entire ring, all atoms must lie in the same plane. Therefore, borazole is planar.
• $\pi$ Electrons: In the delocalised $\pi$ system, each nitrogen atom contributes its lone pair of electrons (2 electrons) from its unhybridised $p$-orbital. There are 3 nitrogen atoms, so they contribute 3 $\times$ 2 = 6 electrons. Each boron atom has an empty $p$-orbital and contributes 0 electrons. The total number of $\pi$ electrons is 6.
• Formal Charges: The formal charges (B$^-$, N$^+$) indicate the polarization of the B-N bonds. Nitrogen is more electronegative and donates its lone pair into the empty $p$-orbital of boron, resulting in a partial negative charge on B and partial positive on N. This does not change the total number of valence electrons or the number of $\pi$ electrons.
• Conclusion: The molecule is planar and has 6 $\pi$ electrons. This matches option D.

Key Takeaways

• Borazole is an inorganic analogue of benzene with alternating B and N atoms.
• Nitrogen contributes lone pairs to the $\pi$ system, while boron contributes empty $p$-orbitals.
• A delocalised $\pi$ system over a six-membered ring with 6 electrons (satisfying Hückel's rule, 4n+2 where n=1) requires planar geometry and $sp^2$ hybridisation.

Common Mistakes

• Assuming boron contributes electrons to the $\pi$ system (it has an empty $p$-orbital, so it contributes 0).
• Counting only one electron per atom (like in benzene) instead of counting the lone pair on nitrogen.
• Assuming the molecule is non-planar due to the formal charges or the different atoms (B vs N), forgetting that $sp^2$ hybridisation and delocalisation enforce planarity.

Things to Be Careful About

• Remember that formal charges (B$^-$, N$^+$) indicate the polarization of the B-N bonds but do not change the number of electrons in the $\pi$ system. The lone pair on N is what makes it N$^+$ (since it's sharing its lone pair) and B$^-$ (since it's accepting electron density).
• Ensure the shape is deduced from the hybridisation and delocalisation requirement, not just VSEPR on individual atoms.
• Borazole is isoelectronic with benzene: 3 $\times$ 5 (from N) + 3 $\times$ 3 (from B) + 6 $\times$ 1 (from H) = 30 valence electrons, same as 6 $\times$ 4 (from C) + 6 $\times$ 1 (from H) = 30.

Background Concept

The ideal gas equation, $pV = nRT$, relates the pressure ($p$), volume ($V$), amount of substance in moles ($n$), and absolute temperature ($T$) of an ideal gas. The gas constant $R$ has a value of $8.31 \text{ J K}^{-1} \text{mol}^{-1}$ when using standard SI units.

When dealing with gases in experiments, it is crucial to ensure that all units are consistent. For $R = 8.31 \text{ J K}^{-1} \text{mol}^{-1}$:

• Pressure must be in Pascals (Pa).
• Volume must be in cubic metres (m³). Note that $1 \text{ m}^3 = 1,000,000 \text{ cm}^3 = 10^6 \text{ cm}^3$, so to convert cm³ to m³, you multiply by $10^{-6}$.
• Temperature must be in Kelvin (K).
• Amount ($n$) is in moles (mol).

The number of moles can be calculated from the mass ($m$) and the molar mass ($M$) using $n = \frac{m}{M}$. The relative molecular mass ($M_r$) is numerically equal to the molar mass in g mol⁻¹, so we can use $n = \frac{m}{M_r}$ where $m$ is in grams.

Understanding the Question

The question asks for the correct algebraic expression to calculate the relative molecular mass ($M_r$) of a volatile liquid, given the mass of the sample, the volume of vapour produced, the temperature, and the pressure. The vapour is assumed to behave as an ideal gas.

Given data:

• Mass of liquid, $m = 0.10 \text{ g}$
• Volume of vapour, $V = 85 \text{ cm}^3$
• Temperature, $T = 400 \text{ K}$
• Pressure, $p = 101\,300 \text{ Pa}$

Goal: Find the correct expression for $M_r$ from the options A, B, C, and D.

Approach

1. Start with the ideal gas equation: $pV = nRT$.
2. Substitute $n = \frac{m}{M_r}$ to introduce the unknown $M_r$.
3. Rearrange the equation to isolate $M_r$.
4. Check the units of the given values against the requirements of the gas constant $R = 8.31 \text{ J K}^{-1} \text{mol}^{-1}$. Specifically, volume must be in m³ and pressure in Pa.
5. Substitute the values (with correct unit conversions) into the rearranged equation to identify the matching option.

Step-by-Step Reasoning

Step 1: Set up the equation
$pV = nRT$
Since $n = \frac{m}{M_r}$:
$pV = \frac{m}{M_r} RT$

Step 2: Rearrange for $M_r$
Multiply both sides by $M_r$ and divide by $pV$:
$M_r = \frac{mRT}{pV}$

Step 3: Check units and convert
Using $R = 8.31 \text{ J K}^{-1} \text{mol}^{-1}$ requires SI units for $p$ and $V$:

• $m = 0.10$ (grams is fine here because $M_r$ is numerically equal to molar mass in g mol⁻¹, and the J/Pa·m³ cancels out to leave g/mol numerically).
• $R = 8.31$
• $T = 400$
• $p = 101\,300$ (already in Pa)
• $V = 85 \text{ cm}^3$. Convert to m³: $85 \div 10^6 = 85 \times 10^{-6} \text{ m}^3$.

Step 4: Construct the expression
Substitute the values into $M_r = \frac{mRT}{pV}$:
$M_r = \frac{0.10 \times 8.31 \times 400}{(85 \times 10^{-6}) \times 101\,300}$

Step 5: Compare with options

• A: Incorrectly calculates $\frac{pV}{mRT}$ (inverse of $M_r$) and fails to convert volume to m³.
• B: Incorrectly calculates $\frac{pV}{mRT}$ and uses inconsistent units ($p$ in kPa but $R$ for Pa).
• C: Matches our derived expression exactly: $\frac{0.10 \times 8.31 \times 400}{(85 \times 10^{-6}) \times 101\,300}$. Correct.
• D: Uses $p = 101.3$ (kPa) but keeps $R = 8.31$ (which requires Pa). This makes the denominator 1000 times too small, giving an incorrect answer.

Key Takeaways

• Always check the units of the gas constant $R$ being used. If $R = 8.31$, volume must be in m³ and pressure in Pa.
• Converting cm³ to m³ is a common pitfall; remember to multiply by $10^{-6}$ (or divide by $10^6$).
• Rearranging $pV = nRT$ to solve for molar mass involves substituting $n = m/M_r$ and isolating $M_r$.

Common Mistakes

• Forgetting to convert volume: Using $V = 85$ directly in the equation with $R = 8.31$ leads to a result that is $10^6$ times too large. This is seen in option A.
• Mixing pressure units: Using $p = 101.3$ kPa with $R = 8.31$ J K⁻¹ mol⁻¹ (option D) is inconsistent. If using kPa, one must use $R = 8.31 \times 10^{-3}$ kJ K⁻¹ mol⁻¹ or convert pressure back to Pa.
• Inverting the formula: Calculating $\frac{pV}{mRT}$ instead of $\frac{mRT}{pV}$ gives $1/M_r$. This is seen in options A and B.

Things to Be Careful About

• Unit consistency: The value $8.31$ is for J K⁻¹ mol⁻¹. Joules are Pa·m³. Therefore, $p$ must be in Pa and $V$ in m³.
• Significant figures: While not the focus of this multiple-choice question, be aware that $0.10$ has 2 sig figs, so the final answer should technically be to 2 sig figs ($M_r \approx 39$).
• State symbols and physical states: Ensure you understand that the liquid has evaporated into a gas occupying the syringe volume. The volume given ($85 \text{ cm}^3$) is the volume of the vapour, not the liquid.

Background Concept

This question tests the relationship between bonding/structure and physical properties. There are four common types of structure in AS chemistry, each with a characteristic set of properties:

• Ionic (e.g. NaF, MgO): high melting and boiling points (strong electrostatic attraction between oppositely charged ions in a giant lattice). The solid does not conduct electricity because the ions are held in fixed positions, but the molten liquid and aqueous solution conduct because the ions are free to move.
• Simple molecular covalent (e.g. $\text{C}_2\text{H}_5\text{Br}$, HCl): low melting and boiling points (weak intermolecular forces between molecules). They do not conduct electricity in any state because there are no free charged particles. Many are insoluble in water, though some (like HCl) dissolve and form conducting solutions.
• Metallic (e.g. Fe, K): high melting points, and good electrical conductivity in both the solid and liquid states because of delocalised electrons that are free to move. Insoluble in water.
• Giant covalent / macromolecular (e.g. $\text{SiO}_2$): very high melting and boiling points (many strong covalent bonds in a giant lattice), but poor electrical conductivity in all states because there are no free electrons or ions. Insoluble in water.

Understanding the Question

We are given a table of physical properties for four substances, W, X, Y and Z, and asked to identify them from four candidate sets. The key information is: melting point, boiling point, electrical conductivity of the solid, of the liquid, and in water. The task is to match each substance's property profile to the correct bonding type, then to the correct compound.

The command is effectively "identify" — a recognition task that requires you to reason from properties to structure. The discriminating features are the conductivity behaviour (solid vs liquid vs aqueous) and solubility in water, more than the exact melting points.

Approach

Classify each substance by its bonding type first, using the properties as evidence. Then match the bonding type to the compounds offered in the options. Where two options share the same bonding type (e.g. both MgO and NaF are ionic), use the other properties (such as solubility of X in water) to eliminate the wrong candidates.

Step-by-Step Reasoning

Substance W — melting point 993 °C, boiling point 1695 °C, poor conductor as a solid, good conductor when molten and in water. This is the classic ionic profile: high melting point from the strong ionic lattice, no conductivity as a solid (ions fixed in the lattice), but conductivity when molten or dissolved (ions free to move). Both MgO and NaF are ionic, so W could be either at this stage.

Substance X — melting point −119 °C, boiling point 39 °C, poor conductor in all states, insoluble in water. Very low melting and boiling points indicate weak intermolecular forces — a simple molecular covalent substance. It does not conduct in any state because there are no free charged particles. The crucial detail is that it is insoluble in water. This eliminates HCl: although HCl is a simple covalent molecule, it is a gas that dissolves readily in water and the resulting solution conducts electricity (it forms $\text{H}^+$ and $\text{Cl}^-$ ions). $\text{C}_2\text{H}_5\text{Br}$ (bromoethane) is a simple covalent molecule that is insoluble in water and non-conducting. So X = $\text{C}_2\text{H}_5\text{Br}$. This eliminates options B and D.

Substance Y — melting point 1535 °C, good conductor as a solid and as a liquid, insoluble in water. Conducting in the solid state is the defining feature of a metal: delocalised electrons are free to move whether solid or liquid. Both Fe and K are metals, so Y could be either. The high melting point is consistent with a metal.

Substance Z — melting point 1610 °C, poor conductor in all states, insoluble in water. Very high melting point with no conductivity at all points to a giant covalent (macromolecular) structure: many strong covalent bonds hold the lattice together, but there are no free electrons or ions. $\text{SiO}_2$ (silicon dioxide) is the classic giant covalent substance. $\text{Al}_2\text{O}_3$ is ionic — it would conduct when molten — so Z cannot be $\text{Al}_2\text{O}_3$. This eliminates option A (and B, already eliminated).

With X = $\text{C}_2\text{H}_5\text{Br}$ and Z = $\text{SiO}_2$, only option C remains: W = NaF, Y = Fe. Both are consistent: NaF is ionic (matches W's profile) and Fe is metallic (matches Y's profile).

Why the distractors fail:

• Option A: Z = $\text{Al}_2\text{O}_3$ is wrong because $\text{Al}_2\text{O}_3$ is ionic and would conduct when molten.
• Options B and D: X = HCl is wrong because HCl is soluble in water and its solution conducts; the table says X is insoluble and non-conducting.
• Options B and D also pair Y = K, but the deciding factor is X and Z, which already settle the answer.

Key Takeaways

• Electrical conductivity in different states is the single most powerful diagnostic: solid only = metallic; liquid/aqueous only = ionic; never = simple molecular or giant covalent.
• Solubility in water helps distinguish simple molecular substances that ionise on dissolving (HCl) from those that do not ($\text{C}_2\text{H}_5\text{Br}$).
• Melting/boiling points separate simple molecular (low) from ionic/metallic/giant covalent (high).
• To identify a substance from properties, classify the structure first, then match to the compound.

Common Mistakes

• Assuming HCl is insoluble: HCl is a gas that dissolves in water to form a conducting acidic solution, so it cannot be the insoluble, non-conducting X.
• Confusing $\text{Al}_2\text{O}_3$ with $\text{SiO}_2$: both are high-melting solids, but $\text{Al}_2\text{O}_3$ is ionic (conducts when molten) whereas $\text{SiO}_2$ is giant covalent (never conducts).
• Ignoring conductivity of the solid: a substance that conducts as a solid must be metallic, which immediately rules out ionic or covalent candidates.
• Stopping after identifying W as ionic: both MgO and NaF are ionic, so you must use the other substances (X and Z) to reach the unique answer.

Things to Be Careful About

• Read the conductivity columns carefully: "poor/good" for solid, liquid and in water are three separate pieces of evidence, and each distinguishes a different structure type.
• The phrase "insoluble" for X is decisive — do not overlook it when comparing HCl and $\text{C}_2\text{H}_5\text{Br}$.
• High melting point alone does not distinguish ionic from metallic or giant covalent; you must combine it with conductivity data.
• In matching options, eliminate systematically: find the substance that locks in one option (here X and Z) rather than trying to decide every entry independently.

Background Concept

Enthalpy of combustion is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions. In a simple calorimetry experiment, the heat released by combustion is assumed to be entirely absorbed by a known mass of water. The heat absorbed is calculated using:

$q = mc\Delta T$

where $m$ is the mass of water (in g), $c$ is the specific heat capacity of water ($4.18 \text{ J g}^{-1} \text{ K}^{-1}$), and $\Delta T$ is the temperature rise (in K or °C, since the scale is the same). The enthalpy change per mole is then:

$\Delta H = \frac{-q}{n}$

The negative sign indicates the reaction is exothermic (heat is released by the system to the surroundings). The result is typically expressed in kJ mol⁻¹, so $q$ must be converted from joules to kilojoules.

Understanding the Question

This is a multiple-choice question asking for the enthalpy of combustion of butan-1-ol from calorimetry data. The image shows a copper can containing 175 cm³ of water being heated by a burner containing butan-1-ol. The table provides:

• Initial and final mass of the burner + fuel (to find mass burned)
• Initial and final temperature of the water (to find ΔT)
• Mr of butan-1-ol = 74

We need to calculate $\Delta H_c$ in kJ mol⁻¹ and match it to one of the four options.

Approach

1. Find the mass of water from the volume using the given density (1.00 g cm⁻³).
2. Calculate the temperature change ΔT.
3. Use $q = mc\Delta T$ to find the heat absorbed by the water in joules, then convert to kJ.
4. Find the mass of butan-1-ol burned from the difference in burner masses.
5. Convert the mass burned to moles using the relative molecular mass.
6. Divide the heat energy by the number of moles to get $\Delta H_c$ per mole, with a negative sign (exothermic).
7. Compare with the options, checking units and significant figures.

Step-by-Step Reasoning

Step 1: Mass of water
$m = 175 \text{ cm}^3 \times 1.00 \text{ g cm}^{-3} = 175 \text{ g}$

Step 2: Temperature change
$\Delta T = 41.1 - 17.6 = 23.5 \ ^\circ\text{C}$

Step 3: Heat absorbed by water
$q = 175 \times 4.18 \times 23.5 = 17190 \text{ J} = 17.19 \text{ kJ}$

Step 4: Mass of butan-1-ol burned
$\Delta m = 58.34 - 57.85 = 0.49 \text{ g}$

Step 5: Moles of butan-1-ol
$n = \frac{0.49}{74} = 6.62 \times 10^{-3} \text{ mol}$

Step 6: Enthalpy of combustion
$\Delta H_c = \frac{-17.19 \text{ kJ}}{6.62 \times 10^{-3} \text{ mol}} = -2596 \text{ kJ mol}^{-1}$

Rounded to three significant figures: $-2600 \text{ kJ mol}^{-1}$.

This matches option C.

Why the other options are wrong:

• A ($-114 \text{ J mol}^{-1}$): This would result from forgetting to convert J to kJ and dividing by moles incorrectly, or from using the mass of fuel instead of moles.
• B ($-17.2 \text{ kJ mol}^{-1}$): This is the total heat released (17.19 kJ) without dividing by the number of moles. A common mistake is to stop at $q$ and report it as the enthalpy change.
• D ($-4540 \text{ kJ mol}^{-1}$): This would result from using the mass of fuel (0.49 g) instead of moles, or from an arithmetic error such as using the wrong ΔT or mass.

Key Takeaways

• In calorimetry, always convert $q$ from joules to kilojoules before dividing by moles to get kJ mol⁻¹.
• Enthalpy of combustion is defined per mole of fuel, not per gram or per total heat released.
• The sign is negative for exothermic reactions (combustion is always exothermic).
• Read all data carefully, including values from diagrams (e.g., volume of water).

Common Mistakes

• Forgetting to convert J to kJ: Reporting $-2.60 \times 10^6 \text{ J mol}^{-1}$ or confusing the scale. Option B ($-17.2 \text{ kJ mol}^{-1}$) is the trap for students who stop at $q$.
• Dividing by mass instead of moles: Giving $-17.19 / 0.49 = -35.1 \text{ kJ g}^{-1}$, which is not the enthalpy of combustion.
• Using the wrong mass of water: Forgetting to use 175 g from the diagram and instead using some other value.
• Sign error: Forgetting that combustion is exothermic and giving a positive value.
• Significant figures: The answer must be given to three significant figures as stated in the question.

Things to Be Careful About

• The specific heat capacity of water ($4.18 \text{ J g}^{-1} \text{ K}^{-1}$) is not given in the question — it is expected knowledge for CIE A-Level Chemistry.
• The volume of water (175 cm³) is in the diagram, not the table — always check both.
• The question asks for the answer to three significant figures: $-2596$ rounds to $-2600$ (3 s.f.). Be careful with trailing zeros — $-2600$ to 3 s.f. is correct; $-2600.0$ would be 5 s.f.
• State symbols are not required for this calculation-based MCQ, but if writing a full answer, $\Delta H_c$ refers to the standard enthalpy change of combustion.

Background Concept

A reaction pathway diagram (or energy profile) plots the energy of the system against the extent of reaction. Key features include:

• The energy level of the reactants and products. If products are higher than reactants, the reaction is endothermic (ΔH > 0). If products are lower, it is exothermic (ΔH < 0).
• The activation energy (Ea) is the minimum energy required for the reaction to occur. It is represented by the energy difference between the reactants and the highest point on the curve (the transition state or activated complex). On a diagram, Ea is always shown as an upward arrow from the reactant energy level to the peak.
• The enthalpy change (ΔH) is the energy difference between the products and the reactants. For an endothermic reaction, ΔH is positive and is shown as an upward arrow from the reactant level to the product level.

Understanding the Question

The question provides a chemical equation with a positive enthalpy change (ΔH° = +90 kJ mol^-1), indicating an endothermic reaction. We must select the diagram that correctly shows:

1. Products at a higher energy level than reactants (endothermic).
2. Ea as the energy difference from reactants to the peak.
3. ΔH° as the energy difference from reactants to products.

Approach

Evaluate each diagram against the three criteria above. Eliminate exothermic diagrams first, then check the labeling of Ea and ΔH° on the remaining endothermic diagrams.

Step-by-Step Reasoning

1. The reaction has ΔH° = +90 kJ mol^-1. A positive ΔH° means the reaction is endothermic, so the energy of the products (NO(g)) must be higher than the energy of the reactants (1/2 N2(g) + 1/2 O2(g)).
2. Diagrams A and D show products at a lower energy level than reactants, representing exothermic reactions. These are incorrect.
3. Diagrams B and C show products at a higher energy level, representing endothermic reactions. We must now check the labeling of Ea and ΔH°.
4. Activation energy (Ea) is the energy barrier from the reactants to the transition state. It must be drawn as an arrow from the reactant energy level up to the peak of the curve.
5. In Diagram B, Ea is incorrectly drawn from the peak down to the product level. This is wrong.
6. In Diagram C, Ea is correctly drawn from the reactant level up to the peak. ΔH° is correctly drawn as an upward arrow from the reactant level to the product level, indicating a positive enthalpy change.
7. Therefore, Diagram C is the correct representation.

Key Takeaways

• Endothermic reactions have products at higher energy than reactants; exothermic have products lower.
• Ea is always measured from reactants to the peak, not from products to the peak.
• ΔH is measured from reactants to products; its direction (up or down arrow) indicates the sign of the enthalpy change.

Common Mistakes

• Confusing the direction of Ea: drawing it from products to the peak instead of reactants to the peak.
• Assuming a positive ΔH means the reaction is exothermic.
• Misidentifying the arrow for ΔH°: it must be between reactant and product levels, not between peak and products.

Things to Be Careful About

• Always check the sign of ΔH to determine if the profile is endothermic or exothermic.
• Ensure Ea is measured from the reactant line, not the product line.
• State symbols and stoichiometric coefficients do not change the shape of the profile, only the relative energy levels and the numerical values of ΔH and Ea.

Background Concept

Oxidation number (oxidation state) is a bookkeeping charge assigned to each atom in a species. It is not a real charge for covalent compounds, but it tracks electron transfer in redox reactions. The key rules are: a free element has oxidation number 0; a monatomic ion has an oxidation number equal to its charge; the sum of oxidation numbers in a neutral compound is 0; the sum in a polyatomic ion equals the ion charge; oxygen is usually $-2$ (except in peroxides); hydrogen is usually $+1$; group 1 metals are $+1$. The more electronegative atom in a bond is assigned the negative oxidation number. Chlorine is more electronegative than hydrogen but less electronegative than oxygen, so in compounds with oxygen chlorine has a positive oxidation number.

Understanding the Question

This multiple-choice question presents four redox equations and asks in which reaction the oxidation number of chlorine changes by the largest amount. The word 'amount' means the absolute magnitude of the change, not whether chlorine is oxidised or reduced. The task is to assign an oxidation number to every chlorine-containing species on both sides of each equation and compare the differences.

Approach

For each option, identify all chlorine-containing species. Use the standard rules: set oxygen to $-2$, hydrogen to $+1$, and use the overall charge of the species to solve for chlorine. Then compute the absolute difference between the reactant oxidation number and the product oxidation number for each chlorine atom. The reaction with the largest such difference is the answer. There is no need to balance the equations because oxidation numbers are determined from the formulas themselves.

Step-by-Step Reasoning

Option A: $\text{KClO}_3$ contains $\text{K}^+$ and three oxygen atoms, so $+1 + x + 3(-2) = 0$, giving $x = +5$. In $\text{KCl}$, chlorine is $-1$. The change is $|+5 - (-1)| = 6$.

Option B: In $\text{ClO}^-$, oxygen is $-2$ and the ion charge is $-1$, so chlorine is $+1$. $\text{Cl}^-$ has chlorine $-1$. In $\text{ClO}_2^-$, the two oxygens contribute $-4$ and the ion charge is $-1$, so chlorine is $+3$. The changes are $|+1 - (-1)| = 2$ and $|+1 - (+3)| = 2$. This is a disproportionation reaction because the same element is both reduced and oxidised.

Option C: $\text{Cl}_2$ is a free element, so chlorine is $0$. In $\text{HCl}$, hydrogen is $+1$, so chlorine is $-1$. In $\text{HClO}$, hydrogen is $+1$ and oxygen is $-2$, so chlorine is $+1$. The changes are $1$ in each case.

Option D: In $\text{NaClO}_2$, sodium is $+1$ and the two oxygens are $-4$, so chlorine is $+3$. $\text{Cl}_2$ has chlorine $0$. In $\text{NaCl}$, chlorine is $-1$. In $\text{ClO}_2$, the two oxygens are $-4$ and the molecule is neutral, so chlorine is $+4$. The changes are $|+3 - (-1)| = 4$, $|0 - (-1)| = 1$, and $|0 - (+4)| = 4$. The largest change in this option is $4$.

Comparing all options, the largest change is $6$, which occurs in option A.

Key Takeaways

Oxidation number assignment is a systematic skill: set known oxidation numbers, use the overall charge, and solve for the unknown. The magnitude of an oxidation number change tells you how many electrons are involved in the redox change for that atom. This question also shows that halogen atoms can have positive oxidation numbers when bonded to more electronegative elements such as oxygen.

Common Mistakes

• Assuming chlorine is always $-1$ because it is a halogen. In oxyanions such as $\text{ClO}^-$ and $\text{ClO}_2^-$, and in oxides such as $\text{ClO}_2$, chlorine is positive.
• Forgetting that oxygen is more electronegative than chlorine, so oxygen is assigned $-2$ and chlorine takes the positive oxidation number.
• In option D, only comparing one product and missing that chlorine appears in three different product species with different oxidation numbers.
• Using the sign of the change instead of its magnitude. The question asks for the largest amount of change, so absolute values must be compared.

Things to Be Careful About

• For a polyatomic ion, the oxidation numbers must sum to the ion charge, not to zero. For example, $\text{ClO}^-$ must sum to $-1$.
• For a neutral molecule such as $\text{ClO}_2$, the oxidation numbers must sum to zero.
• Oxygen is $-2$ in all these species; there are no peroxides here.
• Compare absolute differences, not just whether chlorine is oxidised or reduced.
• The equations do not need to be balanced to determine oxidation number changes.

Background Concept

This question tests Le Chatelier's principle: when a system at equilibrium is subjected to a change in conditions, the equilibrium position shifts in the direction that tends to oppose that change.

For pressure, only gaseous species matter. Increasing pressure favours the side with fewer moles of gas; decreasing pressure favours the side with more moles of gas. For temperature, raising the temperature favours the endothermic direction, while lowering it favours the exothermic direction. The sign of $\Delta H^\ominus$ tells us which direction is endothermic.

Understanding the Question

The equation shows the industrial steam reforming of methane:

All species are gases. The question asks for the conditions that give the highest equilibrium yield of hydrogen, not the fastest rate. We need to decide separately whether low or high pressure, and low or high temperature, push the equilibrium to the right.

Approach

Apply Le Chatelier's principle twice.

1. Count the total moles of gas on each side: left = 2, right = 4. Decide which pressure change favours the side with more gas.
2. Use the sign of $\Delta H^\ominus$ to decide which temperature change favours the forward (endothermic) reaction.
   Combine the two choices and select the option.

Step-by-Step Reasoning

Pressure:
Left-hand side: 1 mol $\text{CH}_4$ + 1 mol $\text{H}_2\text{O}$ = 2 mol gas.
Right-hand side: 1 mol $\text{CO}$ + 3 mol $\text{H}_2$ = 4 mol gas.
A decrease in pressure favours the side with more gas molecules, so low pressure shifts the equilibrium to the right, increasing hydrogen yield.

Temperature:
$\Delta H^\ominus$ is positive, so the forward reaction is endothermic. Adding heat by raising the temperature favours the endothermic direction, so high temperature shifts the equilibrium to the right.

Therefore the highest yield of hydrogen is obtained at low pressure and high temperature, which is option A.

Key Takeaways

• Le Chatelier's principle predicts the direction of shift for pressure, temperature and concentration changes.
• For pressure, count only gaseous moles; shift is toward fewer moles at high pressure and more moles at low pressure.
• For temperature, use the sign of $\Delta H^\ominus$: high temperature favours the endothermic direction.
• Equilibrium yield is a thermodynamic question; rate and catalysts do not change the equilibrium position.

Common Mistakes

• Choosing high pressure because it increases rate. The question asks for equilibrium yield, and high pressure actually favours the 2-mole side (left), reducing hydrogen yield.
• Confusing the direction: $\Delta H^\ominus = +205\text{ kJ mol}^{-1}$ means endothermic, so high temperature favours the forward reaction, not the reverse.
• Forgetting to count all gaseous species. Here water is (g), so it must be included on the left.
• Assuming a catalyst changes the yield. A catalyst only speeds up the attainment of equilibrium; it does not shift the position.

Things to Be Careful About

• Always note the state symbols: only gases count in the pressure effect. Here all species are gases.
• Check the sign of $\Delta H^\ominus$ carefully: positive = endothermic; negative = exothermic.
• The question says "at equilibrium", so answer in terms of equilibrium position, not reaction rate.
• Match the option exactly: low pressure + high temperature is option A.

Background Concept

Disproportionation is a redox reaction in which the same element is simultaneously oxidised and reduced. Here nitrogen is present in only one reactant, $\text{HNO}_2$, and appears in two products with different oxidation states. To identify what happens, we assign oxidation numbers using the usual rules: hydrogen is $+1$, oxygen is $-2$, and the sum of oxidation numbers in a neutral molecule is zero.

For $\text{HNO}_2$: $+1 + x + 2(-2) = 0$, so $x = +3$.
For $\text{HNO}_3$: $+1 + x + 3(-2) = 0$, so $x = +5$.
For $\text{NO}$: $x + (-2) = 0$, so $x = +2$.

Thus nitrogen in $\text{HNO}_2$ is both oxidised from $+3$ to $+5$ and reduced from $+3$ to $+2$. In any redox reaction the total number of electrons lost in oxidation must equal the total number gained in reduction. This electron-balance requirement fixes the stoichiometric coefficients.

Understanding the Question

The question gives a word equation: $W$ moles of $\text{HNO}_2$ disproportionate to $U$ moles of $\text{HNO}_3$ and $V$ moles of $\text{NO}$. No other nitrogen-containing product forms, and nitrogen is the only element whose oxidation state changes. We are asked to choose the set of coefficients $W$, $U$ and $V$ that makes a balanced redox equation.

The options are deliberately chosen so that all of them conserve nitrogen atoms ($W = U + V$). Therefore the deciding factor is not simply counting nitrogen atoms, but balancing the electron transfer and, equivalently, balancing oxygen and hydrogen. The correct coefficients must satisfy a full balanced equation.

Approach

The most reliable method is to use oxidation-number changes and half-equations:

1. Assign oxidation numbers to nitrogen in the reactant and both products.
2. Write the oxidation half-equation: $\text{HNO}_2 \rightarrow \text{HNO}_3$.
3. Write the reduction half-equation: $\text{HNO}_2 \rightarrow \text{NO}$.
4. Balance each half-equation for atoms and charge in acidic conditions.
5. Multiply the half-equations so that the electrons lost equal the electrons gained.
6. Add the half-equations and cancel common species to obtain the overall balanced equation.

This approach is systematic and avoids guessing among the options.

Step-by-Step Reasoning

Oxidation numbers

• In $\text{HNO}_2$, N is $+3$.
• In $\text{HNO}_3$, N is $+5$: oxidation, each N loses $2\text{e}^-$.
• In $\text{NO}$, N is $+2$: reduction, each N gains $1\text{e}^-$.