Chemistry 5070/11 — May/June 2024

Walkthrough

The question provides a heating curve (temperature against time) for a liquid being heated. We need to evaluate three statements about particle behaviour and energy during different stages of this process.

Stage 1: W to X (Liquid Heating)

• The graph slopes upwards, meaning temperature is increasing.
• In a liquid, as temperature rises, the particles gain kinetic energy. This means they move faster and vibrate more vigorously.
• Therefore, Statement 1 is correct.

Stage 2: X to Y (Change of State: Boiling)

• The graph is flat (horizontal), meaning temperature is constant.
• This plateau represents the boiling point where the liquid is turning into a gas.
• During a change of state, the heat energy supplied does not increase the temperature (kinetic energy). Instead, it is used to break or overcome the intermolecular forces of attraction holding the liquid particles together.
• Therefore, Statement 2 is correct.

Stage 3: Y to Z (Gas Heating)

• The graph slopes upwards again, meaning temperature is increasing.
• By point Y, all the liquid has turned into gas. The substance is now a gas being heated.
• In a gas, particles are far apart and move freely. As the gas is heated, the particles gain kinetic energy and move faster. Additionally, gases expand when heated (Charles's Law), so the average distance between particles increases — they move further apart.
• Therefore, Statement 3 is correct.

Since statements 1, 2, and 3 are all correct, the correct option is A.

Key Takeaways

• Heating curves show temperature changes over time. Rising slopes indicate temperature change (kinetic energy change); flat plateaus indicate change of state (potential energy change).
• Temperature is a measure of average kinetic energy. Higher temperature = faster moving particles.
• During a change of state (melting or boiling), temperature remains constant because energy is used to overcome forces of attraction between particles, not to increase speed.
• Gas particles are much further apart than liquid particles. Heating a gas causes it to expand, increasing the distance between particles further.

Common Mistakes

• Thinking temperature rises during boiling: Candidates often assume that adding heat always raises temperature. Remember that during a phase change (plateau on the graph), temperature is constant.
• Confusing particle speed with separation: In the gas phase (Y to Z), particles move faster and move further apart (as the gas expands). Some candidates think particles only move faster in the gas phase and forget about the expansion.
• Misinterpreting the graph axes: Ensure you are reading temperature on the y-axis and time on the x-axis. A rising line means temperature is increasing with time.

Things to Be Careful About

• State symbols and phases: W-X is liquid, X-Y is liquid + gas equilibrium, Y-Z is gas.
• Energy types: Distinguish between kinetic energy (related to temperature and speed) and potential energy (related to particle separation and state change). Energy added during the plateau increases potential energy, not kinetic energy.
• Particle arrangement: In liquids, particles are close together but can move past each other. In gases, particles are far apart and move randomly at high speeds.

Walkthrough

Start with the key difference: a compound has a fixed composition, while a mixture can have any composition. Since X contains only iron and sulfur, any compound of X must contain only Fe and S atoms. The simplest iron sulfide is $\text{FeS}$. Using $A_r(\text{Fe}) = 56$ and $A_r(\text{S}) = 32$, the mass of Fe compared with S in $\text{FeS}$ is $56 : 32$, which simplifies to $7 : 4$. So a compound with a 7 : 4 mass ratio is possible. A mixture of iron and sulfur can be made in exactly that same ratio by weighing out 7 g of iron and 4 g of sulfur, so a 7 : 4 mixture is also possible. Therefore B is the row in which each statement could describe a sample of X.

Row A fails because no iron–sulfur compound has a 4 : 7 mass ratio. Row C fails because a mixture does not have a chemical formula. Row D fails because $\text{FeSO}_4$ contains oxygen, which contradicts the statement that X contains iron and sulfur only.

Key Takeaways

• A compound is a pure substance with a fixed composition; a mixture can have any composition.
• The mass ratio of elements in a compound is found from its formula and the relative atomic masses.
• $\text{FeS}$ gives an Fe : S mass ratio of 7 : 4.
• A mixture can be made with the same mass ratio as a compound.

Common Mistakes

• Choosing A because 4 : 7 is a possible mixture ratio, without checking whether it is a possible compound ratio.
• Thinking a mixture cannot have the same composition as a compound; it can.
• Giving a chemical formula to a mixture; mixtures do not have formulae.
• Overlooking the phrase 'iron and sulfur only' and accepting $\text{FeSO}_4$.
• Reversing the ratio and writing 4 : 7 instead of 7 : 4.

Things to Be Careful About

• Read the ratio order carefully: Fe : S = 7 : 4, not S : Fe.
• A formula gives the atom ratio; convert to mass ratio using relative atomic masses.
• In multiple-choice questions, eliminate any row containing an impossible statement.
• The word 'possible' means each statement could describe some sample of X; it does not require one sample to be both a mixture and a compound at the same time.

Walkthrough

Chlorine is element 17, so its proton number is 17. The nucleon number (mass number) is given as 37. The nucleon number is the total number of protons plus neutrons, so the number of neutrons is found by subtracting the proton number from the nucleon number:

$37 - 17 = 20$

A neutral chlorine atom has the same number of electrons as protons, so it has 17 electrons. These electrons fill the shells in order: the first shell holds 2, the second shell holds 8, and the remaining 7 go into the third shell. So the electron configuration is $2, 8, 7$.

The question asks specifically for the number of electrons in the second shell, which is 8, not the number in the outer shell, which is 7.

Comparing with the table:

• neutrons = 20
• electrons in the second shell = 8

This matches row D.

Key Takeaways

• The nucleon number is the sum of protons and neutrons.
• In a neutral atom, the number of electrons equals the proton number.
• Electron shells fill as 2, 8, 8 for the first three shells.
• Read the question carefully: the second shell is not the same as the outermost shell.

Common Mistakes

• Choosing A or B: using 17 as the number of neutrons. This happens when the proton number is mistaken for the neutron number.
• Choosing C: getting the neutrons correct but writing the second shell as 7. This confuses the outer shell with the second shell.
• Forgetting that chlorine is neutral, so electrons = protons = 17.

Things to Be Careful About

• Nucleon number $=$ protons $+$ neutrons, so neutrons $= 37 - 17$.
• Chlorine has 17 protons, not 18; argon has 18.
• The second shell can hold up to 8 electrons, so it is fully filled in chlorine.
• The electron configuration of chlorine is $2, 8, 7$, not $2, 8, 8$ or $2, 7, 8$.

Walkthrough

Isotopes are atoms of the same element. An element is defined by its proton number, so isotopes must have the same number of protons. They differ in the number of neutrons, which changes the mass number but not the identity of the element.

Look down the proton column in the table:

• W has 25 protons and X has 25 protons, so W and X are the same element. But W and X both have 28 neutrons, so they are actually the same isotope, just with different numbers of electrons (22 and 23). Different electron numbers mean they are ions of the same nuclide, not isotopes.
• Y has 26 protons and Z has 26 protons, so Y and Z are also the same element. Y has 30 neutrons while Z has 28 neutrons — different neutron numbers. This is exactly the definition of isotopes.

Hence the correct pair is Y and Z, option D.

Key Takeaways

• Isotopes are atoms of the same element: same proton number, different neutron number.
• The number of electrons can change without changing the element; that makes an ion, not a different isotope.
• When identifying isotopes from a table, compare the proton numbers first.

Common Mistakes

• Choosing W and X because they have the same proton number but forgetting that they also have the same neutron number — they are the same isotope, not a pair of isotopes.
• Thinking that different electron numbers make isotopes. Electrons do not define the element or the isotope.
• Confusing neutron number with mass number. Mass number = protons + neutrons, but isotopes are distinguished by neutron number alone when proton number is fixed.

Things to Be Careful About

• Read the table headings carefully: electrons, neutrons, protons. Isotopes differ only in neutrons.
• A neutral atom would have electrons equal to protons, but the table shows charged particles, so do not use electrons to decide the element.
• The mark scheme answer is D only; no other pair satisfies the isotope definition.

Walkthrough

This is a 'which statement is correct' multiple-choice question, so check each option in turn.

Option A is the correct statement. In the blast furnace, iron ore is not pure iron oxide; it contains rocky impurities such as sand, which is mainly silicon(IV) oxide, $\text{SiO}_2$. These impurities are removed by reaction with limestone, forming slag. This is a standard fact about iron extraction.

Option B is wrong because the two structures are different. Graphite is made of flat layers in which each carbon atom is bonded to three other carbon atoms, so the bond angles are 120°. Silicon(IV) oxide is a giant covalent structure in which each silicon atom is bonded to four oxygen atoms in a tetrahedral arrangement, so the bond angles are about 109.5°. Therefore the angles are not the same.

Option C is wrong because it gives the wrong reason for the high melting points. Both graphite and silicon(IV) oxide are giant covalent structures. Their melting points are high because a large amount of energy is needed to break many strong covalent bonds. They do not contain ionic bonds, so the statement about ionic bonds being stronger than covalent bonds is irrelevant and incorrect.

Option D is wrong because it confuses the forces that are overcome when graphite is used as a lubricant. Within each layer of graphite, the carbon atoms are joined by strong covalent bonds. Between the layers there are only weak forces of attraction. When graphite lubricates, the layers slide over each other because these weak inter-layer forces are overcome. The covalent bonds between atoms within a layer are not broken.

Key Takeaways

• Graphite and silicon(IV) oxide are both giant covalent structures, so they have very high melting points.
• Graphite has a layered structure: strong covalent bonds within layers and weak forces between layers, which explains its use as a lubricant.
• Silicon(IV) oxide has a tetrahedral structure, so its bond angles are about 109.5°, not 120°.
• Iron ore often contains silicon(IV) oxide as an impurity, which is removed as slag in the blast furnace.

Common Mistakes

• Saying that high melting points are due to ionic bonds when the substance is covalent. Graphite and $\text{SiO}_2$ are not ionic.
• Thinking that graphite lubricates because covalent bonds break. The covalent bonds stay intact; only weak forces between layers are overcome.
• Assuming that all giant covalent structures have the same bond angles. The structure must be considered: graphite is layered with 120° angles, while $\text{SiO}_2$ is tetrahedral with 109.5° angles.

Things to Be Careful About

• Read each statement carefully and look for the reason given, not just the property. Option C mentions a true property (high melting point) but gives a false reason.
• Remember that in a giant covalent structure, melting involves breaking covalent bonds, which requires a lot of energy.
• For the blast furnace, recall that limestone removes the sandy $\text{SiO}_2$ impurity as slag, so the presence of $\text{SiO}_2$ in iron ore is a real and important fact.

Walkthrough

This question tests your understanding of the difference between empirical and molecular formulae.

• The empirical formula shows the simplest whole-number ratio of atoms of each element in a compound.
• The molecular formula shows the actual number of atoms of each element in one molecule of the compound.

Let's evaluate each statement:

A says the empirical and molecular formulae are always different. This is false because for many compounds like water ($\text{H}_2\text{O}$) or carbon dioxide ($\text{CO}_2$), the empirical and molecular formulae are identical. So A is wrong.

B says the empirical formulae of ethyne and benzene are the same. Ethyne is $\text{C}_2\text{H}_2$. Divide both subscripts by 2 to get the simplest ratio, which is $\text{CH}$. Benzene is $\text{C}_6\text{H}_6$. Divide by 6 to get $\text{CH}$. Both simplify to $\text{CH}$, so they have the same empirical formula. This statement is correct.

C says the molecular formula always shows the simplest whole-number ratio. That is actually the definition of an empirical formula, not a molecular formula. The molecular formula shows the actual numbers of atoms. So C is false.

D says the empirical formula always shows the numbers and types of different atoms in one molecule. That is the definition of a molecular formula. The empirical formula only gives the simplest ratio. So D is false.

Therefore, the only correct statement is B.

Key Takeaways

• Know the definitions of empirical and molecular formulae.
• Be able to simplify a molecular formula to its empirical formula by dividing by the highest common factor.
• Understand that the empirical formula and molecular formula can be the same for some compounds.
• Read each statement carefully and test it against the definitions.

Common Mistakes

• Confusing the definitions of empirical and molecular formulae. Remember: empirical = simplest ratio, molecular = actual numbers.
• Thinking that empirical and molecular formulae are always different. They can be the same.
• Forgetting to simplify both subscripts in a molecular formula when finding the empirical formula.

Things to Be Careful About

• When simplifying, ensure you divide all subscripts by the same whole number to get the smallest integer ratio.
• In option B, both compounds contain only carbon and hydrogen, and both simplify to $\text{CH}$. Double-check your arithmetic.
• The term "always" in options A, C, and D is a strong indicator that the statement might be false, as many absolute statements in chemistry have exceptions.

Walkthrough

Aluminium oxide contains the aluminium ion $\text{Al}^{3+}$ and the oxide ion $\text{O}^{2-}$. To balance the charges, the formula has two aluminium ions and three oxide ions: $\text{Al}_2\text{O}_3$.

Next, find the relative formula mass. Use the standard relative atomic masses: aluminium is 27 and oxygen is 16.

$\begin{aligned} M_r &= (2 \times 27) + (3 \times 16) \\ &= 54 + 48 \\ &= 102 \end{aligned}$

So the correct option is C.

Key Takeaways

• Ionic compound formulae are found by balancing positive and negative charges.
• Relative formula mass is the sum of the relative atomic masses of all atoms in the formula.
• For a compound like $\text{Al}_2\text{O}_3$, you must multiply each atomic mass by its subscript.

Common Mistakes

• Forgetting the subscript 2 on aluminium and 3 on oxygen; using the formula AlO would give 43.
• Using aluminium's atomic number (13) instead of its relative atomic mass (27).
• Adding the atomic masses without multiplying by the subscripts.

Things to Be Careful About

• Always write the formula correctly before calculating $M_f$.
• The question asks for $M$, which is a number with no units.
• Option B (75) would come from $27 + 3 \times 16$ if the 2 is missed on aluminium; option A (43) would come from assuming one aluminium and one oxygen.

Walkthrough

The first step is to identify the formula of anhydrous copper sulfate. The word anhydrous means 'without water', so the formula is $\text{CuSO}_4$, not the hydrated form $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$.

Next, calculate the relative formula mass:

So 16.0 g is:

The key point is that the question asks for the number of ions, not the number of formula units. In an ionic compound, a formula unit is the smallest electrically neutral group of ions. $\text{CuSO}_4$ is made of one $\text{Cu}^{2+}$ ion and one $\text{SO}_4^{2-}$ ion, so each formula unit contains 2 ions. Therefore:

Finally, use the Avogadro constant, $6.02 \times 10^{23}$ particles per mole:

This matches option A.

Key Takeaways

• Anhydrous means no water of crystallisation; hydrated copper sulfate would be $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$.
• The mole connects mass to number of particles: $\text{moles} = \frac{\text{mass}}{M_r}$.
• For ionic compounds, count ions per formula unit, not just formula units.
• The Avogadro constant converts moles of any particle to the actual number of particles.

Common Mistakes

• Forgetting to multiply by 2 ions per formula unit. This would give $6.02 \times 10^{22}$ ions, which is not one of the options.
• Using the hydrated formula $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$, which gives a larger $M_r$ and a wrong answer.
• Using the mass as if it were one mole: 160 g would contain $1.20 \times 10^{24}$ ions, which is option C.
• Confusing ions with molecules or atoms; the question specifically asks for ions.

Things to Be Careful About

• Use the correct relative atomic masses (Cu = 64, S = 32, O = 16) to get $M_r = 160$.
• Keep track of units: the answer is a pure number of ions, not a mass or volume.
• Give the final answer to the same number of significant figures as the data (16.0 has 3 significant figures), so $1.20 \times 10^{23}$ is appropriate.
• In the options, all answers are in the form $x \times 10^{23}$ or $x \times 10^{24}$, so check the power of ten carefully.

Walkthrough

The mole is a counting unit: one mole of any substance contains the same number of particles, about $6.02 \times 10^{23}$. So to compare the total number of atoms in different samples, we cannot simply compare the numbers of moles. We must multiply by how many atoms are in one molecule of each substance.

• Water, $\text{H}_2\text{O}$, contains 2 hydrogen atoms and 1 oxygen atom, so 3 atoms per molecule.
  $0.5$ mol of water gives $0.5 \times 3 = 1.5$ mol of atoms.

• Carbon dioxide, $\text{CO}_2$, contains 1 carbon atom and 2 oxygen atoms, so 3 atoms per molecule.
  $1.0$ mol of carbon dioxide gives $1.0 \times 3 = 3.0$ mol of atoms.

• Methane, $\text{CH}_4$, contains 1 carbon atom and 4 hydrogen atoms, so 5 atoms per molecule.
  $1.0$ mol of methane gives $1.0 \times 5 = 5.0$ mol of atoms.

• Hydrogen chloride, $\text{HCl}$, contains 1 hydrogen atom and 1 chlorine atom, so 2 atoms per molecule.
  $2.0$ mol of hydrogen chloride gives $2.0 \times 2 = 4.0$ mol of atoms.

Methane gives 5.0 mol of atoms, which is more than all the other samples, so the correct option is C.

Key Takeaways

• The number of moles alone does not tell you the number of atoms; you must also know how many atoms are in each molecule.
• Total moles of atoms = moles of substance $\times$ number of atoms per molecule.
• Always compare the same thing, for example mol of atoms, before deciding which sample has the most atoms.

Common Mistakes

• Choosing D because 2.0 mol is the largest number of moles, without realising that each $\text{HCl}$ molecule contains only 2 atoms.
• Forgetting to count all atoms in a molecule, such as counting water as 2 atoms instead of 3.
• Comparing moles of molecules instead of moles of atoms.

Things to Be Careful About

• Write the formula before counting: $\text{H}_2\text{O}$ has 3 atoms, $\text{CO}_2$ has 3, $\text{CH}_4$ has 5, and $\text{HCl}$ has 2.
• Remember that the mole is a fixed number of particles, so the comparison is still valid even though the samples are different substances.
• There is no need to convert to an actual number of atoms here because all options can be compared using mol of atoms.

No diagram is needed; the comparison is numerical.

Walkthrough

All four reactions have the other reactant in excess, so the amount of hydrogen is controlled only by the quantity given in each option. At r.t.p. the molar gas volume is $24\text{ dm}^3\text{ / mol}$, so comparing volumes is the same as comparing moles of hydrogen.

For each option, convert the given quantity into moles, then use the balanced equation to find the moles of hydrogen produced.

• A: $1.4$ g of carbon is $1.4/12 = 0.1167$ mol. Since the equation shows 1 mol C $\rightarrow$ 1 mol $\text{H}_2$, this gives $0.1167$ mol $\text{H}_2$.
• B: $2.4$ g of $\text{CaH}_2$ is $2.4/42 = 0.05714$ mol. The equation shows 1 mol $\text{CaH}_2$ $\rightarrow$ 2 mol $\text{H}_2$, so this gives $2 \times 0.05714 = 0.1143$ mol $\text{H}_2$.
• C: $4.0$ g of calcium is $4.0/40 = 0.100$ mol. The equation shows 1 mol Ca $\rightarrow$ 1 mol $\text{H}_2$, so this gives $0.100$ mol $\text{H}_2$.
• D: $50\text{ cm}^3 = 0.050\text{ dm}^3$, so moles of acid $= 0.050 \times 2.0 = 0.100$ mol. The equation shows 1 mol acid $\rightarrow$ 1 mol $\text{H}_2$, so this gives $0.100$ mol $\text{H}_2$.

A gives $0.1167$ mol, which is greater than B's $0.1143$ mol and greater than C and D's $0.100$ mol. Therefore A produces the greatest volume of hydrogen.

Key Takeaways

• To compare gas volumes at r.t.p., compare moles: 1 mol of any gas occupies $24\text{ dm}^3$ at r.t.p.
• Convert masses to moles using $M_r$, and convert $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000 before using concentration.
• Always use the mole ratio from the balanced equation.
• The reactant in excess does not limit the amount of product; the given reactant does.

Common Mistakes

• Comparing the masses directly instead of converting to moles. For example, $2.4$ g of $\text{CaH}_2$ and $4.0$ g of Ca are not compared by mass alone.
• Forgetting the 2:1 ratio in option B: 1 mol $\text{CaH}_2$ gives 2 mol $\text{H}_2$, not 1 mol.
• Using $M_r$ of $\text{CaH}_2$ as 40 instead of 42.
• In option D, forgetting to convert $50\text{ cm}^3$ into $0.050\text{ dm}^3$ before multiplying by concentration.
• Using the molar volume at s.t.p. ($22.4\text{ dm}^3$) instead of r.t.p. ($24\text{ dm}^3$). Here it would not change the answer, but it is a common slip.

Things to Be Careful About

• The question says "measured at room temperature and pressure", so use $24\text{ dm}^3\text{ / mol}$.
• Options A and B are very close: A gives $0.1167$ mol and B gives $0.1143$ mol. A careless rounding could make B look larger, so keep enough decimal places until the final comparison.
• The equations are already balanced, but you still need to read the coefficients carefully, especially the 2 in front of $\text{H}_2$ in option B.
• All other reactants are in excess, so the amount of hydrogen is fixed by the substance named in each option.

Walkthrough

The question asks which method gives a solution of concentration $0.10\text{ mol / dm}^3$.

Concentration is the amount of solute divided by the volume of solution, with volume measured in $\text{dm}^3$:

where $n$ is the number of moles and $V$ is the volume in $\text{dm}^3$.

For NaOH, $M_r = 40$, so 1 mole of NaOH has a mass of 40 g. To find the number of moles in each option, divide the mass by 40.

Option A: $1.6$ g of NaOH is $1.6/40 = 0.040$ mol. The volume is $400\text{ cm}^3 = 0.400\text{ dm}^3$. Dividing moles by volume:

This exactly matches the required concentration, so A is correct.

Checking the other options:

• B: $3.2/40 = 0.080$ mol in $0.250\text{ dm}^3$ gives $0.32\text{ mol / dm}^3$.
• C: $4.0/40 = 0.10$ mol in $0.100\text{ dm}^3$ gives $1.0\text{ mol / dm}^3$.
• D: $8.0/40 = 0.20$ mol in $0.500\text{ dm}^3$ gives $0.40\text{ mol / dm}^3$.

Only A gives the target concentration.

Key Takeaways

• Concentration in $\text{mol / dm}^3$ is calculated as moles of solute divided by volume of solution in $\text{dm}^3$.
• Convert $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000.
• Convert mass to moles using $n = m/M_r$.
• In a multiple-choice question, it is safest to check each option rather than assume the first one is correct.

Common Mistakes

• Forgetting to convert $\text{cm}^3$ to $\text{dm}^3$ before calculating concentration.
• Using the mass in grams as if it were the number of moles.
• Mixing up $250\text{ cm}^3$ and $0.250\text{ dm}^3$; both are correct, but the volume must be in $\text{dm}^3$ when using $\text{mol / dm}^3$.
• Stopping after finding one option that seems close, without checking the exact calculation.

Things to Be Careful About

• Use the given $M_r$ value, 40, for NaOH.
• Write the final concentration with the correct unit, $\text{mol / dm}^3$.
• The answer required is the option letter, A, not the mass or volume.
• In this question there is only one correct option, and the calculation for A is exact, not rounded.

Walkthrough

Electrolysis is the decomposition of a compound using electricity. For electrolysis to happen, the substance must be an electrolyte: an ionic compound that conducts electricity when molten or in aqueous solution because its ions are free to move.

Check each substance:

• Aqueous sodium chloride contains Na+ and Cl- ions free to move in water, so it can be electrolysed.
• Copper is a metal. It conducts electricity, but the charge is carried by delocalised electrons, not ions. Passing electricity through copper does not decompose it, so copper cannot be electrolysed.
• Graphite is a form of carbon. Like a metal, it conducts electricity using delocalised electrons, but it is not an electrolyte and is not decomposed by electricity.
• Molten lead(II) bromide contains Pb2+ and Br- ions free to move, so it can be electrolysed.

Only row B has aqueous sodium chloride and molten lead(II) bromide correct, with copper and graphite marked as not electrolysed.

Key Takeaways

• Electrolysis requires an electrolyte: an ionic compound that is molten or in aqueous solution.
• Electrolytes conduct electricity because they contain free-moving ions.
• Metals and graphite conduct electricity using free-moving electrons, but they are not electrolytes and cannot be electrolysed.
• Solid ionic compounds do not conduct electricity because their ions are held in a fixed lattice.

Common Mistakes

• Thinking that any electrical conductor can be electrolysed. Copper and graphite conduct electricity but are not decomposed by it.
• Confusing electronic conduction (metals and graphite) with ionic conduction (molten or aqueous ionic compounds).
• Thinking that solid ionic compounds can be electrolysed. In the solid state the ions are not free to move.

Things to Be Careful About

• The question asks which substances can be electrolysed, not which substances can conduct electricity.
• Aqueous sodium chloride can be electrolysed, but water also takes part in the electrolysis; this does not change the fact that the solution is an electrolyte.
• Molten lead(II) bromide is a classic example of an electrolyte because its ions are free to move when molten.

Walkthrough

1. Recall the requirements for electroplating: the object to be plated must be the cathode, connected to the negative terminal of the power supply. This ensures that positive metal ions (Cu²⁺) from the electrolyte are attracted to it and gain electrons (reduction) to form a metal coating.
2. The anode (positive electrode) is usually made of the plating metal (copper) so that it dissolves to replenish the metal ions in the solution.
3. The electrolyte must contain ions of the plating metal (aqueous copper(II) sulfate is correct for copper plating).
4. In the given figure, the steel key is connected to the positive terminal (+) and the copper block to the negative terminal (–).
5. Because the key is positive, it acts as the anode. Instead of being coated, the steel would dissolve or oxygen gas would be produced at its surface. The copper block is negative, so copper ions would plate onto the copper block instead of the key.
6. The correct fix is to reverse the connections: connect the key to the negative terminal and the copper electrode to the positive terminal.

Key Takeaways

• In electroplating, the object to be plated is always the cathode (connected to the negative terminal).
• The plating metal is the anode (connected to the positive terminal) so it can replenish the electrolyte.
• Positive ions (cations) in the electrolyte always move to and are reduced at the cathode.

Common Mistakes

• Thinking that the positive terminal attracts metal ions. (Metal ions are positive, so they are repelled by the positive anode and attracted to the negative cathode).
• Confusing the roles of anode and cathode in electrolysis.
• Assuming that changing the concentration or current will fix a polarity error.

Things to Be Careful About

• Remember that in electrolysis, positive ions (cations) move to the cathode (negative electrode) and negative ions (anions) move to the anode (positive electrode).
• The electrolyte must contain the metal ions of the plating metal; replacing it with dilute sulfuric acid (Option C) would not provide copper ions for plating.

Walkthrough

The equation given is the dissolving of ammonium nitrate in water, with a positive enthalpy change. In energetic terms, a positive $\Delta H$ always means endothermic: the process takes in more thermal energy than it gives out. Because the thermal energy must come from somewhere, it is absorbed from the surrounding water. When thermal energy leaves the water, the water becomes colder. So all three statements follow from the single fact that $\Delta H$ is positive.

No calculation is needed. The key is remembering that the sign of $\Delta H$ controls both the direction of energy transfer and whether the surroundings warm up or cool down.

Key Takeaways

• A positive $\Delta H$ means endothermic; a negative $\Delta H$ means exothermic.
• In an endothermic change, thermal energy is absorbed from the surroundings.
• Because the water is the surroundings, absorbing thermal energy from the water makes the water colder.
• All three statements in the question are logically connected, so if one is true, the others are also true.

Common Mistakes

• Thinking a positive $\Delta H$ means exothermic. The sign tells the direction: positive is energy absorbed, negative is energy released.
• Choosing only statement 1 without realising that statements 2 and 3 are direct consequences.
• Thinking that endothermic processes always make their surroundings hotter, which is the opposite of what happens.

Things to Be Careful About

• The unit $\text{kJ / mol}$ is the thermal energy change per mole of ammonium nitrate dissolved, but you do not need to use this value here.
• The water is the surroundings, so if thermal energy is taken from it, the temperature of the water falls.
• When several statements are joined with "and", every statement must be true before you select the option that includes all of them.

Walkthrough

The displayed equation shows two hydrogen molecules reacting with one oxygen molecule to form two water molecules:

In a reaction, bonds are first broken and then new bonds are formed. Bond breaking requires energy (endothermic), while bond making releases energy (exothermic). The enthalpy change is found by subtracting the energy released when new bonds form from the energy absorbed when old bonds break.

Count the bonds broken:

• Two H–H bonds: $2 \times 436 = 872\text{ kJ / mol}$
• One O=O bond: $496\text{ kJ / mol}$

The O–O bond energy given in the table is not used because there is no single O–O bond in this reaction.

Total energy absorbed $= 872 + 496 = 1368\text{ kJ / mol}$.

Count the bonds formed:

• Each water molecule has two O–H bonds, and two water molecules are formed, so $4$ O–H bonds are made.
• Energy released $= 4 \times 463 = 1852\text{ kJ / mol}$.

Therefore:

The negative sign shows the reaction is exothermic, so the correct option is C.

Key Takeaways

• For a bond-energy calculation, use:

• Count every bond in the balanced equation, not just one molecule's bonds.
• A negative $\Delta H$ means the reaction is exothermic; a positive $\Delta H$ means it is endothermic.

Common Mistakes

• Using the O–O bond energy of $146\text{ kJ / mol}$ instead of the O=O bond energy of $496\text{ kJ / mol}$. The reaction contains an oxygen–oxygen double bond, not a single bond.
• Counting only two O–H bonds formed because one water molecule has two O–H bonds, but two water molecules are produced, so four O–H bonds are formed.
• Subtracting in the wrong order and obtaining $+484\text{ kJ / mol}$. Bond breaking absorbs energy and bond making releases it, so the correct order is broken minus formed.
• Forgetting the negative sign, which indicates an exothermic reaction.

Things to Be Careful About

• Use the balanced equation shown in the question: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$.
• Keep the unit $\text{kJ / mol}$ throughout.
• Only use bond energies for bonds that actually appear in the reaction.
• The sign of the enthalpy change matters: energy released is greater than energy absorbed, so the answer must be negative.

Walkthrough

Silicon(IV) chloride, $\text{SiCl}_4$, is a covalent molecular substance. Each molecule is held together by strong covalent bonds between the silicon atom and the chlorine atoms. However, the molecules are only attracted to each other by weak intermolecular forces.

When $\text{SiCl}_4$ boils, it changes from a liquid to a gas. This is a change of state, not a change of substance. The particles gain enough energy to overcome the weak intermolecular forces and move apart, but the covalent bonds inside each molecule remain intact. Since no new substance is formed, the change is physical.

Now look at the options:

• A says chemical and intermolecular forces break. Boiling is not chemical, so A is wrong.
• B says chemical and Si–Cl covalent bonds break. Boiling is not chemical, and the covalent bonds are not broken, so B is wrong.
• C says physical and intermolecular forces break. This is correct.
• D says physical and Si–Cl covalent bonds break. The change is physical, but the covalent bonds are not broken, so D is wrong.

Therefore the correct answer is C.

Key Takeaways

• A physical change is a change of state or shape in which no new substance is formed.
• A chemical change produces one or more new substances.
• Simple molecular substances, such as $\text{SiCl}_4$, have weak intermolecular forces between molecules.
• Boiling or melting only breaks the weak intermolecular forces; the covalent bonds inside the molecule stay intact.
• The strength of intermolecular forces explains why simple molecular substances have low melting and boiling points.

Common Mistakes

• Choosing D because the change is physical, but incorrectly saying that covalent bonds break. Boiling does not break covalent bonds.
• Choosing A or B because the student thinks bubbles or a gas formed means a chemical change. Boiling is still a physical change.
• Confusing intermolecular forces with covalent bonds. Intermolecular forces act between molecules; covalent bonds act between atoms inside a molecule.

Things to Be Careful About

• The word “break” in the explanation means the forces are overcome or overcome. In the mark scheme, either “break” or “overcome” is usually accepted for intermolecular forces.
• Do not say that covalent bonds break during boiling; this would give the wrong explanation.
• Remember that silicon(IV) chloride is a simple molecular compound, not a giant covalent structure. Its low boiling point is evidence of weak intermolecular forces.

Walkthrough

This question asks which change makes the reaction go more slowly, so we need the factor that decreases the rate.

For a reaction to happen, particles must collide with enough energy. This is collision theory.

• Increasing acid concentration puts more acid particles in the same volume, so collisions happen more often and the reaction is faster.
• Increasing temperature gives the particles more kinetic energy, so more collisions have enough energy to react and the reaction is faster.
• Increasing the surface area of the solid exposes more calcium carbonate to the acid, so there are more collisions per second and the reaction is faster.
• Increasing the size of the solid particles does the opposite: larger lumps have a smaller surface area for the same mass, so fewer acid particles can collide with the solid each second and the reaction is slower.

Therefore the correct option is B.

Key Takeaways

• The rate of a reaction depends on how often successful collisions happen.
• Concentration, temperature, surface area and catalysts all affect the rate.
• For a fixed mass of solid, larger particles mean smaller surface area.
• "Increasing particle size" and "increasing surface area" are opposite changes.

Common Mistakes

• Choosing C because surface area is mentioned, without noticing that the question asks for a slower reaction.
• Thinking that increasing particle size increases surface area. In fact, larger particles have less surface area for the same mass.
• Selecting a factor that speeds up the reaction instead of slowing it down.

Things to Be Careful About

• Read whether the question asks for faster or slower.
• In rate questions, the precise explanation should mention successful collisions per second.
• For a solid reactant, "surface area" and "particle size" are inversely related for a fixed mass of solid.

Walkthrough

The table gives the total volume of hydrogen collected. The reaction stops when the acid is used up because aluminium is in excess. Therefore the final volume, $143\text{ cm}^3$, tells us how much aluminium reacted, not how much was added.

Convert the volume to moles using the molar gas volume at r.t.p., $24\text{ dm}^3 / \text{mol}$:

The equation shows $2\text{ mol Al}$ gives $3\text{ mol } \text{H}_2$, so:

Option A quotes $0.107\text{ g}$, but this is only the mass that reacted. Since the aluminium was in excess, the mass added is larger and unknown, so B is correct.

For the rate options, calculate the rate in each two-minute interval:

• $0-2$ min: $(3-0)/2 = 1.5$
• $2-4$ min: $(53-3)/2 = 25$
• $4-6$ min: $(103-53)/2 = 25$
• $6-8$ min: $(131-103)/2 = 14$
• later intervals are smaller.

The highest rate is $25\text{ cm}^3 / \text{min}$, not $71.5$, so C is false. Also, the highest rate is not at the start, even though the acid concentration is highest at the start; the data show the rate increases after the first two minutes, so D is false.

Key Takeaways

• In a reaction with one reactant in excess, the amount of product is limited by the other reactant; the amount of the excess reactant added cannot be found from product volume.
• Molar gas volume at r.t.p. is $24\text{ dm}^3 / \text{mol}$.
• Rates are calculated from differences between consecutive readings, not by dividing the final total by the total time.
• A higher concentration usually gives a higher rate, but other factors, such as a surface oxide layer, can make the highest rate occur later.

Common Mistakes

• Quoting $0.107\text{ g}$ as the mass of aluminium added. It is the mass of aluminium reacted.
• Calculating the rate as $143 / 2 = 71.5\text{ cm}^3 / \text{min}$; rates must be found between consecutive readings.
• Assuming D is correct because concentration is highest at the start; the data contradict this.
• Forgetting to convert $\text{cm}^3$ to $\text{dm}^3$ before using the molar gas volume.

Things to Be Careful About

• Read "excess aluminium" carefully: it means some aluminium is left over.
• Use the balanced equation ratio $2\text{ Al} : 3\text{ H}_2$.
• Use $A_r(\text{Al}) = 27$.
• The volume stops changing after 12 min, showing the acid has been used up.
• The units of rate are $\text{cm}^3 / \text{min}$.

Walkthrough

Start by finding the concentration of the bismuth(III) chloride solution. The relative formula mass is $209 + 3 \times 35.5 = 315.5$. The mass used is $157.75$ g, so the amount is $157.75 / 315.5 = 0.500$ mol. Since $500\ \text{cm}^3$ is $0.500\ \text{dm}^3$, concentration is $0.500 / 0.500 = 1.0\ \text{mol / dm}^3$. Statement 1 is true.

Statement 2 concerns equilibrium. In any reversible reaction at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This is the definition of dynamic equilibrium; it does not mean concentrations of reactants and products are equal. Statement 2 is true.

Statement 3: the equation has $\text{HCl}$ on the right. Adding more $\text{HCl}$ increases the concentration of a product. According to Le Chatelier's principle, the equilibrium shifts in the direction that reduces this increase, i.e. it uses up $\text{HCl}$, so it shifts to the left, forming more $\text{BiCl}_3$ and $\text{H}_2\text{O}$. Statement 3 is true.

Since all three statements are correct, option A is the answer. Option B misses statement 3, option C misses statement 2, and option D misses statement 1.

Key Takeaways

• Concentration is calculated as moles divided by volume in $\text{dm}^3$.
• At equilibrium, forward and reverse rates are equal.
• Adding a product to a system at equilibrium shifts the position to the left (towards reactants).

Common Mistakes

• Forgetting to convert $500\ \text{cm}^3$ to $0.500\ \text{dm}^3$; using $500\ \text{dm}^3$ would give the wrong concentration.
• Thinking that at equilibrium the amounts or concentrations of reactants and products are equal; only the rates are equal.
• Thinking that adding $\text{HCl}$ shifts equilibrium to the right because it is a reactant; $\text{HCl}$ is a product here, so the shift is to the left.

Things to Be Careful About

• $\text{BiCl}_3$ contains three chlorine atoms, so $M_r = 209 + 3(35.5)$, not $209 + 35.5$.
• The volume must be in $\text{dm}^3$ for concentration in $\text{mol / dm}^3$: $500\ \text{cm}^3 = 0.500\ \text{dm}^3$.
• The solid $\text{BiOCl}$ appears in the equation but does not change the reasoning here; the added $\text{HCl}$ is aqueous and affects the equilibrium.
• In a multiple-choice question, check each statement independently before choosing the option.

Walkthrough

The flow chart illustrates the industrial preparation of ammonia. The final product is ammonia, which is formed via the Haber process. The Haber process combines nitrogen gas and hydrogen gas under high temperature, high pressure, and in the presence of an iron catalyst. Therefore, the two inputs to the ammonia box must be nitrogen and hydrogen.

Looking at the flow chart, 'substance 2' and 'hydrogen' combine to form 'ammonia'. This means 'substance 2' must be nitrogen. The arrow leading to 'substance 2' comes from 'air'. Air is approximately 78% nitrogen, and nitrogen is separated from air by fractional distillation of liquid air. This confirms substance 2 is nitrogen.

The other reactant is hydrogen. The flow chart shows 'substance 1' going through 'process X' to produce 'hydrogen'. In the organic chemistry syllabus, hydrogen is often associated with the cracking of petroleum fractions. Cracking is the process of breaking down long-chain alkanes (substance 1) into shorter, more useful molecules, including shorter alkanes, alkenes, and hydrogen gas. Therefore, process X is cracking and substance 1 is long-chain alkanes.

Matching these findings to the options:

• process X: cracking
• substance 1: long-chain alkanes
• substance 2: nitrogen

This corresponds to option A.

Key Takeaways

• The Haber process requires nitrogen and hydrogen.
• Nitrogen is sourced from air.
• Hydrogen can be sourced from the cracking of long-chain alkanes (petroleum fractions) in simplified industrial flow charts.
• Cracking breaks large hydrocarbon molecules into smaller ones.

Common Mistakes

• Confusing the sources of nitrogen and hydrogen. Nitrogen comes from air; hydrogen comes from hydrocarbons (natural gas or petroleum fractions).
• Misidentifying process X. Fractional distillation separates mixtures (like crude oil into fractions) but does not chemically break down alkanes to produce hydrogen. Cracking is the chemical breakdown process.
• Assuming substance 1 is an alkene. Alkenes are products of cracking, not the starting material (substance 1) for producing hydrogen in this context.

Things to Be Careful About

• Ensure you identify the reactants for the final product (ammonia) first. This is the anchor point (nitrogen and hydrogen).
• Remember that air separation gives nitrogen (and oxygen, but oxygen is not needed for ammonia). Substance 2 is nitrogen, not oxygen.
• Cracking produces alkenes and shorter alkanes, but in the context of hydrogen production for ammonia, it is the process that breaks down the long-chain alkanes (substance 1).

Walkthrough

Look at each statement in turn.

• A: Acidified manganate(VII) ions are purple. When reduced they are changed to a colourless or very pale pink solution. The statement says the opposite, so A is incorrect.
• B: Redox reactions always involve both oxidation and reduction: one species gives electrons while another gains them. This is correct.
• C: An oxidising agent takes electrons from another substance, so it gains electrons and is itself reduced. Stating that it loses electrons is wrong.
• D: Reduction is gaining hydrogen, not losing hydrogen. Loosing hydrogen is a sign of oxidation, so D is incorrect.

Key Takeaways

• Remember oil rig: Oxidation Is Loss, Reduction Is Gain (of electrons).
• Oxidation and reduction always occur together in a redox reaction.
• An oxidising agent gains electrons; a reducing agent loses electrons.
• In a redox reaction, the oxidising agent is also reduced, and the reducing agent is also oxidised.

Common Mistakes

• Mistaking the colour change of acidified manganate(VII): it is purple and goes colourless or pale pink when reduced, not the reverse.
• Saying an oxidising agent loses electrons — it actually gains electrons.
• confusing reduction with the loss of hydrogen. Reduction is the gain of hydrogen, while oxidation can be the loss of hydrogen.

Things to Be Careful About

• Always know the direction of electron transfer: use OIL RIG.
• Do not confuse oxidising agent with reducing agent. The oxidising agent is reduced, the reducing agent is oxidised.
• When a question talks about hydrogen, remember that gaining hydrogen is reduction, losing hydrogen is oxidation.

Walkthrough

This question tests a halogen displacement reaction. Chlorine is more reactive than bromine, so chlorine displaces bromine from potassium bromide. Chlorine is converted into chloride ions by gaining an electron. Gaining electrons is reduction, so chlorine is reduced. The substance that brings about oxidation by removing electrons from another substance is the oxidising agent. Because chlorine is reduced, chlorine is the oxidising agent.

The bromide ions lose an electron and are oxidised to bromine. Losing electrons is oxidation, and the substance that provides the electrons is the reducing agent. Hence bromide ions are the reducing agent.

Option A is not correct because bromide ions are the reducing agent, not the oxidising agent. Option B is not correct because bromine is the product, not a reactant acting as a reducing agent. Option C is not correct because chloride ions are also products and have not acted as a reducing agent.

Key Takeaways

• In a halogen displacement, the more reactive halogen oxidises the less reactive halide ion.
• The oxidising agent is reduced; the reducing agent is oxidised.
• The identity of the oxidising and reducing agents must be decided from electron transfer, not from the names of the substances.

Common Mistakes

• Confusing "oxidising agent" with"being oxidised". The oxidising agent is the substance that gets reduced, not oxidised.
• Calling bromine the reducing agent when bromine is a product, not a reactant.
• Confusing chloride ions with bromide ions. The ions are involved in the electron transfer.

Things to Be Careful About

• Use the terms "oxidising agent" and "reducing agent" in the correct order: the agent accepts/ donates electrons accordingly.
• Make sure the half-equations are balanced in charge and in atoms.
• This question relies on the old Group VII displacement trend: chlorine is more reactive than bromine.

Walkthrough

The question asks for the row that correctly describes an aqueous alkali in two ways: its pH, and which ion, $\text{H}^+$ or $\text{OH}^-$, is present in the greatest concentration.

An alkali is a soluble base. When dissolved in water it produces hydroxide ions, $\text{OH}^-$. Because of these hydroxide ions, an alkali is not acidic, so its pH must be greater than 7. The pH scale runs from below 7 for acids to above 7 for alkalis, with 7 being neutral.

In any aqueous solution there are always some $\text{H}^+$ and $\text{OH}^-$ ions, but their relative amounts decide whether the solution is acidic or alkaline. In an alkali, the concentration of $\text{OH}^-$ ions is greater than the concentration of $\text{H}^+$ ions. That is exactly what makes the solution alkaline.

Row B is the only row that has both pH greater than 7 and $\text{OH}^-$ as the ion with the greatest concentration.

Key Takeaways

• An alkali is a soluble base that releases hydroxide ions, $\text{OH}^-$, in water.
• Alkalis have pH greater than 7.
• The ion present in the greatest concentration in an alkali is $\text{OH}^-$, not $\text{H}^+$.

Common Mistakes

• Choosing A: it correctly says pH greater than 7 but incorrectly states that $\text{H}^+$ has the greatest concentration. That would describe an acid, not an alkali.
• Choosing C or D: these say pH less than 7, which describes an acid, not an alkali.

Things to Be Careful About

The question asks for the ion with the greatest concentration, not the ion that is present at all. Both $\text{H}^+$ and $\text{OH}^-$ are present in water, but their relative amounts decide whether the solution is acidic or alkaline.

Walkthrough

This question asks how to neutralise acid lake water rather than merely dilute it. Neutralisation means removing H⁺ ions using a base or carbonate.

Calcium carbonate is a basic carbonate. It reacts with acids to form a salt, water and carbon dioxide. Because it is insoluble, any calcium carbonate still present after the acid has reacted just remains as a solid at the bottom of the lake; it does not keep raising the pH into the alkaline range. This makes it suitable to add in excess.

Phosphoric acid is itself an acid, so it would lower the pH still further and poison the fish even more. Potassium hydroxide is a strong, soluble alkali; if added in excess it would neutralise the acid first but then make the water much too alkaline. Sodium nitrate is a neutral salt, but it has no acid-neutralising ability, so the lake would remain acidic.

The equation shows the key reaction between hydrogen ions and calcium carbonate:

For each mole of calcium carbonate, two moles of H⁺ are removed, so the pH rises towards 7.

Key Takeaways

• Acid neutralisation can be done with an alkaline, a base or a carbonate.
• Calcium carbonate is an insoluble carbonate, so it neutralises acid without making the environment strongly alkaline when used in excess.
• Supplying only acidic or neutral compounds cannot neutralise acidic water.

Common Mistakes

• Choosing potassium hydroxide because it is an alkali, and failing to notice the word “excess” in the question. Excess potassium hydroxide would leave pH too high.
• Choosing sodium nitrate because it is neutral. A neutral salt does not react significantly with the acid, so it does not neutralise it.
• Choosing phosphoric acid, which is an acid and would add more H⁺ ions.

Things to Be Careful About

• “In excess” is an important condition: an alkali that is completely soluble and strong would overshoot pH7 if present in excess.
• Calcium carbonate is insoluble in water, so the excess remains as an insoluble solid; this is why it is the best choice for lake treatment.
• If the equation were needed, for two H⁺ ions are present, giving Ca²⁺ and CO₂, balance all atoms and charges. State symbols should be included: solid carbonate, aqueous hydrogen ions, aqueous salt, carbon dioxide gas and liquid water.

Walkthrough

Barium sulfate is an insoluble salt. In the school laboratory, an insoluble salt is made by precipitation: mix two soluble salts, one providing the positive ion and the other providing the negative ion, and the insoluble salt appears as a solid.

• Barium chloride is soluble, so it provides $\text{Ba}^{2+}$ ions in solution.
• Sodium sulfate is soluble, so it provides $\text{SO}_4^{2-}$ ions.
• When the two solutions are mixed, $\text{BaSO}_4$ precipitates and sodium chloride stays dissolved.

This makes option C correct. The solid is then separated by filtration, washed with distilled water and dried, giving a pure sample.

Why the other options are wrong:

• A: barium carbonate is insoluble, so it does not give barium ions in solution. This is not a clean precipitation method.
• B: barium carbonate is insoluble and sodium sulfate is soluble, so no reaction occurs.
• D: barium hydroxide is not a suitable soluble source of barium ions here, and concentrated sulfuric acid is not appropriate for a school laboratory.

Key Takeaways

• Insoluble salts are prepared by precipitation from two soluble reagents.
• Solubility rules: all sodium salts are soluble, most chlorides are soluble, and barium sulfate is insoluble.
• The precipitate is collected by filtration, washed and dried.

Common Mistakes

• Choosing A because a carbonate and an acid react: barium carbonate is insoluble, so it cannot supply barium ions in solution.
• Forgetting that both reagents must be soluble for a clean precipitation.
• Using concentrated sulfuric acid: this is not the school-laboratory method.

Things to Be Careful About

• Know the solubility rules: sulfates are soluble except barium sulfate (lead sulfate is also insoluble, and calcium sulfate is slightly soluble).
• The equation must be balanced and include state symbols.
• The precipitate must be washed with distilled water and dried to remove soluble impurities.

Walkthrough

For a neutral atom, the number of electrons equals the proton number. So Q has proton number 17 and R has proton number 19.

• Q with 17 electrons has the electron configuration 2,8,7. This places it in Group VII of the Periodic Table; it is chlorine, a non-metal.
• R with 19 electrons has the electron configuration 2,8,8,1. This places it in Group I; it is potassium, a metal.

Now check each statement.

• A: A metal and a non-metal usually form an ionic compound. Potassium chloride is ionic, not covalent, so A is wrong.
• B: Non-metals in Group VII gain one electron to form negative ions such as $\text{Cl}^-$. Q does not form positive ions, so B is wrong.
• C: Outer-shell electrons are the electrons in the highest occupied shell. Q has 7 outer-shell electrons, while R has only 1, so R does not have more outer-shell electrons than Q. C is wrong.
• D: Metallic character is associated with elements that lose electrons and form positive ions. Potassium is a metal, while chlorine is a non-metal, so R is more metallic than Q. D is correct.

Key Takeaways

• In a neutral atom, the number of electrons equals the proton number.
• The electron configuration can be written in shells, e.g. 2,8,7 or 2,8,8,1.
• For main-group elements, the group number often tells you the number of outer-shell electrons.
• Metals tend to form positive ions; non-metals tend to form negative ions.
• A metal and a non-metal generally form an ionic compound, not a covalent compound.

Common Mistakes

• Assuming that an element with 17 electrons must form positive ions because it is an element. In fact, chlorine is a non-metal and forms a negative ion.
• Confusing the outer-shell electron counts: Q has 7 outer-shell electrons, not fewer than R.
• Thinking that any reaction between two elements produces a covalent compound. Metal + non-metal usually gives an ionic compound.

Things to Be Careful About

• The phrase “total number of electrons” only gives the proton number if the atom is neutral.
• Write the electron configuration shell by shell (2,8,7 and 2,8,8,1) before comparing outer-shell electrons.
• Metallic character is a periodic trend: Group I metals are strongly metallic, while Group VII elements are non-metals.
• In multiple-choice questions, eliminate each wrong statement using a clear reason before selecting the correct option.