Chemistry 9701/12 — October/November 2024

Background Concept

The number of electrons in a neutral atom equals its proton number (atomic number, $Z$). For an ion, the electron count is adjusted by the charge: a negative charge means extra electrons (anions gain electrons), while a positive charge means fewer electrons (cations lose electrons). For a molecule or polyatomic ion, you sum the atomic numbers of all constituent atoms and then adjust for the overall charge. This question tests that fundamental counting skill across four different species.

Understanding the Question

The question asks which of four species contains a different number of electrons from the other three. Three of the species share the same electron count (50 electrons), while one differs (54 electrons). The answer is D, $\text{Te}^{2-}$. The question requires you to know the atomic numbers of Cl, O, H, S, and Te, and to correctly apply the charge correction for each ionic species.

Approach

For each species, follow the same procedure:

1. Identify the atomic number of each element present.
2. Multiply each atomic number by the number of atoms of that element in the formula.
3. Sum the contributions.
4. Adjust for the overall charge: add electrons for a negative charge, subtract for a positive charge.
5. Compare the final totals.

This systematic approach avoids missing the charge adjustment or miscounting atom numbers.

Step-by-Step Reasoning

A: $\text{ClO}_4^-$

• Cl: atomic number 17
• O: atomic number 8, four O atoms: $4 \times 8 = 32$
• Sum: $17 + 32 = 49$
• Charge $-1$: add 1 electron → $49 + 1 = 50$ electrons

B: $\text{H}_2\text{SO}_4$

• H: atomic number 1, two H atoms: $2 \times 1 = 2$
• S: atomic number 16
• O: atomic number 8, four O atoms: $4 \times 8 = 32$
• Sum: $2 + 16 + 32 = 50$ electrons (neutral molecule, no charge adjustment)

C: $\text{SO}_4^{2-}$

• S: atomic number 16
• O: atomic number 8, four O atoms: $4 \times 8 = 32$
• Sum: $16 + 32 = 48$
• Charge $-2$: add 2 electrons → $48 + 2 = 50$ electrons

D: $\text{Te}^{2-}$

• Te: atomic number 52 (tellurium, Group 16, Period 5)
• Charge $-2$: add 2 electrons → $52 + 2 = 54$ electrons

Comparing the totals: A, B, and C all have 50 electrons; D has 54. Therefore, D is the odd one out.

Why the distractors are tempting:

• A student who forgets the charge adjustment on A would get 49, or on C would get 48, and might incorrectly pick one of those as the odd one.
• A student who confuses Te (tellurium, Z = 52) with Se (selenium, Z = 34) or misremembers its atomic number could make an error.
• Forgetting to multiply O by 4 in the polyatomic ions would also give wrong totals.

Key Takeaways

• The electron count of any species = sum of atomic numbers of all atoms ± charge.
• Always adjust for ionic charge: anions gain electrons, cations lose electrons.
• Know the atomic numbers of common elements, especially those in polyatomic ions like $\text{ClO}_4^-$ and $\text{SO}_4^{2-}$.
• Work systematically through each species to avoid careless arithmetic errors.

Common Mistakes

• Forgetting the charge adjustment: e.g., counting $\text{ClO}_4^-$ as 49 electrons or $\text{SO}_4^{2-}$ as 48 electrons instead of 50. The negative charge means extra electrons.
• Miscounting atom numbers: forgetting to multiply O by 4 in $\text{ClO}_4^-$ or $\text{SO}_4^{2-}$.
• Confusing elements: Te is tellurium (Z = 52), not selenium (Z = 34) or tellurium's neighbour in the periodic table.
• Applying the charge correction in the wrong direction: adding electrons for a positive charge or subtracting for a negative charge.

Things to Be Careful About

• The charge is written as a superscript after the formula; make sure you read it correctly (e.g., $\text{SO}_4^{2-}$ has a $-2$ charge, not $-1$).
• Neutral molecules like $\text{H}_2\text{SO}_4$ need no charge adjustment.
• Always double-check your arithmetic when multiplying atomic numbers by the number of atoms.
• No state symbols or units are needed for this type of question — it is purely a counting exercise.

Background Concept

First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions. For hydrogen:

H(g) → H⁺(g) + e⁻

For helium:

He(g) → He⁺(g) + e⁻

The size of the first ionisation energy depends mainly on three factors:

• the nuclear charge (the number of protons in the nucleus);
• the distance of the outer electron from the nucleus;
• the shielding of the outer electron by inner electrons.

A higher nuclear charge pulls the outer electrons more strongly, so more energy is needed to remove them. Greater distance and greater shielding both make removal easier.

Understanding the Question

This is a multiple-choice question asking for the cause of helium’s higher first ionisation energy compared with hydrogen. The question is not asking what is true about helium — all four statements may be partly true — but which factor actually explains the trend. You must compare hydrogen and helium directly.

Hydrogen has one proton and one electron, electron configuration 1s¹. Helium has two protons and two electrons, electron configuration 1s². Both electrons in helium are in the same 1s orbital, so both atoms have the same outermost principal quantum shell.

Approach

Start by writing down the relevant factors that determine first ionisation energy: nuclear charge, distance of the outer electron, and shielding. Then compare hydrogen and helium using each factor.

• Nuclear charge: H has +1, He has +2.
• Distance: both outer electrons are in the 1s orbital, so the distance is essentially the same.
• Shielding: neither atom has an inner shell; helium’s second electron provides very little shielding, so the shielding is not lower in helium than in hydrogen.

The only factor that clearly differs in the direction needed to explain helium’s higher ionisation energy is nuclear charge.

Step-by-Step Reasoning

Consider each option in turn.

A — In the 1s orbital in helium, electrons are paired.
This is true: helium’s two electrons occupy the same 1s orbital and are therefore paired. However, pairing causes electron–electron repulsion, which makes an electron slightly easier to remove, not harder. So this factor would tend to lower helium’s first ionisation energy relative to what it would be without the repulsion. It cannot be the reason helium’s first ionisation energy is higher than hydrogen’s.

B — The lowest energy level in helium is filled.
Helium’s 1s sub-shell is full, but simply having a filled lowest energy level is not a cause of high ionisation energy. All atoms fill their lowest available energy levels. A fully filled sub-shell can add stability, but here the deciding comparison between H and He is not about electron configuration being filled; it is about how strongly the nucleus holds the outer electrons.

C — The nuclear charge in helium is higher than in hydrogen.
This is the correct explanation. Helium has two protons, so the effective attraction felt by each 1s electron is much greater than the attraction felt by hydrogen’s lone electron. More energy must be supplied to overcome that attraction, so helium’s first ionisation energy is higher. In fact, hydrogen’s first ionisation energy is about 1310 kJ mol⁻¹, while helium’s is about 2370 kJ mol⁻¹.

D — There is less shielding of the outer shell in helium.
This statement is not correct for helium compared with hydrogen. Hydrogen has no inner electrons, so its outer electron experiences no inner-shell shielding. Helium has a second electron in the same 1s orbital, which provides a very small amount of shielding to each electron from the other. So helium does not have less shielding than hydrogen; if anything, it has very slightly more. The dominant factor remains the doubled nuclear charge.

Therefore the correct option is C.

Key Takeaways

• First ionisation energy is controlled by nuclear charge, atomic radius, and shielding.
• When comparing elements in the same period, increasing nuclear charge is usually the dominant factor causing an increase in first ionisation energy.
• Hydrogen and helium are a useful pair because both use the 1s orbital and have no inner shell, isolating nuclear charge as the main variable.
• Be careful to distinguish a genuine fact about an atom from the explanation of a trend.

Common Mistakes

• Choosing A because “paired electrons” sounds like greater stability. Paired electrons cause repulsion, which tends to reduce ionisation energy, not raise it.
• Choosing D because “less shielding” seems to follow from having a higher nuclear charge. The opposite is true here: helium has no inner shell, so there is very little shielding to begin with; the second electron provides only slight shielding, not less than hydrogen.
• Forgetting that first ionisation energy refers to gaseous atoms forming gaseous ions. State symbols matter in written answers.

Things to Be Careful About

• Compare like with like: helium’s first ionisation energy removes one electron from neutral He, not from He⁺. The second ionisation energy of helium is much higher because it removes an electron from a 1s¹ ion, where the nuclear charge is no longer balanced by a second electron.
• Use the phrase “nuclear charge attracts electrons more strongly” rather than “more electrons means higher ionisation energy”, because adding electrons in the same shell can introduce repulsion as well.
• In any question about ionisation energy trends, identify the shell of the outer electron first; if the outer shells are the same, shielding differences are usually small and nuclear charge dominates.

Background Concept

At room temperature and pressure (rtp), one mole of any gas occupies 24 dm³ = 24000 cm³. This is the molar gas volume. Stoichiometry links the amount of a substance (in moles) to its mass, and to the volume of a gas.

Understanding the Question

We have 0.216 g of aluminium carbide (unknown formula AlₓCᵧ) reacting with excess water to produce methane (CH₄) as the only carbon-containing product. The methane is then burned completely to give CO₂ and H₂O. The volume of CO₂ at rtp is 108 cm³. We need to deduce which formula matches.

Approach

1. Convert the volume of CO₂ to moles using the molar gas volume.
2. Since each carbon atom in the carbide ends up as one CO₂ molecule, moles of C atoms = moles of CO₂.
3. Use the mass of the carbide and the candidate formulas to find which gives the correct number of moles of carbon atoms.

Step-by-Step Reasoning

• Moles of CO₂ = 108 cm³ / 24000 cm³ mol⁻¹ = 0.00450 mol.
• Therefore moles of C atoms = 0.00450 mol.
• For each formula, calculate the molar mass, then moles of carbide, then moles of C atoms.
• Only Al₄C₃ gives exactly 0.00450 mol of C atoms.

Key Takeaways

• The molar gas volume at rtp is 24 dm³ mol⁻¹ (24000 cm³ mol⁻¹).
• In combustion analysis, all carbon in the original compound ends up as CO₂, so moles of C = moles of CO₂.
• When deducing a formula, test each candidate using the given data.

Common Mistakes

• Using the wrong molar gas volume (e.g., 22.4 dm³ at STP vs 24 dm³ at rtp).
• Forgetting that each mole of carbide contains y moles of carbon atoms.
• Rounding intermediate values prematurely.

Things to Be Careful About

• Ensure units are consistent: cm³ vs dm³.
• Use the correct relative atomic masses: Al = 27, C = 12.
• The reaction with water produces methane, but we don't need to write the equation; we only need the carbon balance.

Background Concept

A catalytic converter in a petrol-engined car cleans up the exhaust gases. Two important pollutants are carbon monoxide (CO, a toxic product of incomplete combustion) and nitrogen oxides (NO and NO2, formed when the high temperature of the engine makes nitrogen and oxygen in the air react). The converter carries out redox reactions that convert these pollutants into harmless gases: carbon monoxide is oxidised to carbon dioxide, and nitrogen monoxide is reduced to nitrogen gas.

The key reaction is:

$2\text{NO} + 2\text{CO} \rightarrow \text{N}_2 + 2\text{CO}_2$

Understanding the Question

The question tells us that in this catalytic-converter reaction, four reactant molecules produce three product molecules. We are asked to identify which pair of gases from the options could be the two reactants.

The trick is that the question is purely about counting molecules in a balanced equation — we must write a balanced equation for each candidate pair and see which one has 4 reactant molecules on the left and 3 product molecules on the right.

Approach

1. Recall the main reaction happening in a catalytic converter.
2. Write the balanced equation for the pair in each option.
3. Count the total molecules on each side.
4. Select the option where the left side has 4 molecules and the right side has 3.

Step-by-Step Reasoning

Option C — nitrogen monoxide and carbon monoxide:

$2\text{NO} + 2\text{CO} \rightarrow \text{N}_2 + 2\text{CO}_2$

Left side: $2\text{NO} + 2\text{CO}$ = 4 molecules.
Right side: $\text{N}_2 + 2\text{CO}_2$ = 3 molecules.

This exactly matches the condition. So C is the answer.

Why the other options fail:

• Option A — nitrogen and carbon dioxide: Nitrogen is extremely unreactive and does not react with carbon dioxide in a catalytic converter. Also, carbon dioxide is a product of the converter, not a reactant. No sensible reaction fits the 4-to-3 molecule count.

• Option B — nitrogen monoxide and carbon dioxide: Carbon dioxide is a product, not a reactant, in this system. Nitrogen monoxide would need a reducing agent to become nitrogen, and CO2 is not one. No reaction here gives 4 reactants and 3 products.

• Option D — nitrogen dioxide and carbon monoxide: Balancing this gives, for example, $2\text{NO}_2 + 4\text{CO} \rightarrow \text{N}_2 + 4\text{CO}_2$, which is 6 reactant molecules and 5 product molecules — not 4 and 3. Even the simpler $\text{NO}_2 + \text{CO} \rightarrow \text{NO} + \text{CO}_2$ gives 2 reactants and 2 products. Neither fits.

Key Takeaways

• The central catalytic converter reaction is $2\text{NO} + 2\text{CO} \rightarrow \text{N}_2 + 2\text{CO}_2$.
• CO is oxidised (C from +2 to +4) and NO is reduced (N from +2 to 0).
• When a question specifies molecule counts, always write a balanced equation and count the coefficients.

Common Mistakes

• Forgetting that nitrogen gas is diatomic (N2), which changes the product count.
• Assuming nitrogen dioxide is the main nitrogen oxide removed, when the standard converter reaction uses nitrogen monoxide.
• Confusing products and reactants: carbon dioxide is produced, not consumed, in this reaction.

Things to Be Careful About

• Count molecules, not atoms. A molecule is one unit of a substance (one formula unit of a covalent gas).
• Make sure the equation is fully balanced before counting — an unbalanced equation gives a wrong molecule count.
• Remember that nitrogen is inert, so any option using N2 as a reactant is chemically implausible in this context.

Background Concept

VSEPR theory says that electron pairs around a central atom repel one another and adopt positions that minimise this repulsion. Both bonding pairs and lone pairs count as electron domains. For a central atom with four electron domains, the ideal arrangement is tetrahedral with bond angles of 109.5°. If two of those four domains are lone pairs, the molecule or ion is bent, and the bond angle is smaller than 109.5°, typically about 104.5–105°, because lone pair–lone pair repulsion is stronger than lone pair–bonding pair or bonding pair–bonding pair repulsion.

Here the central atom is nitrogen, which has 5 valence electrons. Each hydrogen in an N–H bond contributes 1 electron to the shared pair. The total number of valence electrons around N therefore depends on the charge of the ion: a negative ion has extra electrons, while a positive ion has fewer.

Understanding the Question

The ion contains exactly one nitrogen and two hydrogens, so its formula is $\text{NH}_2$ with an unknown charge. You are told that the H–N–H bond angle is approximately 105° and asked to identify the correct combination of:

• the number of lone pairs around N, and
• the overall charge on the ion.

The bond angle is the key clue: it tells you how many electron domains are around N and therefore how many lone pairs there are. Once the number of lone pairs is known, the charge follows from counting valence electrons.

Approach

Try the two possible signs of the charge using valence-electron totals.

For a –1 ion:

• N contributes 5 electrons.
• Two H atoms contribute 1 each, giving 2.
• The –1 charge adds one electron.
• Total = 8 electrons.

Two N–H bonds use 4 electrons. The remaining 4 electrons form two lone pairs. This gives 4 electron domains: 2 bonding pairs plus 2 lone pairs, which produce a bent shape with an angle of about 105°.

For a +1 ion:

• Total = 5 + 2 – 1 = 6 electrons.
• Two N–H bonds use 4 electrons, leaving only one lone pair.
• This gives only 3 electron domains, so the shape would be trigonal planar with an angle near 120°, not near 105°.

Therefore the observed angle identifies the –1 ion with two lone pairs.

Step-by-Step Reasoning

1. Build the Lewis structure of $\text{NH}_2^-$:
   • Total valence electrons = $5 + 2 + 1 = 8$.
   • Two N–H sigma bonds use $2 \times 2 = 4$ electrons.
   • Remaining electrons = $8 - 4 = 4$, which form two lone pairs on N.
2. Count the electron domains around N: two bonding pairs plus two lone pairs = 4 domains.
3. The 4 domains point to the corners of a tetrahedron. With two atoms bonded and two lone pairs, the observed shape is bent.
4. Lone pairs compress the bond angle from 109.5° to about 105°, exactly as stated.
5. Compare with $\text{NH}_2^+$: only 6 valence electrons, so after the two bonds it would have one lone pair and only 3 electron domains; the bond angle would be close to 120°, not 105°.

So the correct row is D: 2 lone pairs and charge –1.

Key Takeaways

• The charge on an ion changes the total number of valence electrons: negative ions have extra electrons, positive ions have fewer.
• VSEPR counts lone pairs as well as bonding pairs when predicting shape.
• Four electron domains with two lone pairs produce a bent shape with a bond angle near 104.5–105°, while three electron domains with one lone pair give a bond angle near 120°.
• A bond angle can be used to distinguish between otherwise similar formulas such as $\text{NH}_2^-$ and possible positive analogues.

Common Mistakes

• Forgetting to count the extra electron that comes from a –1 charge. If you forget it, you get only 7 valence electrons and may incorrectly predict one lone pair.
• Confusing $\text{NH}_2^-$ with ammonia, $\text{NH}_3$. Ammonia has one lone pair and an H–N–H angle of about 107°, whereas the ion here has two lone pairs and an angle of about 105°.
• Ignoring the bond-angle clue and simply guessing the charge from vague ideas about nitrogen's usual bonding. The angle is the decisive piece of evidence.
• Thinking a positive charge on N means less electron repulsion, without actually counting VSEPR domains.n

Things to Be Careful About

• For a negative ion, add an electron when counting valence electrons; for a positive ion, subtract one.
• The question asks for lone pairs around N, so do not count lone pairs on hydrogen; hydrogen here has no lone pairs.
• Make sure the charge sign is written correctly: the correct answer is –1, not +1.
• The pair/tetra notation: four electron domains with two lone pairs is the only arrangement that gives an H–N–H angle of about 105°.

Background Concept

Molecules are held together in liquids and solids by intermolecular forces. The strength of these forces determines volatility and boiling point. The main types are London (dispersion) forces, permanent dipole–dipole forces, and hydrogen bonds. London forces arise from instantaneous fluctuations in electron distribution and increase with the number of electrons and the polarisability of the molecule. Permanent dipole–dipole forces occur between polar molecules that have a permanent separation of charge.

Understanding the Question

The question asks why ICl has a higher boiling point than Br2. Both are simple covalent molecules, so the key is to compare the intermolecular forces. We need to identify which molecule is polar and which is not, and then relate that to the strength of intermolecular forces.

Approach

1. Determine the polarity of each molecule.
2. Identify the types of intermolecular forces present.
3. Compare the strengths of these forces.
4. Relate the strength of forces to boiling point.

Step-by-Step Reasoning

1. Polarity: ICl is a diatomic molecule with two different atoms. Iodine and chlorine have different electronegativities (I ≈ 2.5, Cl ≈ 3.0), so the bond is polar, and the molecule has a permanent dipole. Br2 is a diatomic molecule with two identical atoms, so the bond is non-polar and the molecule has no permanent dipole.
2. Intermolecular forces: ICl, being polar, experiences permanent dipole–dipole forces in addition to London forces. Br2, being non-polar, experiences only London forces.
3. Strength: Dipole–dipole forces are generally stronger than London forces for molecules of similar size and electron count. Additionally, ICl has a larger, more polarisable electron cloud due to the iodine atom, which enhances London forces as well. However, the key difference is the permanent dipole.
4. Boiling point: More energy is required to overcome the stronger intermolecular forces in ICl, so it has a higher boiling point.

Key Takeaways

• Boiling point reflects the strength of intermolecular forces.
• Polar molecules have permanent dipole–dipole forces in addition to London forces.
• Non-polar molecules only have London forces.
• For molecules with similar electron counts, polarity can be the deciding factor.

Common Mistakes

• Assuming that relative molecular mass is the only factor. While ICl has a slightly higher Mr than Br2, the number of electrons is the same, and the main reason is polarity.
• Confusing bond energy with intermolecular forces. Boiling point does not depend on the strength of covalent bonds within the molecule.
• Overlooking that London forces also exist in polar molecules.

Things to Be Careful About

• Check whether the molecule is polar by considering bond polarity and molecular shape (though for diatomic molecules, any difference in electronegativity makes it polar).
• Remember that London forces increase with the number of electrons, but here both have the same number, so polarity is the differentiator.
• In exam answers, explicitly mention permanent dipole–dipole forces for ICl and only London forces for Br2.

Background Concept

The ideal gas equation

$$pV = nRT$$

relates the pressure, volume, amount and temperature of an ideal gas. Here $R$ is the gas constant, $8.31 \text{ J K}^{-1} \text{ mol}^{-1}$. The equation is only used directly when the units are consistent: pressure in Pa, volume in m$^3$, amount in mol and temperature in K.

Nitrogen is a diatomic molecule, $\text{N}_2$, so its molar mass is $2 \times 14.0 = 28.0 \text{ g mol}^{-1}$. The question tells us to assume nitrogen behaves ideally, so we can confidently apply $pV = nRT$.

Understanding the Question

We are asked to find the volume occupied by $1.0 \text{ g}$ of nitrogen at $50^\circ\text{C}$ and a pressure of $2.0$ atmospheres. The mass is given, not the amount in moles, and the pressure and temperature are not in the SI units required by the ideal gas equation. The task is therefore to convert the given data into usable units, find the number of moles, substitute into the rearranged ideal gas equation, and then express the volume in $\text{cm}^3$ to match the options.

Approach

1. Convert the mass of nitrogen to moles using $n = \frac{m}{M_r}$.
2. Convert the temperature to kelvin: $T(\text{K}) = T(^\circ\text{C}) + 273$.
3. Convert the pressure to pascals: $1$ atm $= 101 \text{ kPa} = 101000 \text{ Pa}$, so $2.0$ atm $= 202000 \text{ Pa}$.
4. Rearrange $pV = nRT$ to give $V = \frac{nRT}{P}$ and substitute.
5. Convert the volume from $\text{m}^3$ to $\text{cm}^3$ using $1 \text{ m}^3 = 10^6 \text{ cm}^3$, then choose the closest option.

Step-by-Step Reasoning

First find the amount of nitrogen:

$$n = \frac{1.0}{28.0} = 0.0357 \text{ mol}$$

Next convert the temperature:

$$T = 50 + 273 = 323 \text{ K}$$

Convert the pressure:

$$P = 2.0 \times 101000 = 202000 \text{ Pa}$$

Now substitute into the rearranged ideal gas equation:

$$V = \frac{nRT}{P} = \frac{0.0357 \times 8.31 \times 323}{202000}$$

The numerator is approximately $95.8 \text{ J}$, so

$$V = \frac{95.8}{202000} = 4.74 \times 10^{-4} \text{ m}^3$$

Since $1 \text{ m}^3 = 10^6 \text{ cm}^3$:

$$V = 4.74 \times 10^{-4} \times 10^6 = 474 \text{ cm}^3$$

This is closest to option C, $470 \text{ cm}^3$.

A common slip would be to use $1.0$ atm instead of $2.0$ atm, which would give about $950 \text{ cm}^3$, matching option D. Forgetting to convert $^\circ\text{C}$ to K would also change the answer substantially.

Key Takeaways

• The ideal gas equation requires SI units: Pa, m$^3$, mol and K.
• Always convert mass to moles using the molar mass of the actual substance present, remembering that nitrogen is $\text{N}_2$.
• Volume in $\text{m}^3$ must be converted to $\text{cm}^3$ when the answer choices are given in $\text{cm}^3$.
• When options are rounded, calculate accurately and select the closest value.

Common Mistakes

• Using the molar mass of a nitrogen atom, $14.0$, instead of the nitrogen molecule, $28.0$. This would double the calculated amount and give the wrong volume.
• Using $1.0$ atm instead of $2.0$ atm, which gives roughly twice the correct volume.
• Forgetting to add 273 to convert $50^\circ\text{C}$ to kelvin.
• Leaving the volume in $\text{m}^3$ and selecting an option without converting to $\text{cm}^3$.
• Using $R = 0.0821$ with pressure in atm and volume in dm$^3$ without being consistent with the other units.

Things to Be Careful About

• The gas constant $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$ works with pressure in Pa and volume in m$^3$.
• $1$ atmosphere is exactly $101 \text{ kPa} = 101000 \text{ Pa}$ in this question.
• $1 \text{ m}^3 = 10^6 \text{ cm}^3$, not $10^3$.
• Nitrogen exists as $\text{N}_2$, so its molar mass is $28.0 \text{ g mol}^{-1}$.
• The final volume is approximately $474 \text{ cm}^3$, so the nearest option is $470 \text{ cm}^3$.

Background Concept

Different types of bonding give different physical properties because of the way particles are arranged and how they can move.

In ionic compounds, oppositely charged ions are held in a giant three-dimensional lattice by strong electrostatic attractions. In the solid state those ions are fixed in place, so the solid cannot conduct electricity. When melted or dissolved in water, the ions become mobile and the substance conducts.

In metals, positive ions are surrounded by a sea of delocalised electrons. These electrons are free to move throughout the structure, so metals conduct electricity in the solid state as well as when molten.

Covalent substances can be either molecular or giant covalent. Molecular covalent compounds contain discrete molecules held together by weak intermolecular forces, so they usually have low melting points. Giant covalent structures, such as diamond and silica, have a network of strong covalent bonds throughout the whole structure, giving very high melting points.

Hydrogen bonding is a special type of intermolecular force. It only occurs when a hydrogen atom is covalently bonded to a highly electronegative atom with a lone pair, typically nitrogen, oxygen or fluorine, and that hydrogen is attracted to a lone pair on another N, O or F atom. It is not enough for a molecule simply to contain oxygen and hydrogen.

Understanding the Question

This is a multiple-choice question asking which single statement about the properties associated with different types of bonding is correct. The four options make claims about hydrogen bonding, the coexistence of ionic and covalent bonds, the electrical conductivity of ionic solids compared with metals, and the melting points of covalent compounds.

You need to test each statement against the fundamental bonding-property relationships. The correct statement must be true without exception.

Approach

The most efficient approach is to recall the key property for each type of bonding and then examine each option.

• For A, recall the precise condition for hydrogen bonding.
• For B, think of a compound that contains both ionic and covalent bonds.
• For C, compare how charge carriers move in ionic solids and in metals.
• For D, think of a covalent substance with a high melting point that does not use hydrogen bonding.

Only one statement should survive this check.

Step-by-Step Reasoning

Option A is false. A molecule containing both oxygen and hydrogen does not automatically form hydrogen bonds. Hydrogen bonding requires a hydrogen atom attached to N, O or F, and a lone pair on another N, O or F atom. Many organic molecules contain O and H atoms but have the hydrogen atoms attached to carbon, not to oxygen, so they cannot form hydrogen bonds. Even when a molecule has an O-H bond, it must also be able to interact with a lone pair on an N, O or F atom of another molecule. The statement is too broad and is incorrect.

Option B is false. Ionic and covalent bonding can coexist in the same compound. For example, ammonium salts contain covalent bonds within the ammonium ion and ionic bonds between the ammonium ion and the anion. Similarly, ionic compounds containing polyatomic ions such as sulfate or nitrate have covalent bonds within the ion and ionic bonds between the ions. The statement is therefore incorrect.

Option C is true. In an ionic solid, the ions occupy fixed positions in a lattice and cannot move, so the solid does not conduct electricity. In a metal, the delocalised electrons are mobile even in the solid state, so metals do conduct electricity when solid. This is a correct distinction between ionic compounds and metals.

Option D is false. High melting points in covalent substances are not limited to hydrogen-bonded compounds. Giant covalent structures such as diamond and silica have very high melting points because of the extensive network of strong covalent bonds, even though they contain no hydrogen bonding. Hydrogen-bonded molecular solids, such as ice, actually have relatively low melting points compared with giant covalent structures.

Since C is the only correct statement, the answer is C.

Key Takeaways

• Ionic solids do not conduct electricity because their ions are fixed in the lattice; they conduct when molten or in aqueous solution.
• Metals conduct electricity in the solid state because of delocalised electrons.
• Hydrogen bonding has a specific condition: H bonded to N, O or F, with a lone pair on N, O or F.
• High melting points in covalent substances can arise from giant covalent structures, not only from hydrogen bonding.
• Ionic and covalent bonding can occur together in the same compound.

Common Mistakes

• Assuming that any compound containing O and H forms hydrogen bonds. This ignores the need for H to be bonded directly to N, O or F and for a lone pair to be available.
• Thinking ionic solids conduct electricity because they contain charged particles. The ions must be mobile, which requires melting or dissolving.
• Believing covalent compounds always have low melting points. This only applies to simple molecular substances, not giant covalent structures.
• Forgetting that ionic and covalent bonding can coexist, for example in ammonium salts or compounds with polyatomic ions.

Things to Be Careful About

• Read each statement as a universal claim. A statement such as "any covalent compound that contains both oxygen and hydrogen forms hydrogen bonds" is false if even one exception exists.
• Be precise about the difference between a simple molecular covalent structure and a giant covalent structure.
• Remember that electrical conductivity depends on the presence of mobile charged particles, not just on the existence of charged particles.
• When answering multiple-choice questions, test every option rather than stopping at the first plausible one.

Background Concept

The standard enthalpy change of formation, $\Delta H_f^\ominus$, is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states, under standard conditions (usually 298 K and 100 kPa).

The standard state of an element is its most stable physical form under those conditions. For example:

• carbon: graphite, $\text{C(s)}$
• hydrogen: $\text{H}_2\text{(g)}$
• nitrogen: $\text{N}_2\text{(g)}$
• oxygen: $\text{O}_2\text{(g)}$
• chlorine: $\text{Cl}_2\text{(g)}$

Fractional coefficients are allowed in the equation because the definition refers to forming exactly one mole of the compound, even if that requires a fraction of a mole of an element.

Understanding the Question

The question asks which of four reactions has an enthalpy change that is an enthalpy change of formation. This is a definition-recognition question: you must test each equation against the three key requirements of the definition.

Approach

Check every equation against all three criteria:

1. Does it form exactly one mole of a single compound?
2. Are all reactants elements?
3. Are those elements in their standard states?

Any equation that fails even one criterion cannot represent an enthalpy change of formation.

Step-by-Step Reasoning

Option A
$\text{C(g)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}$

This forms one mole of methane from its elements, but carbon is shown as $\text{C(g)}$. The standard state of carbon is graphite, $\text{C(s)}$, not gaseous carbon atoms. Therefore this is not an enthalpy change of formation.

Option B
$\frac{1}{2}\text{N}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{NO(g)}$

Here nitrogen and oxygen are both in their standard states, $\text{N}_2\text{(g)}$ and $\text{O}_2\text{(g)}$, and exactly one mole of NO is formed. Fractional coefficients are perfectly acceptable. This fits the definition exactly.

Option C
$\text{Na}_2\text{O(s)} + \text{SO}_3\text{(g)} \rightarrow \text{Na}_2\text{SO}_4\text{(s)}$

The reactants are compounds, not elements. Even though one mole of sodium sulfate is formed, the reaction does not start from the constituent elements, so it is not a formation reaction.

Option D
$\text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow \text{PCl}_5\text{(g)}$

Again, $\text{PCl}_3$ is a compound, not an element. The reaction forms one mole of $\text{PCl}_5$, but not from phosphorus and chlorine in their standard states, so it is not an enthalpy change of formation.

Therefore the correct option is B.

Key Takeaways

• An enthalpy change of formation must involve formation of one mole of a compound.
• The reactants must be elements.
• Each element must be in its standard state.
• Fractional coefficients are allowed in the formation equation.
• Always check state symbols carefully when applying this definition.

Common Mistakes

• Choosing A because methane is formed from its elements, while overlooking that $\text{C(g)}$ is not the standard state of carbon.
• Choosing C or D because a compound is formed, while ignoring that the reactants are not elements.
• Thinking that coefficients in a formation equation must be whole numbers; they do not, because the definition specifies one mole of product.
• Confusing enthalpy change of formation with enthalpy change of reaction or combustion.

Things to Be Careful About

• The standard state of carbon is graphite, not diamond or gaseous carbon.
• State symbols are essential: $\text{C(s)}$, $\text{H}_2\text{(g)}$, $\text{N}_2\text{(g)}$, $\text{O}_2\text{(g)}$.
• Standard conditions are usually 298 K and 100 kPa.
• The definition requires exactly one mole of the compound formed, so the coefficient of the product must be 1.

Background Concept

The standard enthalpy change of formation, $\Delta H_f^\ominus$, is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. Hess's law states that the enthalpy change of a reaction is independent of the path taken — it depends only on the initial and final states. This allows us to calculate the enthalpy change of a reaction by combining formation reactions (or their reverses) and summing the enthalpy changes.

Understanding the Question

We are given the standard enthalpies of formation of VCl₂ (−452 kJ mol⁻¹) and VCl₃ (−573 kJ mol⁻¹), and asked for the enthalpy change of the reaction:

$$3\text{VCl}_2 \rightarrow 2\text{VCl}_3 + \text{V}$$

Vanadium metal is an element in its standard state, so its enthalpy of formation is zero and it does not contribute to the calculation. The reaction converts three moles of VCl₂ into two moles of VCl₃ and one mole of elemental vanadium.

Approach

Use Hess's law. Write the formation reactions for both vanadium chlorides, then combine them — reversing and scaling as needed — so that they sum to the target reaction. The stoichiometric coefficients in the target reaction (3 for VCl₂, 2 for VCl₃) tell us how many times to use each formation reaction.

Step-by-Step Reasoning

1. Write the formation reactions:

   $$\text{V} + \text{Cl}_2 \rightarrow \text{VCl}_2, \quad \Delta H_f^\ominus = -452 \text{ kJ mol}^{-1}$$ $$\text{V} + \tfrac{3}{2}\text{Cl}_2 \rightarrow \text{VCl}_3, \quad \Delta H_f^\ominus = -573 \text{ kJ mol}^{-1}$$

2. We need 3VCl₂ on the reactant side. Reverse the VCl₂ formation and multiply by 3:

   $$3\text{VCl}_2 \rightarrow 3\text{V} + 3\text{Cl}_2, \quad \Delta H = 3 \times (+452) = +1356 \text{ kJ}$$

   Reversing changes the sign; multiplying by 3 scales the enthalpy.

3. We need 2VCl₃ on the product side. Multiply the VCl₃ formation by 2:

   $$2\text{V} + 3\text{Cl}_2 \rightarrow 2\text{VCl}_3, \quad \Delta H = 2 \times (-573) = -1146 \text{ kJ}$$

4. Add the two steps:

   $$3\text{VCl}_2 \rightarrow 3\text{V} + 3\text{Cl}_2 \quad (\Delta H = +1356 \text{ kJ})$$ $$2\text{V} + 3\text{Cl}_2 \rightarrow 2\text{VCl}_3 \quad (\Delta H = -1146 \text{ kJ})$$

   Net: 3VCl₂ + 2V + 3Cl₂ → 3V + 3Cl₂ + 2VCl₃

   Cancel the 3Cl₂ on both sides; 2V on the left and 3V on the right leaves a net 1V on the right:

   $$3\text{VCl}_2 \rightarrow 2\text{VCl}_3 + \text{V}$$

   This matches the target reaction.

5. Combine the enthalpy changes:

   $$\Delta H = +1356 + (-1146) = +210 \text{ kJ mol}^{-1}$$

   The positive value indicates the reaction is endothermic — energy is absorbed.

Key Takeaways

• Hess's law lets us calculate reaction enthalpies from formation enthalpies without performing the reaction directly.
• Reversing a reaction changes the sign of ΔH.
• Multiplying a reaction by a coefficient multiplies ΔH by the same coefficient.
• Elements in their standard states have ΔHf° = 0.
• Always verify that the combined reactions cancel correctly to give the target reaction.

Common Mistakes

• Forgetting to change the sign when reversing a formation reaction — this would give −210 kJ mol⁻¹ (option A), a very tempting wrong answer.
• Forgetting to multiply the ΔH by the stoichiometric coefficient (e.g. using +452 instead of +1356).
• Not verifying that the net reaction matches the target, leading to an incorrect combination of steps.
• Confusing the sign convention: a positive ΔH means endothermic (energy absorbed), not a mistake.

Things to Be Careful About

• Apply the stoichiometric coefficients (3 and 2) to the ΔH values, not just to the formulae.
• The sign of the final answer: +210 kJ mol⁻¹, not −210 kJ mol⁻¹.
• V is an element in its standard state, so its formation enthalpy is zero and it does not appear in the enthalpy calculation.
• Units: kJ mol⁻¹ — the "per mole" refers to the reaction as written (3 mol VCl₂ → 2 mol VCl₃ + 1 mol V).

Background Concept

Redox reactions are electron-transfer reactions. The oxidation number of an atom is a bookkeeping charge used to decide whether it has been oxidised (oxidation number increases, electrons lost) or reduced (oxidation number decreases, electrons gained). The oxidising agent is the species that accepts electrons and is itself reduced; the reducing agent is the species that donates electrons and is itself oxidised.

In $\text{H}_2\text{O}_2$, each hydrogen is $+1$, so the two oxygens together are $-2$ and each oxygen is $-1$. This is unusual: in most compounds oxygen is $-2$, but the O–O bond in a peroxide makes $-1$ possible. Because the oxygen in $\text{H}_2\text{O}_2$ can be reduced to $-2$ in water or oxidised to $0$ in oxygen gas, hydrogen peroxide can act as either an oxidising agent or a reducing agent.

Understanding the Question

The question gives three balanced redox equations and asks in which of them $\text{H}_2\text{O}_2$ is acting as a reducing agent. A reducing agent is itself oxidised, so we need the reactions in which oxygen in $\text{H}_2\text{O}_2$ changes from $-1$ to $0$, forming $\text{O}_2$. The command word is implicit: identify the correct combination of reactions from the options.

Approach

For each equation, assign oxidation numbers to the oxygen atoms in $\text{H}_2\text{O}_2$ and in the oxygen-containing products. If the oxygen goes from $-1$ in $\text{H}_2\text{O}_2$ to $0$ in $\text{O}_2$, the peroxide has been oxidised and is the reducing agent. If it goes from $-1$ to $-2$ in $\text{H}_2\text{O}$, the peroxide has been reduced and is the oxidising agent. Checking the other reactant (iron or permanganate) confirms the same conclusion.

Step-by-Step Reasoning

Reaction 1:
$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$: iron is oxidised from $+2$ to $+3$. Therefore $\text{H}_2\text{O}_2$ must be reduced. The oxygen in $\text{H}_2\text{O}_2$ is $-1$ and becomes $-2$ in $\text{H}_2\text{O}$. So $\text{H}_2\text{O}_2$ is the oxidising agent, not the reducing agent.

Reaction 2:
In $\text{MnO}_4^-$, manganese is $+7$; in $\text{Mn}^{2+}$ it is $+2$. Manganese is reduced, so $\text{H}_2\text{O}_2$ is oxidised. Oxygen changes from $-1$ in $\text{H}_2\text{O}_2$ to $0$ in $\text{O}_2$. Hence $\text{H}_2\text{O}_2$ is the reducing agent.

Reaction 3:
$\text{Fe}^{3+} \rightarrow \text{Fe}^{2+}$: iron is reduced. The oxygen in $\text{H}_2\text{O}_2$ changes from $-1$ to $0$ in $\text{O}_2$, so $\text{H}_2\text{O}_2$ is oxidised and is the reducing agent.

Therefore reactions 2 and 3 are correct, which is option C. Option A is wrong because reaction 1 has $\text{H}_2\text{O}_2$ acting as an oxidising agent; option B misses reaction 3; option D misses reaction 3 as well.

Key Takeaways

• A reducing agent is oxidised during a redox reaction.
• The oxidation number of oxygen in a peroxide is $-1$, not $-2$.
• $\text{H}_2\text{O}_2$ is a reducing agent when it forms $\text{O}_2$, and an oxidising agent when it forms $\text{H}_2\text{O}$.
• Always check every reaction before choosing a combination option.

Common Mistakes

• Treating oxygen in $\text{H}_2\text{O}_2$ as $-2$; this leads to the wrong conclusion about whether it is oxidised or reduced.
• Confusing the reducing agent with the species being reduced. The reducing agent is oxidised.
• Stopping after reaction 2 and forgetting that reaction 3 also shows $\text{H}_2\text{O}_2$ acting as a reducing agent.
• Thinking that because $\text{Fe}^{3+}$ is reduced in reaction 3, $\text{H}_2\text{O}_2$ must be the oxidising agent; actually $\text{Fe}^{3+}$ is the oxidising agent and $\text{H}_2\text{O}_2$ is the reducing agent.

Things to Be Careful About

• Use oxidation numbers, not memorised roles, to decide oxidising/reducing agents.
• In $\text{O}_2$, each oxygen atom has oxidation number $0$.
• In $\text{H}_2\text{O}$, each oxygen atom has oxidation number $-2$.
• The oxidation number of hydrogen in $\text{H}_2\text{O}_2$ is $+1$, so the total for two oxygens is $-2$, giving $-1$ per oxygen.
• Balanced equations do not change the oxidation-number analysis; they only ensure electron conservation.

Background Concept

This question is about redox reactions and balancing them by oxidation number. Oxidation number is a bookkeeping charge assigned to an atom in a species, assuming all bonding electrons are given to the more electronegative atom. An increase in oxidation number is oxidation (loss of electrons); a decrease is reduction (gain of electrons). In a balanced redox equation, the total number of electrons lost by the reducing agent must equal the total number of electrons gained by the oxidising agent.

For polyatomic ions, the sum of the oxidation numbers of all atoms equals the charge on the ion. Oxygen is almost always $-2$ in these ions, and hydrogen is $+1$.

Understanding the Question

The equation involves thiosulfate ions, $\text{S}_2\text{O}_3^{2-}$, being converted to tetrathionate ions, $\text{S}_4\text{O}_6^{2-}$, while dioxovanadium(V) ions, $\text{VO}_2^+$, are reduced to oxovanadium(IV) ions, $\text{VO}^{2+}$. The coefficients $x$, $y$ and $z$ are unknown, and the question asks which row correctly states both the relationship between $x$, $y$ and $z$, and the change in oxidation number of vanadium.

This is a redox balancing problem: you must find how many electrons are released by the sulfur-containing species and how many are absorbed by vanadium. The ratio of those electron counts fixes $x$ and $z$, and the water molecules fix $y$.

Approach

1. Find the oxidation number of sulfur in thiosulfate and in tetrathionate.
2. Work out how many electrons are lost when two thiosulfate ions form one tetrathionate ion.
3. Find the oxidation number of vanadium in $\text{VO}_2^+$ and in $\text{VO}^{2+}$.
4. Use the electron balance to determine $x$ and $z$.
5. Use the water molecules to determine $y$.
6. Compare $x$, $y$ and $z$ and identify the correct option.

Step-by-Step Reasoning

Oxidation number of sulfur in thiosulfate

In $\text{S}_2\text{O}_3^{2-}$:

• Total charge $= -2$.
• Oxygen contributes $3 \times (-2) = -6$.
• Therefore the two sulfur atoms together must contribute $+4$.
• Average oxidation state of S $= +4/2 = +2$.

Oxidation number of sulfur in tetrathionate

In $\text{S}_4\text{O}_6^{2-}$:

• Total charge $= -2$.
• Oxygen contributes $6 \times (-2) = -12$.
• Therefore the four sulfur atoms together must contribute $+10$.
• Average oxidation state of S $= +10/4 = +2.5$.

Electrons lost by sulfur

Two thiosulfate ions contain four sulfur atoms with total oxidation number $4 \times (+2) = +8$.

One tetrathionate ion contains four sulfur atoms with total oxidation number $4 \times (+2.5) = +10$.

The total oxidation number increases by $2$, so $2$ electrons are lost:

$$2\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{e}^-$$

Oxidation number of vanadium

In $\text{VO}_2^+$:

• Total charge $= +1$.
• Oxygen contributes $2 \times (-2) = -4$.
• Therefore V $= +5$.

In $\text{VO}^{2+}$:

• Total charge $= +2$.
• Oxygen contributes $-2$.
• Therefore V $= +4$.

So each vanadium ion gains one electron:

$$\text{VO}_2^+ + \text{e}^- \rightarrow \text{VO}^{2+}$$

Balancing electrons

Two electrons are released by the sulfur reaction, and each vanadium ion accepts one electron. Therefore two vanadium ions are needed:

$$x = z = 2$$

Finding $y$

The product side contains $2\text{H}_2\text{O}$, which has $4$ hydrogen atoms. The only source of hydrogen on the left is $y\text{H}^+$, so:

$$y = 4$$

Thus $x = z = 2$ and $y = 4$, so $x$ and $z$ are half the value of $y$. The vanadium oxidation number changes from $+5$ to $+4$. This matches option C.

A quick charge check confirms the balanced equation:

Left: $2(-2) + 2(+1) + 4(+1) = -4 + 2 + 4 = +2$

Right: $-2 + 2(+2) = -2 + 4 = +2$

Key Takeaways

• Oxidation numbers can be averaged over identical atoms in a polyatomic ion when the structure is not specified.
• In a redox equation, the total electrons lost must equal the total electrons gained.
• Coefficients can be deduced from oxidation number changes, then checked by balancing atoms and charge.
• The oxidation number of vanadium in oxo-ions follows from the usual rules: oxygen is $-2$ and the sum equals the ion charge.

Common Mistakes

• Assigning vanadium an oxidation number of $+4$ in $\text{VO}_2^+$ by forgetting that the ion has a $+1$ charge; it is actually $+5$.
• Thinking the sulfur oxidation number changes from $+2$ to $+2.5$ per sulfur atom and then incorrectly counting electrons; the correct approach is to use the total change for all sulfur atoms.
• Confusing the relationship between coefficients: $x$ and $z$ are equal to each other and are half of $y$, not a quarter of $y$.
• Forgetting to balance electrons before assigning coefficients, which can lead to $x = z = 1$.

Things to Be Careful About

• Use average oxidation numbers for sulfur when the exact structural oxidation states are not required; this is sufficient for balancing redox.
• Always check that the total charge is the same on both sides of the final balanced equation.
• Remember that $\text{H}^+$ and water are linked: the number of hydrogen ions is fixed by the number of water molecules formed.
• The oxidation number change of vanadium is a reduction, from $+5$ to $+4$, because the ion gains an electron.

Background Concept

The equation describes a sparingly soluble ionic solid dissolving to form its constituent ions. The solid is in equilibrium with aqueous ions; the solid's concentration is constant and is not included in an equilibrium expression, but the position of equilibrium still responds to changes in the concentrations of the aqueous ions. Le Chatelier's principle states that if a change is imposed on an equilibrium, the position shifts to oppose that change. Increasing the concentration of a product ion shifts the equilibrium towards reactants (more solid); decreasing a product ion concentration shifts it towards products (more dissolving).

Here the product ions are $\text{Ca}^{2+}$, $\text{PO}_4^{3-}$ and $\text{OH}^-$. The options must be judged by whether they change the concentration of any of these ions, especially $\text{OH}^-$.

Understanding the Question

The question gives the dissolution equilibrium of hydroxyapatite and asks which added compound increases the amount of solid $\text{Ca}_5(\text{PO}_4)_3\text{OH(s)}$. This is a Le Chatelier problem: we need the additive that pushes the equilibrium to the left. The options are a weak base ($\text{NH}_3$), an ammonium salt ($\text{NH}_4\text{Cl}$), a weak acid ($\text{CH}_3\text{CO}_2\text{H}$) and a neutral salt ($\text{NaCl}$).

Approach

Check what each compound does to $[\text{OH}^-]$ in water. A base produces $\text{OH}^-$ and should increase the amount of solid. An acid or an acidic salt consumes $\text{OH}^-$ and should decrease the amount of solid. A salt with no common ion should have no effect. Then apply Le Chatelier to select the compound that raises $[\text{OH}^-]$.

Step-by-Step Reasoning

Option A: $\text{NH}_3$ is a weak base. In water it establishes:

$$\text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)}$$

This increases $[\text{OH}^-]$, a product ion in the dissolution equilibrium. By Le Chatelier, the equilibrium shifts to the left, forming more $\text{Ca}_5(\text{PO}_4)_3\text{OH(s)}$. So A is correct.

Option B: $\text{NH}_4\text{Cl}$ dissociates into $\text{NH}_4^+$ and $\text{Cl}^-$. The ammonium ion is a weak acid and reacts with hydroxide ions:

$$\text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightleftharpoons \text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)}$$

This lowers $[\text{OH}^-]$, so the equilibrium shifts to the right and less solid remains.

Option C: $\text{CH}_3\text{CO}_2\text{H}$ is a weak acid. It neutralises $\text{OH}^-$:

$$\text{CH}_3\text{CO}_2\text{H(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{CH}_3\text{CO}_2^-\text{(aq)} + \text{H}_2\text{O(l)}$$

Again $[\text{OH}^-]$ falls and the equilibrium shifts right, decreasing the solid.

Option D: $\text{NaCl}$ provides $\text{Na}^+$ and $\text{Cl}^-$, neither of which appears in the equilibrium. It does not change $[\text{Ca}^{2+}]$, $[\text{PO}_4^{3-}]$ or $[\text{OH}^-]$, so the amount of solid is essentially unchanged.

Therefore the only compound that increases the amount of solid is $\text{NH}_3$, option A.

Key Takeaways

• Le Chatelier's principle can be applied to heterogeneous equilibria involving a solid and aqueous ions.
• Increasing the concentration of a product ion shifts the equilibrium towards the solid; decreasing it shifts the equilibrium towards dissolution.
• Recognise which species are acids and which are bases: $\text{NH}_3$ is a base, $\text{NH}_4^+$ is an acidic ion, and $\text{CH}_3\text{CO}_2\text{H}$ is an acid.
• A salt such as $\text{NaCl}$ has no effect unless it supplies an ion common to the equilibrium.

Common Mistakes

• Thinking that $\text{NH}_3$ is acidic because it contains hydrogen; it is a base and produces $\text{OH}^-$ in water.
• Forgetting that $\text{NH}_4^+$ is acidic and can remove $\text{OH}^-$, so $\text{NH}_4\text{Cl}$ shifts the equilibrium in the opposite direction.
• Assuming all salts are neutral; ammonium salts are acidic in solution.
• Confusing "increases the amount of solid" with "speeds up the reaction"; the question is about equilibrium position, not rate.
• Ignoring the common-ion effect: $\text{NaCl}$ does not supply any ion in the equilibrium, so it cannot shift it.

Things to Be Careful About

• The solid is not included in the equilibrium expression, but the equilibrium position still depends on the concentrations of the aqueous ions.
• Use state symbols when writing the equations; the aqueous and liquid states matter in acid-base equilibria.
• $\text{NH}_3$ is a weak base, so the increase in $[\text{OH}^-]$ is small, but it is still enough to shift the equilibrium left.
• When comparing options, identify the ion each compound adds or removes and connect that to the equilibrium, rather than relying on memory of the compounds.

Background Concept

For a general reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$, the equilibrium constant is:

$$K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$$

where square brackets denote concentration in mol dm$^{-3}$ at equilibrium. For this reaction:

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

Because the container volume is 1.0 dm$^3$, the concentration of each gas is numerically equal to its amount in moles. This lets us work directly in moles.

Understanding the Question

We start with 2.0 mol of $\text{H}_2$ and 2.0 mol of $\text{I}_2$ in a 1.0 dm$^3$ container. The reaction reaches equilibrium with $K_c = 9.0$. We need the equilibrium amount of $\text{HI}$. This is a standard equilibrium calculation: write initial amounts, define the change, substitute into $K_c$, and solve.

Approach

Let $x$ be the number of moles of $\text{H}_2$ that react. Since $\text{H}_2$ and $\text{I}_2$ react in a 1:1 ratio, $x$ mol of $\text{I}_2$ also react, and $2x$ mol of $\text{HI}$ are produced. Then:

• $[\text{H}_2] = 2.0 - x$
• $[\text{I}_2] = 2.0 - x$
• $[\text{HI}] = 2x$

Substitute into $K_c$. The numerator and denominator are both perfect squares, so taking the positive square root gives a simple linear equation. Solve for $x$, then multiply by 2 to find the amount of $\text{HI}$.

Step-by-Step Reasoning

1. Initial concentrations: $[\text{H}_2] = [\text{I}_2] = 2.0$ mol dm$^{-3}$, $[\text{HI}] = 0$.
2. Change: $-x$ for $\text{H}_2$ and $\text{I}_2$, $+2x$ for $\text{HI}$.
3. Equilibrium concentrations: $[\text{H}_2] = [\text{I}_2] = 2.0 - x$, $[\text{HI}] = 2x$.
4. Write $K_c$:

$$9.0 = \frac{(2x)^2}{(2.0 - x)^2}$$

1. Take the positive square root:

$$3.0 = \frac{2x}{2.0 - x}$$

1. Rearrange:

$$3.0(2.0 - x) = 2x \Rightarrow 6.0 - 3x = 2x \Rightarrow 6.0 = 5x \Rightarrow x = 1.2$$

1. Therefore $[\text{HI}] = 2(1.2) = 2.4$ mol dm$^{-3}$, and since the volume is 1 dm$^3$, the amount of $\text{HI}$ is 2.4 mol.

Check: at equilibrium, $[\text{H}_2] = [\text{I}_2] = 0.8$ mol dm$^{-3}$ and $[\text{HI}] = 2.4$ mol dm$^{-3}$.

$$K_c = \frac{(2.4)^2}{(0.8)(0.8)} = \frac{5.76}{0.64} = 9.0$$

This confirms the answer. Option D is correct. Option B (1.2 mol) is the value of $x$, not the amount of $\text{HI}$. Option A (0.57 mol) is what you get if $K_c$ is inverted. Option C (1.5 mol) is not supported by the equilibrium expression.

Key Takeaways

• Set up an ICE table: Initial, Change, Equilibrium.
• Relate changes using stoichiometric coefficients.
• Substitute equilibrium amounts/concentrations into $K_c$.
• When the expression is a perfect square, taking the square root can avoid a quadratic.
• Multiply the reacted amount by the stoichiometric factor to get the product amount.

Common Mistakes

• Stopping at $x = 1.2$ mol and choosing B, forgetting that 2 mol of $\text{HI}$ are formed for every 1 mol of $\text{H}_2$ consumed.
• Forgetting that volume is 1 dm$^3$ and treating concentrations incorrectly; in general convert moles to concentrations using $c = n/V$.
• Taking the negative square root; concentrations cannot be negative.
• Inverting $K_c$: $K_c = [\text{products}]/[\text{reactants}]$, not the reverse.
• Using initial amounts instead of equilibrium amounts in the $K_c$ expression.

Things to Be Careful About

• Include the square on $[\text{HI}]$ and on each reactant concentration.
• For this reaction, the units of $K_c$ cancel because $\Delta n = 0$, so $K_c$ is dimensionless.
• If the volume were not 1 dm$^3$, the volume would appear in each concentration term; here it cancels because the total number of moles of gas is the same on both sides.
• Always check the final value by substituting back into the $K_c$ expression.