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A link to a video of this paper can be found in OCR A Level Chemistry Paper 1 (2023) – Step-by-Step Question Breakdown
19
This question is about oxides of nitrogen.
(a)
An investigation is carried out on the equilibrium system shown below.
2NO₂(g) ⇌ N₂O₄(g)
∆H = –57.4 kJmol⁻¹
(i)
A sealed flask containing 6.00 moles of NO2(g) is heated to a constant temperature and allowed to reach equilibrium
The equilibrium mixture contains 5.40 mol of NO2(g), and the total pressure is 5.00 atm.
Determine the value of Kp and give your answer to 3 significant figures.
Include an expression for Kp and the units of Kp in your answer.
Kp is given by the equation:
$$\small k_p = \frac{p(N_2O_4(g))}{p(NO_2(g))^2}$$
As the partial pressures are given in atm, the units are atm-1.
We need to find the partial pressures of the N2O4 and the NO2 in equilibrium. These can be found by multiplying the total pressure by the molar fraction of each element, i.e.:
p(NO2) = molar fraction NO2 × total pressure
p(N2O4) = molar fraction N2O4 × total pressure
To find each molar fraction, we need to figure out the number of moles in equilibrium of NO2 and N2O4. We do it by using an ICE (Initial, Change and Equilibrium) table for the reaction:
2NO2(g) ⇄ N2O4(g)
We start by adding the information that we have from the problem
| NO2 | N2O4 | |
|---|---|---|
| Initial | 6.00 | 0.00 |
| Change | ||
| Equilibrium | 5.40 |
To complete the table, we need to check what has changed. We know that for each mole of N2O4 formed, we have lost 2 moles of NO2. Let’s call x to the number of moles of N2O4 formed, then, we have lost 2x moles of NO2 (i.e.-2x). I fill the table with these details:
| NO2 | N2O4 | |
|---|---|---|
| Initial | 6.00 | 0.00 |
| Change | -2x | x |
| Equilibrium | 5.40 |
In the last step, I add the Initial and Change rows and put them in the Equilibrium row.
| NO2 | N2O4 | |
|---|---|---|
| Initial | 6.00 | 0.00 |
| Change | -2x | x |
| Equilibrium | 5.40 = 6.00 – 2x | x[+0.00] |
The value of x can be calculated from the number of moles in equilibrium of NO2:
5.40 = 6.00-2x
x = (6 – 5.4) ÷ 2 = 0.3
Then, the table is:
| NO2 | N2O4 | |
|---|---|---|
| Initial | 6.00 | 0.00 |
| Change | -0.60 | 0.30 |
| Equilibrium | 5.40 | 0.30 |
Therefore, the total moles is 5.40 + 0.30 = 5.70 moles. Now, we can calculate the partial pressures:
p(NO2) = molar fraction NO2 × total pressure
p(NO2) = (5.40 ÷ 5.70) × 5.00 = 4.737 atm
and
p(N2O4) = molar fraction N2O4 × total pressure
p(N2O4) = (0.30 ÷ 5.70) × 5.00 = 0.263 atm
Replacing these values in kp:
$$\small k_p = \frac{p(N_2O_4(g))}{p(NO_2(g))^2} = \frac{0.263}{4.737^2} = 1.17 \times 10^{-2}$$
(rounded to 3 significant figures, as requested)
Kp = ………………………1.17×10-2……………….. units …………atm-1……….
[5]
(ii)
The sealed flask in (a)(i) is then heated to a higher temperature at an increased pressure. The system is allowed to reach equilibrium again.
Explain why it is difficult to predict how these changes in reaction conditions affect the amount of N2O4(g) formed at equilibrium.
When the sealed flask is heated to a higher temperature, the equilibrium position is affected because the forward reaction (2NO₂ → N₂O₄) is exothermic, as indicated by ∆H = -57.4 kJmol⁻¹. According to Le Chatelier’s Principle, increasing the temperature favours the endothermic direction, which is the reverse reaction (N₂O₄ → 2NO₂). Consequently, this shift reduces the amount of N₂O₄ formed at equilibrium.
At the same time, increasing the pressure favours the direction with fewer gas molecules. In this equilibrium system, the forward reaction reduces the total number of gaseous molecules from 2NO₂ (2 molecules) to N₂O₄ (1 molecule). Hence, higher pressure shifts the equilibrium to the right, increasing the yield of N₂O₄.
These two opposing factors—temperature favouring the reverse reaction and pressure favouring the forward reaction—make it difficult to predict the overall effect on the amount of N₂O₄ formed at equilibrium. The relative impact of each factor depends on the magnitude of change in temperature and pressure, as well as the specific properties of the equilibrium system.
[3]
(b)
N2O4 reacts fully with oxygen to form a different oxide of nitrogen, oxide A, as the only product.
Oxide A is collected and cooled to 75.0°C at a pressure of 101kPa.
Under these conditions, oxide A is a gas that occupies a volume of 74.0cm3 and has a mass of 0.280g.
Calculate the molar mass of oxide A and suggest its molecular formula.
To determine the number of moles of the oxide A, we use the ideal gas equation:
$$\small \mathrm{n} = \frac{\mathrm{P}\mathrm{V}}{\mathrm{R}\mathrm{T}}$$
Remember to use the right units for each variable:
P = 101 kPa = 101000 Pa
V = 74 cm3 = 7.4 × 10-5 m3
T = 75°C = 348 K
R = 8.314 J/(mol K)
$$\small \mathrm{n} = \frac{\mathrm{PV}}{\mathrm{RT}} = \frac{(101000 \times 7.4 \times 10^{-5})}{(8.314 \times 348)} = 2.58 \times 10^{-3} , \mathrm{mol}$$
Using the mass of the gas, we can find it molecular mass:
0.280 g ÷ (2.58 × 10-3 mol) = 108.5 g/mol
Now, it is question to play around with the atomic masses of N (14 g/mol) and O (16 g/mol) to find that the combination that gives a molar mass of 108. The only solution is N2O5, any other combination of N and O will not get closer.
molar mass = …………………108……………………….. g mol–1
molecular formula = ……………N2O5…………………………………………
[5]
20
This question is about acids and bases.
(a)
Table 20.1 shows the ionic product, Kw, of water at 25°C and 40°C.
Table 20.1
| Temperature/°C | Kw /mol2dm–6 |
|---|---|
| 25 | 1.00 × 10−14 |
| 40 | 2.92 × 10−14 |
(i)
Calculate the pH of water at 40°C.
Give your answer to 2 decimal places.
The ionic product of water is given by the expression:
Kw = [H+][OH–]
In pure water, [H+]=[OH–], then we can write:
Kw = [H+]2
Replacing the value that we have at 40°C:
2.92×10-14 = [H+]2
$$\small \mathrm{[H^+]} = \sqrt{2.92 \times 10^{-14}} = 1.7088 \times 10^{-7}$$
As pH = -log10[H+], we get that:
pH = -log10[H+] = – log10(1.7088×10-7) = 6.77 (rounded to 2 decimal places)
pH = …………………………………6.77………………. [2]
(ii)
Table 20.1 shows different Kw values at 25°C and at 40°C. A student suggests that water is neutral at these temperatures.
Explain why this student is correct.
In pure water [H+] = [OH–]
[1]
(b)
A student reacts strontium metal with water to make a 250.0 cm3 solution of aqueous strontium hydroxide, Sr(OH)2. The solution contains 0.145g of strontium hydroxide.
- Write an equation for the reaction of strontium with water.
- Calculate the pH of this 250.0 cm3 solution of strontium hydroxide at 40°C.
You should refer back to Table 20.1 at the start of (a).
Give your answer to 2 decimal places.
Equation ………………………………Sr + 2H2O → Sr(OH)2 + H2…………………………………………………………………………………..
The molecular mass of Sr(OH)2 is 87.6 + 2×(16.0 + 1.0) = 121.6 g/mol
The number of moles of 0.145 g of Sr(OH)2 is:
$$\small \mathrm{0.145 \, g \, Sr(OH)_2 \times \frac{1 \, mol \, Sr(OH)_2}{121.6 \, g \, Sr(OH)_2} = 1.1924 \times 10^{-3} \, mol \, Sr(OH)_2}$$
And its molar concentration is:
$$\small \mathrm{\frac{1.1924 \times 10^{-3} \, mol \, Sr(OH)_2}{0.250 \, dm^3 \, solution} = 4.7696 \times 10^{-3} \, molar \, Sr(OH)_2}$$
Sr(OH)2 is a strong base, completely dissociating in water:
Sr(OH)2 → Sr2+ + 2OH–
Each mol of Sr(OH)2 produces 2 moles of OH–, then, the concentration of [OH–] is 2× 4.7696 ×10-3 = 9.5392 ×10-3
Using the value of Kw from table 20.1 and this concentration of [OH–]
Kw = [H+][OH–]
2.92 × 10-14 = [H+] × 9.5392 ×10-3
[H+] = 2.92 × 10-14 ÷ (9.5392 ×10-3) = 3.061 ×10-12
Using the definition of pH:
pH = -log10[H+] = – log10(3.061 ×10-12) = 11.51 (rounded to 2 decimal places)
pH = ……………………………………11.51……………. [5]
(c)
A student reacts 1.00 g of strontium carbonate, SrCO3, with an excess of dilute nitric acid, HNO3. A gas is produced.
(i)
Construct the equation for this reaction.
………………………………………SrCO3 + 2HNO3 → Sr(NO3)2 + CO2 + H2O…………………………………………………………………………….. [1]
(ii)
The student then reacts 1.00 g of calcium carbonate, CaCO3, with an excess of dilute nitric acid, HNO3.
Explain why the student’s two reactions produce different volumes of gas.
The molar mass of SrCO₃ is greater than that of CaCO₃: The molar mass of SrCO₃ is approximately 147.6 g/mol, while the molar mass of CaCO₃ is approximately 100 g/mol. Since the same mass (1.00 g) is used, fewer moles of SrCO₃ will react compared to CaCO₃. The reaction with CaCO₃ produces more moles of CO₂ gas than the reaction with SrCO₃, resulting in a greater volume of gas.
[2]
(d)
A student reacts an excess of magnesium with 25.0 cm3 of 0.500 mol dm–3 hydrochloric acid, HCl.
The student also reacts an excess of magnesium with 25.0 cm3 of 0.500 mol dm–3 ethanoic acid, CH3COOH.
(i)
Construct an ionic equation for the reaction of magnesium with an acid.
The magnesium displaces hydrogen ions from the acid, releasing hydrogen gas. The ionic equation is:
Mg + 2H⁺ → Mg²⁺ + H₂
Remember that we are asked by the ionic equation, not the full molecular equation.
…………………………………………………Mg + 2H⁺ → Mg²⁺ + H₂………………………………………………………………….. [1]
(ii)
Explain why these two reactions of magnesium produce the same volume of gas but at different rates.
Both reactions produce the same volume of hydrogen gas because hydrochloric acid (HCl) and ethanoic acid (CH₃COOH) provide the same number of moles of hydrogen ions overall. However, the reaction rates differ due to the following key factors:
- Strength of the Acid: Hydrochloric acid is a strong acid and completely dissociates in solution, instantly providing a higher concentration of H⁺ ions. Ethanoic acid is a weak acid and only partially dissociates, releasing H⁺ ions gradually.
- H⁺ Concentration and Collision Frequency: The higher initial concentration of H⁺ ions in hydrochloric acid results in more frequent collisions with magnesium atoms, leading to a faster reaction rate. Ethanoic acid has a lower initial H⁺ concentration, slowing down the reaction.
- Final Number of Moles of H⁺ Ions: Over time, more ethanoic acid molecules dissociate to release enough H⁺ ions to match the total moles of H⁺ provided by hydrochloric acid. This ensures that both reactions produce the same volume of gas by the end.
Exam Tip: Since this part of the question is worth 3 marks, the examiner is likely looking for three distinct points. Using bullet points helps to clearly identify these key ideas, making it easier to score full marks.
[3]
(e)
Butanoic acid, CH3CH2CH2COOH, is a weak monobasic acid
(i)
Explain what is meant by the term monobasic acid.
It means that one mole of acid donates one mole of protons.
Note: Be sure to specify that this applies per molecule or mole of the acid, as failing to clarify this point could result in losing marks.
(ii)
A buffer solution is prepared by dissolving 3.39 g of potassium hydroxide in 250 cm3 of 0.376 moldm–3 butanoic acid.
This buffer solution has a pH of 5.07 at 25°C.
Calculate the acid dissociation constant, Ka, of butanoic acid at 25°C.
Assume that the volume of the solution remains constant at 250cm3 when the potassium hydroxide is dissolved.
Exam Tips:
- Write down what you’re calculating at each step. This helps the examiner follow your logic, especially if you make a mistake and need to claim “error carried forward” marks.
- Break it down into clear sections like [H⁺], [A⁻], [HA], and Ka.
- Watch out for common mistakes like using [H⁺] as [HA] or forgetting to subtract the moles of KOH from the initial moles of butanoic acid.
To simplify the nomenclature, instead of CH3CH2CH2COOH I use AH and instead of CH3CH2CH2COO– I use A–
The reaction that we have is:
AH ⇄ A– + H+
The expression of Ka is:
$$\small \mathrm{K_a = \frac{[A^-][H^+]}{[AH]}}$$
[H] is given by pH. [AH] and [A-] are given by the reaction of butanoic acid and the potassium hydroxide.
Find [H⁺] using the pH.
Since the pH is 5.07, we calculate [H⁺] as:
[H⁺] = 10-5.07 = 8.51 × 10⁻⁶ mol/dm³.
Work out [A⁻]
The butanoate ions come from the potassium hydroxide reacting with the butanoic acid. As the acid is a weak acid and the potassium hydroxide is a strong base, we can assume that all the KOH is neutralized by the acid:
AH + OH– → A– + H2O
The moles of KOH can be calculated by using its molar mass (39.1 + 1.0 + 16.0 = 56.9 g/mol)
$$\small \mathrm{Moles \, of \, KOH = 3.39 \, g \, KOH \times \frac{1 \, mol \, KOH}{56.9 \, g \, KOH} = 0.0604 \, mol \, KOH}$$
Since each KOH molecule produces one butanoate ion, the moles of A⁻ in the buffer are also 0.0604 mol. Dividing by the volume (0.25 dm³), we get the concentration:
[A⁻] = 0.0604 ÷ 0.25 = 0.242 mol/dm³.
Work out [HA] (butanoic acid):
The initial number of moles of AH in the 250 cm3 (0.25 dm3) of the 0.376 mol/dm³ of butanoic acid solution is:
$$\small \mathrm{Initial \, moles \, AH = \frac{0.376 \, mol \, AH}{1 \, dm^3} \times 0.25 \, dm^3 = 0.094 \, mol \, AH}$$
But 0.0604 of these moles had reacted with KOH to form A–, then in equilibrium, we have:
0.094 − 0.0604 = 0.0336 mol of AH
As we assume that the volume remains constant, we have that
[AH] = 0.0336 mol ÷ (0.25 dm3) = 0.134 mol/dm³
Calculate Ka:
We can now uise the expression of Ka with these values:
$$\small \mathrm{K_a = \frac{[A^-][H^+]}{[AH]} = \frac{(0.242 \times 8.51 \times 10^{-6})}{0.134} = 1.53 \times 10^{-5}}$$
Ka = …………… 1.53×10-3 ……………………… moldm–3 [4]
(f)
A buffer solution has a pH of 4.50.
When a small volume of water is added to this buffer solution, the pH does not change.
Explain why the pH does not change.
When a small volume of water is added to a buffer solution with a pH of 4.50, the pH stays the same because the ratio between the concentrations of the weak acid (HA) and its conjugate base (A⁻) doesn’t change. In a buffer solution, this balance is what maintains the pH. Adding water dilutes everything equally, so the proportions of HA and A⁻ remain consistent, and the pH doesn’t budge.
[1]
21
Some grass fertilisers contain compounds of iron.
During heavy rain, a fertiliser is washed into a nearby river causing the water to be polluted with a mixture of iron(II) and iron(III) ions.
(a)
A student determines the concentration of iron(II) ions in a sample of river water by titration with potassium manganate(VII).
25.0 cm3 portions of river water are acidified with dilute sulfuric acid. Each portion is titrated with 0.00250 moldm–3 potassium manganate(VII) until a colour change is seen.
MnO4–(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
(i)
State the colour change seen at the end point of the titration.
When carrying out the titration with potassium manganate (VII), the colour change you’ll see at the endpoint is from colourless to (pale) pink or pale purple (the mark scheme accepts pale purple as an answer, but not just purple). This happens because all the iron(II) ions have reacted, leaving a tiny excess of potassium manganate (VII), which gives that light pink tint. It’s worth noting that potassium manganate (VII) is originally purple, but when diluted, it can appear lighter in colour.
This was definitely a bit of a tricky question, as many students get caught out by mixing up the colour change or providing incorrect colours. To make life easier, check out A Level Chemistry A – Colours of Inorganic Ions and Complexes, as it’s super helpful for OCR exams.
from …………colourless………………………… to ……(pale) pink OR pale purple………………………………
[1]
(ii)
The student’s titration results are shown in the table below.
The trial titre has been omitted.
| 1 | 2 | 3 | |
|---|---|---|---|
| Final volume/cm3 | 12.65 | 25.60 | 38.35 |
| Initial volume/cm3 | 0.00 | 12.65 | 25.60 |
| Titre volume/cm3 |
Complete the table above and calculate the mean titre that the student should use to determine the concentration of iron(II) ions in the river water.
To complete the table and calculate the mean titre, you first work out the titre volumes for each titration by subtracting the initial volume from the final volume. This gives you:
| 1 | 2 | 3 | |
|---|---|---|---|
| Final volume/cm3 | 12.65 | 25.60 | 38.35 |
| Initial volume/cm3 | 0.00 | 12.65 | 25.60 |
| Titre volume/cm3 | 12.65 | 12.95 | 12.75 |
Now, here’s the important part—titres need to be concordant. That means you only use values that are close together (within 0.10 cm³ of each other). In this case, titres 1 and 3 are concordant, so you ignore titre 2 when calculating the mean.
To find the mean, simply add the concordant titres together and divide by 2:
(12.65 + 12.75) ÷ 2 = 12.70 cm³
mean titre = …………………………12.70……………….. cm3 [2]
(iii)
Determine the concentration, in moldm–3, of iron (II) ions in the river water.
Moles of MnO4⁻ used in the titration.
$$
\small \mathrm{\frac{0.00250 \, mol}{dm^3} \times 12.7 \, cm^3 \times \left(\frac{1 \, dm}{10 \, cm}\right)^3 = 3.175 \times 10^{-5} \, mol}
$$
From the stoichiometry of the reaction, each mole of MnO4⁻ is going to react with 5 moles of Fe2+, then:
$$\small \mathrm{3.175 \times 10^{-5} \, mol \, MnO_4^- \times \frac{5 \, mol \, Fe^{2+}}{1 \, mol \, MnO_4^-} = 1.5875 \times 10^{-4} \, mol \, Fe^{2+}}$$
The molar concentration of Fe(II) is given by dividing the number of moles of iron (II) by the volume of the portion in dm-3
1.5875 ×10-4 ÷ (0.025 dm3) = 6.35×10-3 mol dm-3
concentration = ……………………6.35×10-3……………… mol dm–3 [3]
(b)
The student modifies the experiment in (a) to determine the combined concentration of iron (II) and iron (III) ions in the river water.
The student’s method is shown below.
Step 1
Add excess zinc to a 250.0 cm3 sample of river water and warm gently.
Step 2
Cool the solution and remove excess zinc by filtration.
Step 3
Acidify 25.0 cm3 portions of the filtrate from Step 2. Then titrate each portion with 0.00250 moldm–3 potassium manganate (VII) until a colour change is seen.
The table below shows information about three redox systems.
| Redox system | Half-equation | Eө/V |
|---|---|---|
| 1 | Zn2+(aq) + 2e– ⇄ Zn(s) | –0.76 |
| 2 | Fe3+(aq) + e– ⇄ Fe2+(aq) | +0.77 |
| 3 | MnO4–(aq) + 8H+(aq) + 5e– ⇄ Mn2+(aq) + 4H2O(l) | +1.51 |
Use the information in the table above to explain the reasons for Step 1 and Step 2.
The information provided gives us a clear signal: we need to pay attention to the possible reactions happening within the system, especially those involving positive electrode potentials, that will favour the forward direction. Remember, a more positive potential value indicates that the forward reaction is favoured, meaning the species involved is more likely to be reduced or to act as an oxidising agent.
Step 1 (Adding zinc and warming gently): Zinc is a reducing agent, and it’s used to convert all the iron (III) ions in the river water sample into iron (II) ions. This process works because zinc has a more negative standard electrode potential. Specifically:
Zn(s) ⇄ Zn2+(aq) + 2e–
Eө = +0.76 V
Fe3+(aq) + e– ⇄ Fe2+(aq)
Eө = +0.77 V
Zn(s) + 2Fe3+(aq) ⇄ Zn2+(aq) + 2Fe2+(aq)
Eө = +1.53 V
Step 2 (Filtration to remove excess zinc): After step 1, there is still leftover zinc metal that needs to be removed, otherwise it will react with the potassium manganate (VII) during titration because the zinc has a more negative standard electrode potential. Specifically:
Zn(s) ⇄ Zn2+(aq) + 2e–
Eө = +076 V
MnO4–(aq) + 8H+(aq) + 5e– ⇄ Mn2+(aq) + 4H2O(l)
Eө = +1.51 V
5Zn(s) + 2MnO4–(aq) + 16H+(aq) ⇄ 5Zn2+(aq) + 2Mn2+(aq) + 8H2O(l)
Eө = +2.27 V
[4]
22
This question is about the d-block elements in Period 4 of the periodic table (Sc to Zn).
(a)*
Explain, with examples from Period 4, what is meant by the terms d-block element and transition element.
Explain why some d-block elements are not transition elements.
Use electron configurations to support your explanations.
A d-block element is one whose highest-energy, or valence, electrons are found in the d-sub-shell. In other words, the d orbitals are being filled with electrons. For example, iron (Fe) has the electron configuration [Ar] 4s² 3d⁶, so it’s clearly a d-block element because its d-sub-shell contains electrons.
A transition element is a special kind of d-block element. It forms ions with an incomplete d-sub-shell. For example: Iron(II) ions (Fe²⁺) have the configuration [Ar] 3d⁶. The d-sub-shell isn’t full, which makes iron a transition element.
Not all d-block elements fit the definition of a transition element because they don’t form ions with a partially filled d-sub-shell. For example, scandium and zinc:
- Scandium forms Sc³⁺ ions, which have the electron configuration [Ar]. The 3d sub-shell is empty, so scandium isn’t a transition element.
- Zinc forms Zn²⁺ ions, which have the electron configuration [Ar] 3d¹⁰. Here, the 3d sub-shell is completely full, so zinc also isn’t a transition element.
[6]
(b)
(i)
Describe precipitation reactions using either copper or chromium ions as examples.
Include equations.
When copper(II) ions react with sodium hydroxide or ammonia, copper(II) hydroxide (Cu(OH)₂) forms as a pale blue precipitate:
Cu²⁺ + 2OH⁻ → Cu(OH)₂
Note that an excess of ammonia will dissolve the precipitate, forming a deep blue solution, as we will see in the next question.
When chromium(III) ions react with sodium hydroxide or ammonia, chromium(III) hydroxide (Cr(OH)₃) is formed as a dark grey-green precipitate:
Cr³⁺ + 3OH⁻ → Cr(OH)₃
Note that an excess of ammonia or sodium hydroxide will dissolve the precipitate, as we will see in the next question.
[2]
(ii)
Describe ligand substitution reactions using either copper or chromium ions as examples.
Include equations.
In ligand substitution reactions, one or more ligands in a transition metal complex are replaced by different ligands, leading to noticeable colour changes due to modifications in the metal ion’s electronic structure.
Copper(II) (Cu²⁺) Ligand Substitution
Copper (II) aqua ions can undergo ligand exchange with chloride ions (Cl⁻) or excess of ammonia molecules (NH₃), resulting in distinct colour transformations.
- Reaction with Ammonia (NH₃)
- Equation: [Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O
- Colour Change: Pale blue → Deep blue
- Reaction with Chloride Ions (Cl⁻)
- Equation: [Cu(H₂O)₆]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 6H₂O
- Colour Change: Blue → Yellow (appears green due to mixing of blue and yellow complexes)
- Structural Note: Since chloride ions are larger than water and ammonia ligands, the coordination number decreases from six to four, forming a tetrahedral-shaped complex.
Chromium(III) (Cr³⁺) Ligand Substitution
Chromium(III) ions can undergo substitution reactions with excess hydroxide (OH⁻) or ammonia (NH₃), leading to noticeable shifts in colour.
- Reaction with Excess Ammonia (NH₃)
- Equation: [Cr(H₂O)₆]³⁺ + 6NH₃ → [Cr(NH₃)₆]³⁺ + 6H₂O
- Colour Change: Violet or pale purple → Purple
- Reaction with Hydroxide Ions (OH⁻)
- Equation: [Cr(H₂O)₆]³⁺ + 6OH⁻ → [Cr(OH)₆]³⁻ + 6H₂O
- Colour Change: Violet or pale purple → Dark green
[2]
(c)
The ethanedioate ion, C2O42–, is a bidentate ligand.
A complex ion of cobalt(III) contains two ethanedioate ligands and two water ligands.
Determine the charge of this complex ion and the coordination number of cobalt in the complex ion.
Determining the Charge
- Cobalt(III) has a +3 charge.
- Each ethanedioate ligand has a –2 charge, contributing –4 overall.
- Water ligands are neutral, adding no charge.
The total charge of the complex is:
(+3) + (–4) + (0) = –1
Thus, the charge of the complex ion is –1.
Determining the Coordination Number
- Ethanedioate is a bidentate ligand, meaning each molecule forms two bonds with cobalt.
- Two ethanedioate ligands contribute four bonds.
- Two water ligands contribute two more bonds.
The total number of bonds to cobalt is 6, meaning the coordination number of cobalt is 6.
Charge of complex ion ……………………………………-1…………………………………………………………..
Coordination number of cobalt ………………………6………………………………………………………..
[2]
(d)
An acidified solution containing Cr2O72– ions reacts with vanadium(III) ions in a redox reaction to form a solution containing Cr3+ ions and VO2+ ions.
Construct the overall equation for this reaction.
We have a redox reaction with all the species provided. As it is in acidic solution, we might need to use H+. I use half equations.
1. Identify the elements that are reduced and oxidised.
Oxidation: V3+ → VO2+
Reduction: Cr2O72- → Cr3+
2. Write their half-reaction balancing the elements except for the hydrogen (H) and the oxygen (O)
Oxidation: V3+ → VO2+
Reduction: Cr2O72- → 2Cr3+
3. Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation
Oxidation: V3+ + 2H2O→ VO2+
Reduction: Cr2O72- → 2Cr3+ + 7H2O
4. Balance the hydrogen atoms (including those added in the previous step) by adding protons (H+) to the opposite side of the equation.
Oxidation: V3+ + 2H2O→ VO2+ + 4H+
Reduction: Cr2O72- + 14H+→ 2Cr3+ + 7H2O
5. Add up the charges on each side and add electrons (e–) to equalize the charges in both sides of each half-equation.
Oxidation: V3+ + 2H2O→ VO2+ + 4H+ + 2e–
Reduction: Cr2O72- + 14H+ + 6e–→ 2Cr3+ + 7H2O
6. Compare the number of electrons in each half-equation, if the number is different, multiply by a convenient number (it helps to find the lowest common multiple) to make them equal
Oxidation: 3×[V3+ + 2H2O→ VO2+ + 4H+ + 2e–]
Reduction: Cr2O72- + 14H+ + 6e–→ 2Cr3+ + 7H2O
Oxidation: 3V3+ + 6H2O→ 3VO2+ + 12H+ + 6e–
Reduction: Cr2O72- + 14H+ + 6e–→ 2Cr3+ + 7H2O
7. Add both half-equations
3V3+ + 6H2O + Cr2O72- + 14H+ + 6e–→ 3VO2+ + 12H+ + 6e– + 2Cr3+ + 7H2O
8. Cancel the electrons (that should be the same number on both sides) and any common terms in reactants and in products.
3V3+ + 6H2O + Cr2O72- + 14H+ + 6e–→ 3VO2+ + 12H+ + 6e– + 2Cr3+ + 7H2O
3V3+ + Cr2O72- + 2H+ → 3VO2+ + 2Cr3+ + H2O
The answer is:
3V3+ + Cr2O72- + 2H+ → 3VO2+ + 2Cr3+ + H2O
[2]
END OF QUESTION PAPER