Question 1
Which isolated gaseous atom has a total of five electrons occupying spherically shaped orbitals?
(B) Fluorine
(C) Boron
(D) Potassium
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Spherically shaped orbitals are s-orbitals.
Electron configurations:
- Sodium: \( \mathrm{1s^2\,2s^2\,2p^6\,3s^1} \) → \(2+2+1=5\) electrons in s-orbitals.
- Fluorine: \( \mathrm{1s^2\,2s^2\,2p^5} \) → \(4\) electrons in s-orbitals.
- Boron: \( \mathrm{1s^2\,2s^2\,2p^1} \) → \(4\) electrons in s-orbitals.
- Potassium: \( \mathrm{1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1} \) → \(2+2+2+1=7\) electrons in s-orbitals.
Therefore, the isolated gaseous atom with a total of five electrons occupying spherically shaped orbitals is sodium, so the correct answer is (A).
Question 2
Diagram 1 shows the reaction pathway for a reaction.
Diagram 2 shows a graph of concentration of reactant against time for the same reaction.
Which diagrams, drawn to the same scale, show the effect of increased temperature on this reaction?
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
Increasing the temperature does not change the activation energy, \( \mathrm{E_a} \), or the enthalpy change of the reaction. It only increases the number of particles with sufficient energy to react.
As a result, the reaction proceeds faster, so the concentration of the reactant decreases more rapidly with time.
Therefore, the correct diagrams must show:
- The same activation energy, \( \mathrm{E_a} \), on the energy profile.
- A steeper concentration-time curve, indicating a faster reaction.
Only option C satisfies both conditions. Therefore, the correct answer is (C).
Question 3
The equation shows the overall reaction between an amine and an acid chloride.
\( \mathrm{2CH_3CH_2NH_2 + CH_3CH_2COCl \rightarrow CH_3CH_2CONHCH_2CH_3 + CH_3CH_2NH_3Cl} \)
The four steps in the mechanism are as follows.
Which steps are Brønsted-Lowry acid-base reactions?
(B) 3 and 4
(C) 3 only
(D) 4 only
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
A Brønsted-Lowry acid-base reaction involves the transfer of a proton, \( \mathrm{H^+} \).
Steps 1 and 2 involve nucleophilic attack and elimination of chloride, not proton transfer.
In step 3, the positively charged intermediate loses a proton to chloride ion, producing \( \mathrm{HCl} \).
In step 4, ethylamine accepts a proton from \( \mathrm{HCl} \) to form \( \mathrm{CH_3CH_2NH_3^+Cl^-} \).
Therefore, the Brønsted-Lowry acid-base reactions are steps 3 and 4, so the correct answer is (B).
Question 4
Chromium is present in compound X.
- Two moles of compound X react with exactly three moles of silicon.
- The only products of this reaction are four moles of chromium and three moles of a silicon compound in which the oxidation state of the silicon is \(+4\).
- Chromium and silicon are the only elements that change their oxidation states in this reaction.
What could be the identity of compound X?
(B) \( \mathrm{Cr_2H_6} \)
(C) \( \mathrm{Cr_2H_4} \)
(D) \( \mathrm{Cr_2O_4} \)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
The silicon product has silicon in the oxidation state \(+4\), so silicon is oxidised from \(0\) to \(+4\), losing 4 electrons per atom.
For \(3\) moles of silicon:
\( \mathrm{3Si \rightarrow 3Si^{4+} + 12e^-} \)
These \(12\) electrons reduce the chromium ions in \(2\) moles of X to \(4\) moles of chromium atoms.
Therefore, each chromium atom gains \(3\) electrons, so chromium must have an oxidation state of \(+3\) in compound X.
In \( \mathrm{Cr_2O_3} \), oxygen is \(-2\):
\( \mathrm{2x + 3(-2)=0 \;\Rightarrow\; x=+3} \)
Hence, compound X is \( \mathrm{Cr_2O_3} \), so the correct answer is (A).
Question 5
The reversible reaction shown is in equilibrium at a temperature of \(450^\circ\mathrm{C}\).
\( \mathrm{H_2 + I_2 \rightleftharpoons 2HI} \qquad \Delta H = -9.5\,\mathrm{kJ\,mol^{-1}} \)
The table shows the equilibrium concentrations in the reaction mixture.
| \( \mathrm{H_2} \) | \( \mathrm{I_2} \) | \( \mathrm{HI} \) |
|---|---|---|
| \(1.02\,\mathrm{mol\,dm^{-3}}\) | \(0.02\,\mathrm{mol\,dm^{-3}}\) | \(0.98\,\mathrm{mol\,dm^{-3}}\) |
The temperature of the reaction mixture is increased at constant volume.
The concentration of one of the components falls to \(0.80\,\mathrm{mol\,dm^{-3}}\) at equilibrium under these conditions.
What is the concentration of \( \mathrm{H_2} \) in the new equilibrium mixture?
(B) \(0.93\,\mathrm{mol\,dm^{-3}}\)
(C) \(1.11\,\mathrm{mol\,dm^{-3}}\)
(D) \(1.20\,\mathrm{mol\,dm^{-3}}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
The forward reaction is exothermic (\( \Delta H < 0 \)). Increasing the temperature shifts the equilibrium to the left, producing more \( \mathrm{H_2} \) and \( \mathrm{I_2} \), and less \( \mathrm{HI} \).
The component that decreases to \(0.80\,\mathrm{mol\,dm^{-3}}\) is \( \mathrm{HI} \):
Decrease in \( \mathrm{HI} = 0.98 – 0.80 = 0.18\,\mathrm{mol\,dm^{-3}} \)
From
\( \mathrm{H_2 + I_2 \rightleftharpoons 2HI} \)
an increase of \(0.09\,\mathrm{mol\,dm^{-3}}\) occurs in both \( \mathrm{H_2} \) and \( \mathrm{I_2} \).
New concentration of \( \mathrm{H_2} \):
\(1.02 + 0.09 = 1.11\,\mathrm{mol\,dm^{-3}}\)
Therefore, the correct answer is (C).
Question 6
The first stage in the industrial production of nitric acid from ammonia can be represented by the following equation.
\( \mathrm{4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)} \)
Using the following standard enthalpy change of formation data, what is the value of the standard enthalpy change, \( \Delta H^\circ_{\mathrm{r}} \), for this reaction?
| Compound | \( \Delta H^\circ_{\mathrm{f}}/\mathrm{kJ\,mol^{-1}} \) |
|---|---|
| \( \mathrm{NH_3(g)} \) | \(-46.1\) |
| \( \mathrm{NO(g)} \) | \(+90.3\) |
| \( \mathrm{H_2O(g)} \) | \(-241.8\) |
(B) \(-105.4\,\mathrm{kJ\,mol^{-1}}\)
(C) \(-905.2\,\mathrm{kJ\,mol^{-1}}\)
(D) \(-1274.0\,\mathrm{kJ\,mol^{-1}}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
Use the equation:
\( \Delta H^\circ_{\mathrm{r}}=\sum \Delta H^\circ_{\mathrm{f}}(\text{products})-\sum \Delta H^\circ_{\mathrm{f}}(\text{reactants}) \)
Products:
\(4(+90.3)+6(-241.8)=361.2-1450.8=-1089.6\,\mathrm{kJ}\)
Reactants:
\(4(-46.1)+5(0)=-184.4\,\mathrm{kJ}\)
Therefore,
\( \Delta H^\circ_{\mathrm{r}}=-1089.6-(-184.4)=-905.2\,\mathrm{kJ\,mol^{-1}} \)
Therefore, the correct answer is (C).
Question 7
The gaseous compound Z decomposes on heating.
In the diagram, Boltzmann distributions for Z at two different temperatures, P and Q, are shown.
The lines X and Y indicate activation energies for the decomposition of Z with and without a catalyst.
Which curve and which line describe the decomposition of Z at a higher temperature and with a catalyst present?
(B) P, Y
(C) Q, X
(D) Q, Y
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
Increasing the temperature makes the Maxwell-Boltzmann distribution broader and lower, with the peak shifted to higher molecular energies. Therefore, curve Q represents the higher temperature.
A catalyst lowers the activation energy of a reaction. Hence, the activation energy in the presence of a catalyst is represented by the line further to the left, X.
Therefore, the correct combination is:
- Higher temperature: Q
- Catalyst present: X
Therefore, the correct answer is (C).
Question 8
The structure of a molecule found in food is shown.
Four statements about the molecule are listed.
1. This molecule contains only \( \sigma \) bonds.
2. This molecule contains both \( \sigma \) and \( \pi \) bonds.
3. The six-membered ring is planar.
4. The six-membered ring is non-planar.
Which two statements are correct?
(B) 1 and 4
(C) 2 and 3
(D) 2 and 4
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
The molecule contains only single covalent bonds, so all bonds are \( \sigma \) bonds. There are no double or triple bonds, so there are no \( \pi \) bonds.
Six-membered rings in carbohydrates adopt chair or boat conformations to minimise strain and are therefore non-planar.
Thus:
- Statement 1 is correct.
- Statement 2 is incorrect.
- Statement 3 is incorrect.
- Statement 4 is correct.
Therefore, the correct answer is (B).
Question 9
In which substance are covalent bonds broken as it melts?
(B) Ice
(C) Iodine
(D) Ethanol
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Silicon(IV) oxide has a giant covalent structure. Melting requires breaking some of the strong covalent bonds in the three-dimensional network.
Ice, iodine and ethanol are simple molecular substances. When they melt, only the intermolecular forces (hydrogen bonding or van der Waals’ forces) are overcome, while the covalent bonds within the molecules remain intact.
Therefore, the correct answer is (A).
Question 10
Why is the second ionisation energy of sodium larger than the second ionisation energy of magnesium?
(B) The nuclear charge of \( \mathrm{Na^+} \) is greater than that of \( \mathrm{Mg^+} \).
(C) The outer electron of \( \mathrm{Na^+} \) is more shielded than the outer electron of \( \mathrm{Mg^+} \).
(D) The outer electron of \( \mathrm{Na^+} \) is in the same orbital as the outer electron of \( \mathrm{Mg^+} \).
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
After losing one electron:
- \( \mathrm{Na^+} \): \( \mathrm{1s^22s^22p^6} \) (neon configuration)
- \( \mathrm{Mg^+} \): \( \mathrm{1s^22s^22p^63s^1} \)
The second ionisation energy of sodium removes an electron from the inner \(2p\) shell, whereas the second ionisation energy of magnesium removes the remaining \(3s\) valence electron.
An inner-shell electron experiences a much stronger attraction to the nucleus because it is closer to the nucleus and less shielded.
Therefore, the attraction between the nucleus and the outermost electron is greater in \( \mathrm{Na^+} \) than in \( \mathrm{Mg^+} \), so the correct answer is (A).
Question 11
\( \mathrm{BF_3} \) reacts with \( \mathrm{NH_3} \) to form a single compound with a simple molecular structure.
\( \mathrm{BF_3 + NH_3 \rightarrow F_3B{-}NH_3} \)
Which row describes how the bond angles change during the reaction?
| \( \mathrm{F{-}B{-}F} \) bond angle | \( \mathrm{H{-}N{-}H} \) bond angle | |
|---|---|---|
| A | decreases | decreases |
| B | decreases | increases |
| C | increases | decreases |
| D | increases | increases |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
In \( \mathrm{BF_3} \), boron is trigonal planar with bond angles of \(120^\circ\).
When \( \mathrm{NH_3} \) donates its lone pair to boron, a coordinate bond forms and boron becomes tetrahedral. The \( \mathrm{F{-}B{-}F} \) bond angle decreases from \(120^\circ\) to approximately \(109.5^\circ\).
In \( \mathrm{NH_3} \), the \( \mathrm{H{-}N{-}H} \) bond angle is about \(107^\circ\) because of the lone pair on nitrogen. After donation of the lone pair, nitrogen becomes tetrahedral with four bonding pairs, so the bond angle increases to approximately \(109.5^\circ\).
Therefore, the \( \mathrm{F{-}B{-}F} \) bond angle decreases and the \( \mathrm{H{-}N{-}H} \) bond angle increases, so the correct answer is (B).
Question 12
The equation for the reaction of nitrogen with hydrogen is shown.
\( \mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)} \qquad \Delta H^\circ_{\mathrm{r}}=-92.2\,\mathrm{kJ\,mol^{-1}} \)
Which statement is correct?
(B) \( \Delta H^\circ_{\mathrm{r}} \) is measured at \(101\,\mathrm{kPa}\).
(C) \( \Delta H^\circ_{\mathrm{r}} \) represents the standard enthalpy change for the formation of ammonia gas.
(D) \( \Delta H^\circ_{\mathrm{r}} \) represents the enthalpy change when \(1.0\,\mathrm{mol}\) of \( \mathrm{N_2(g)} \) reacts with \(1.0\,\mathrm{mol}\) of \( \mathrm{H_2(g)} \).
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
The standard enthalpy change of reaction, \( \Delta H^\circ_{\mathrm{r}} \), is measured under standard conditions, which include:
- Pressure of \(101\,\mathrm{kPa}\).
- Temperature of \(298\,\mathrm{K}\) (not \(298^\circ\mathrm{C}\)).
- All substances in their standard states.
Option (C) is incorrect because the standard enthalpy of formation refers to the formation of one mole of a compound from its constituent elements in their standard states, whereas the given equation forms two moles of ammonia.
Option (D) is incorrect because the enthalpy change applies to the reaction as written, requiring \(1\) mole of \( \mathrm{N_2} \) and \(3\) moles of \( \mathrm{H_2} \).
Therefore, the correct answer is (B).
Question 13
Hydrogen peroxide, \( \mathrm{H_2O_2} \), decomposes into water and oxygen when a suitable catalyst is added.
\(20.0\,\mathrm{cm^3}\) of aqueous hydrogen peroxide decomposes to produce \(600\,\mathrm{cm^3}\) of oxygen at room conditions.
What is the concentration of the aqueous hydrogen peroxide?
(B) \(2.50\,\mathrm{mol\,dm^{-3}}\)
(C) \(1.25\,\mathrm{mol\,dm^{-3}}\)
(D) \(0.625\,\mathrm{mol\,dm^{-3}}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
The decomposition reaction is:
\( \mathrm{2H_2O_2 \rightarrow 2H_2O + O_2} \)
At room conditions, \(1\,\mathrm{mol}\) of gas occupies \(24\,000\,\mathrm{cm^3}\).
Moles of \( \mathrm{O_2} \):
\( \dfrac{600}{24000}=0.025\,\mathrm{mol} \)
From the equation, moles of \( \mathrm{H_2O_2} \):
\(2 \times 0.025=0.050\,\mathrm{mol}\)
Volume of solution:
\(20.0\,\mathrm{cm^3}=0.0200\,\mathrm{dm^3}\)
Concentration:
\( \dfrac{0.050}{0.0200}=2.50\,\mathrm{mol\,dm^{-3}} \)
Therefore, the correct answer is (B).
Question 14
When \(0.15\,\mathrm{g}\) of an organic compound is vaporised, it occupies a volume of \(65.0\,\mathrm{cm^3}\) at \(405\,\mathrm{K}\) and \(1.00\times10^5\,\mathrm{Pa}\).
Using the expression \( \mathrm{pV=nRT} \), which expression should be used to calculate the relative molecular mass, \(M_{\mathrm{r}}\), of the compound?
(B) \( \displaystyle \frac{0.15\times8.31\times405}{1\times10^{5}\times65\times10^{-3}} \)
(C) \( \displaystyle \frac{0.15\times65\times10^{-3}\times1\times10^{5}}{8.31\times405} \)
(D) \( \displaystyle \frac{0.15\times8.31\times405}{1\times10^{5}\times65\times10^{-6}} \)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Using the ideal gas equation:
\( \mathrm{pV=nRT} \)
Since
\( \mathrm{n=\dfrac{m}{M_r}} \)
Substituting gives:
\( \mathrm{M_r=\dfrac{mRT}{pV}} \)
The volume must be converted to SI units:
\(65.0\,\mathrm{cm^3}=65.0\times10^{-6}\,\mathrm{m^3}\)
Hence the required expression is:
\( \displaystyle \frac{0.15\times8.31\times405}{1\times10^{5}\times65\times10^{-6}} \)
Therefore, the correct answer is (D).
Question 15
A washing powder contains sodium hydrogencarbonate, \( \mathrm{NaHCO_3} \), as one of the ingredients.
In a titration, a solution containing \(1.00\,\mathrm{g}\) of this washing powder requires \(7.15\,\mathrm{cm^3}\) of \(0.100\,\mathrm{mol\,dm^{-3}}\) sulfuric acid for complete reaction. The sodium hydrogencarbonate is the only ingredient that reacts with the acid.
What is the percentage by mass of sodium hydrogencarbonate in the washing powder?
(B) 6.0%
(C) 12.0%
(D) 24.0%
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
The balanced equation is:
\( \mathrm{H_2SO_4 + 2NaHCO_3 \rightarrow Na_2SO_4 + 2CO_2 + 2H_2O} \)
Moles of \( \mathrm{H_2SO_4} \):
\(0.100 \times \dfrac{7.15}{1000}=7.15\times10^{-4}\,\mathrm{mol}\)
From the equation:
Moles of \( \mathrm{NaHCO_3}=2\times7.15\times10^{-4}=1.43\times10^{-3}\,\mathrm{mol}\)
Mass of \( \mathrm{NaHCO_3} \):
\(1.43\times10^{-3}\times84.0=0.120\,\mathrm{g}\)
Percentage by mass:
\( \dfrac{0.120}{1.00}\times100=12.0\% \)
Therefore, the correct answer is (C).
Question 16
The flow chart shows some reactions of nitrogen compounds.
Which row identifies gas 1 and salt 2?
| Gas 1 | Salt 2 | |
|---|---|---|
| A | Ammonia | Ammonium nitrate |
| B | Ammonia | Sodium nitrate |
| C | Nitrogen dioxide | Ammonium nitrate |
| D | Nitrogen dioxide | Sodium nitrate |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
Ammonia reacts with an acid to form an ammonium salt (salt 1).
When an ammonium salt is treated with an alkali, ammonia is released and a salt of the alkali metal is formed. For example:
\( \mathrm{NH_4NO_3 + NaOH \rightarrow NH_3 + NaNO_3 + H_2O} \)
Thus, gas 1 is ammonia and salt 2 is sodium nitrate.
Therefore, the correct answer is (B).
Question 17
Mixing aqueous silver nitrate and aqueous sodium chloride produces a precipitate.
Addition of which reagent to the mixture gives a colourless solution?
(B) Aqueous potassium iodide
(C) Dilute hydrochloric acid
(D) Dilute nitric acid
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Mixing aqueous silver nitrate and sodium chloride forms a white precipitate of silver chloride:
\( \mathrm{AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)} \)
Silver chloride dissolves in aqueous ammonia because it forms the soluble diamminesilver(I) complex:
\( \mathrm{AgCl(s) + 2NH_3(aq) \rightleftharpoons [Ag(NH_3)_2]^+(aq) + Cl^-(aq)} \)
The other reagents do not dissolve the silver chloride precipitate.
Therefore, the correct answer is (A).
Question 18
Which statement is correct?
(B) Calcium oxide does not react with water at room temperature.
(C) Magnesium hydroxide reacts with steam to form magnesium oxide and hydrogen.
(D) Strontium oxide reacts with water at room temperature to form strontium hydroxide only.
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Group 2 metal oxides react with water to form the corresponding hydroxides:
\( \mathrm{MO + H_2O \rightarrow M(OH)_2} \)
Strontium oxide reacts readily with water to form strontium hydroxide only. No hydrogen gas is produced because the reaction involves the oxide, not the metal.
Option (A) is incorrect because hydrogen is not produced. Option (B) is incorrect because calcium oxide reacts readily with water to form calcium hydroxide. Option (C) is incorrect because magnesium hydroxide does not react with steam to produce magnesium oxide and hydrogen.
Therefore, the correct answer is (D).
Question 19
Which statement about the properties of halogens and hydrogen halides is correct?
(B) Chlorine is more volatile than bromine because chlorine has stronger intermolecular forces.
(C) Hydrogen bromide is more thermally stable than hydrogen iodide because it has a stronger covalent bond.
(D) Iodine is less reactive than bromine because iodine has weaker covalent bonds.
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
The thermal stability of hydrogen halides decreases down Group 17 because the H–X covalent bond becomes weaker as the halogen atom becomes larger.
The bond strength follows:
\( \mathrm{H{-}F > H{-}Cl > H{-}Br > H{-}I} \)
Therefore, hydrogen bromide has a stronger covalent bond and is more thermally stable than hydrogen iodide.
Option (A) is incorrect because iodide ions are stronger reducing agents than chloride ions. Option (B) is incorrect because chlorine is more volatile due to its weaker intermolecular forces. Option (D) is incorrect because weaker covalent bonds make iodine more reactive in bond-breaking reactions, not less.
Therefore, the correct answer is (C).
Question 20
Three statements about the chemical periodicity of Period 3 oxides are listed.
1. The maximum oxidation state of the Period 3 elements in their oxides increases from sodium to phosphorus, then decreases from phosphorus to sulfur.
2. The oxides from sodium to aluminium dissolve in water without hydrolysis; from silicon to sulfur they are hydrolysed by water.
3. The structure and bonding changes from giant ionic to giant covalent to simple molecular.
Which statements are correct?
(B) 1 only
(C) 2 and 3 only
(D) 3 only
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Statement 1 is incorrect because the maximum oxidation state continues to increase from phosphorus (\(+5\)) to sulfur (\(+6\)) in their highest oxides.
Statement 2 is incorrect because aluminium oxide is largely insoluble in water and does not simply dissolve without hydrolysis.
Statement 3 is correct. Across Period 3, the oxides change from giant ionic structures (e.g. \( \mathrm{Na_2O} \), \( \mathrm{MgO} \)) to giant covalent \( \mathrm{SiO_2} \), and then to simple molecular oxides such as \( \mathrm{P_4O_{10}} \), \( \mathrm{SO_2} \), and \( \mathrm{SO_3} \).
Therefore, only statement 3 is correct, so the correct answer is (D).
Question 21
Which dot-and-cross diagrams represent molecules that can react with unburned hydrocarbons to form peroxyacetyl nitrate (PAN), a component of photochemical smog?
(B) 1 and 4
(C) 2 and 3
(D) 2 and 4
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Photochemical smog forms when nitrogen dioxide and ozone react with unburned hydrocarbons in the presence of sunlight, producing pollutants such as peroxyacetyl nitrate (PAN).
Diagram 2 represents nitrogen dioxide, \( \mathrm{NO_2} \), and diagram 4 represents ozone, \( \mathrm{O_3} \).
These two molecules are involved in the chain reactions that produce PAN in photochemical smog.
Therefore, the correct answer is (D).
Question 22
Which set of three elements contains a single element that has both the highest melting point and the lowest electrical conductivity of the three elements in the set?
(B) Aluminium, silicon and phosphorus
(C) Sodium, magnesium and aluminium
(D) Silicon, phosphorus and chlorine
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Compare the properties of the elements in each set:
- Magnesium and aluminium are metals with relatively high electrical conductivity.
- Silicon has a giant covalent structure, giving it the highest melting point of the three and a much lower electrical conductivity than the metals.
Thus, in set A, silicon is the single element that has both the highest melting point and the lowest electrical conductivity.
The other sets do not contain one element satisfying both conditions simultaneously.
Therefore, the correct answer is (A).
Question 23
In which row do the particles increase in size?
| Smallest | → | Largest | |
|---|---|---|---|
| A | N | O | F |
| B | \( \mathrm{N^{3-}} \) | \( \mathrm{O^{2-}} \) | \( \mathrm{F^-} \) |
| C | \( \mathrm{Na^+} \) | \( \mathrm{Mg^{2+}} \) | \( \mathrm{Al^{3+}} \) |
| D | \( \mathrm{Na^+} \) | Ne | \( \mathrm{F^-} \) |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
The particles in option D are isoelectronic; they all have 10 electrons.
For isoelectronic species, ionic or atomic radius increases as nuclear charge decreases.
Nuclear charge:
- \( \mathrm{Na^+} \): 11 protons
- \( \mathrm{Ne} \): 10 protons
- \( \mathrm{F^-} \): 9 protons
Hence the sizes increase in the order:
\( \mathrm{Na^+ < Ne < F^-} \)
Therefore, the correct answer is (D).
Question 24
W, X, Y and Z are Group 2 elements barium, calcium, magnesium and strontium, but not in that order.
- W reacts faster with dilute hydrochloric acid than Y.
- The hydroxide of X is less soluble in water than the hydroxide of W.
- The sulfate of Y is the most soluble of the sulfates of W, X, Y and Z.
- The carbonate of Z decomposes more slowly than the carbonate of W at the same temperature.
What is the \(M_{\mathrm{r}}\) of the nitrate of W?
(B) 164.1
(C) 211.6
(D) 261.3
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
Use the Group 2 trends:
- Reactivity increases down the group.
- Hydroxide solubility increases down the group.
- Sulfate solubility decreases down the group.
- Carbonate thermal stability increases down the group.
Since Y has the most soluble sulfate, Y must be magnesium.
W is more reactive than Y, so W is either calcium, strontium or barium.
Z has a carbonate more thermally stable than W, so Z must be below W in the group. This eliminates barium as W.
If W were calcium, X would have to be magnesium to satisfy the hydroxide solubility trend, leaving no element below calcium for Z. Therefore, W is strontium.
The nitrate of strontium is:
\( \mathrm{Sr(NO_3)_2} \)
\( M_{\mathrm{r}} = 87.6 + 2(14.0 + 3 \times 16.0) = 87.6 + 124.0 = 211.6 \)
Therefore, the correct answer is (C).
Question 25
Polymer J has repeat unit \( \mathrm{-[CH(C_2H_5)CH(CH_3)]-} \).
Polymer K has repeat unit \( \mathrm{-[CH(CH_3)CH(CH_3)]-} \).
Which row is correct?
| Monomer from which polymer J is produced | Monomer from which polymer K is produced | |
|---|---|---|
| A | pent-1-ene | but-1-ene |
| B | pent-1-ene | but-2-ene |
| C | pent-2-ene | but-1-ene |
| D | pent-2-ene | but-2-ene |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
In addition polymerisation, the groups attached to the two carbon atoms of the double bond remain attached to the polymer backbone.
For polymer J, the repeat unit has substituents \( \mathrm{C_2H_5} \) and \( \mathrm{CH_3} \), so the monomer is:
\( \mathrm{CH_3CH_2CH=CHCH_3} \) (pent-2-ene).
For polymer K, each backbone carbon carries a \( \mathrm{CH_3} \) group, so the monomer is:
\( \mathrm{CH_3CH=CHCH_3} \) (but-2-ene).
Therefore, the correct answer is (D).
Question 26
\(0.200\,\mathrm{mol}\) ethanenitrile reacts with an excess of dilute sodium hydroxide. The reaction produces organic compound S and ammonia gas only.
The reaction has an \(80.0\%\) yield.
Which mass of S is produced?
(B) \(13.1\,\mathrm{g}\)
(C) \(15.4\,\mathrm{g}\)
(D) \(16.4\,\mathrm{g}\)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
Ethanenitrile undergoes alkaline hydrolysis:
\( \mathrm{CH_3CN + 2H_2O + NaOH \rightarrow CH_3COONa + NH_3} \)
Thus, compound S is sodium ethanoate, \( \mathrm{CH_3COONa} \).
The mole ratio is \(1:1\), so the theoretical amount of S is:
\(0.200\,\mathrm{mol}\)
Applying the \(80.0\%\) yield:
\(0.200 \times 0.80 = 0.160\,\mathrm{mol}\)
\(M_{\mathrm{r}}(\mathrm{CH_3COONa}) = 23 + 24 + 32 + 23 = 82\)
Mass produced:
\(0.160 \times 82 = 13.12\,\mathrm{g} \approx 13.1\,\mathrm{g}\)
Therefore, the correct answer is (B).
Question 27
The structure of butenedioic acid is shown.
When an excess of \( \mathrm{NaOH(aq)} \) is added to butenedioic acid, which organic product is formed?
(B) B
(C) C
(D) D
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Excess aqueous sodium hydroxide neutralises both carboxylic acid groups of butenedioic acid to form the corresponding dicarboxylate salt.
The carbon-carbon double bond is not affected by aqueous \( \mathrm{NaOH} \), so no addition reaction occurs.
Therefore, the product contains:
- Two \( \mathrm{-COO^-} \) groups.
- The original \( \mathrm{C=C} \) double bond unchanged.
Only option A matches this structure. Therefore, the correct answer is (A).
Question 28
Four organic compounds are shown.
A mixture of these four compounds reacts with hot aqueous sodium hydroxide.
The products of this reaction are acidified.
Which organic acids are present in the products?
(B) Butanoic, ethanoic and methanoic acids
(C) Ethanoic, 3-methylbutanoic and propanoic acids
(D) Methanoic, 3-methylbutanoic and propanoic acids
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
Hot aqueous sodium hydroxide hydrolyses esters to form a carboxylate salt and an alcohol. Acidification converts the carboxylate salts into the corresponding carboxylic acids.
From the four structures shown:
- The first ester produces butanoic acid.
- The second compound is an ether and does not undergo hydrolysis.
- The third ester is a methanoate ester, producing methanoic acid.
- The fourth ester is an ethanoate ester, producing ethanoic acid.
Therefore, the organic acids present after hydrolysis and acidification are:
\( \mathrm{butanoic\ acid,\ ethanoic\ acid,\ and\ methanoic\ acid} \)
Therefore, the correct answer is (B).
Question 29
Ethanal reacts with KCN dissolved in liquid HCN. This reaction involves the formation of an intermediate.
Which statement is correct?
(B) The \( \mathrm{O^{\delta-}} \) of \( \mathrm{C{=}O} \) attacks \( \mathrm{K^+} \) in a nucleophilic attack.
(C) A proton from HCN is transferred to the intermediate.
(D) The reaction is a nucleophilic substitution.
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
The reaction of ethanal with \( \mathrm{HCN} \) is a nucleophilic addition reaction.
First, the cyanide ion attacks the partially positive carbon atom of the carbonyl group to form a negatively charged alkoxide intermediate.
The intermediate is then protonated by HCN to produce the hydroxynitrile.
Thus, a proton is transferred from HCN to the intermediate.
Option (A) is incorrect because HCN contains a lone pair on nitrogen. Option (B) is incorrect because \( \mathrm{CN^-} \), not \( \mathrm{O^{\delta-}} \), is the nucleophile. Option (D) is incorrect because the reaction is nucleophilic addition, not substitution.
Therefore, the correct answer is (C).
Question 30
What is formed when propanone is heated under reflux with a solution of \( \mathrm{NaBH_4} \)?
(B) Propan-1-ol
(C) Propanal
(D) Propane
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
Sodium borohydride, \( \mathrm{NaBH_4} \), is a reducing agent that reduces aldehydes and ketones to alcohols.
Propanone is a ketone:
\( \mathrm{CH_3COCH_3} \)
Reduction of the carbonyl group produces the corresponding secondary alcohol:
\( \mathrm{CH_3COCH_3 + 2[H] \rightarrow CH_3CH(OH)CH_3} \)
The product formed is propan-2-ol.
\( \mathrm{NaBH_4} \) is not strong enough to reduce the ketone further to an alkane.
Therefore, the correct answer is (A).
Question 31
The compounds shown are all produced by plants.
Each compound is warmed with acidified \( \mathrm{K_2Cr_2O_7} \).
Which compound will give a different observation to the other three?
(B) B
(C) C
(D) D
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Acidified potassium dichromate(VI) oxidises primary and secondary alcohols, changing from orange to green.
Compounds A, B and C contain primary alcohol groups, so they are readily oxidised.
Compound D contains a tertiary alcohol, which is not oxidised by acidified \( \mathrm{K_2Cr_2O_7} \) under these conditions because the carbon bearing the \( \mathrm{-OH} \) group has no hydrogen atom attached.
Therefore, compound D gives a different observation, with no colour change.
Therefore, the correct answer is (D).
Question 32
Structural and stereoisomerism should be considered when answering this question.
How many alcohols with the molecular formula \( \mathrm{C_5H_{12}O} \) give a yellow precipitate with alkaline \( \mathrm{I_2(aq)} \)?
(B) 3
(C) 4
(D) 5
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
The iodoform test is given by compounds containing the group:
\( \mathrm{CH_3CH(OH)-} \)
Among the alcohols with molecular formula \( \mathrm{C_5H_{12}O} \), the alcohols that satisfy this condition are:
- Pentan-2-ol
- 3-Methylbutan-2-ol
- 2-Methylbutan-1-ol? No (primary alcohol)
- 3-Methylbutan-2-ol has one chiral centre and exists as two stereoisomers.
Counting both structural and stereoisomers gives:
\(1\) (pentan-2-ol) \(+\;1\) (pentan-3-ol does not react) \(+\;2\) enantiomers of 3-methylbutan-2-ol \(=\;4\) alcohols that give a positive iodoform test.
Therefore, the correct answer is (C).
Question 33
An amine is produced in the following reaction.
\( \mathrm{C_2H_5I + 2NH_3 \rightarrow C_2H_5NH_2 + NH_4I} \)
What is the mechanism?
(B) Free-radical substitution
(C) Nucleophilic addition
(D) Nucleophilic substitution
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Ammonia acts as a nucleophile and attacks the carbon atom bonded to iodine in the haloalkane.
The iodide ion leaves as the leaving group, so one group is substituted by another.
This is an \( \mathrm{S_N2} \) nucleophilic substitution reaction.
Therefore, the correct answer is (D).
Question 34
The reaction shown can be used to lengthen a carbon chain.
\( \mathrm{CH_3CH_2CH_2Br + CN^- \rightarrow CH_3CH_2CH_2CN + Br^-} \)
Which row shows the correct reagent and conditions for this reaction?
| Reagent | Conditions | |
|---|---|---|
| A | KCN | Heat under reflux in dilute sulfuric acid |
| B | KCN | Heat under reflux in ethanol |
| C | HCN | Heat under reflux in dilute sulfuric acid |
| D | HCN | Heat under reflux in ethanol |
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
The cyanide ion, \( \mathrm{CN^-} \), acts as a nucleophile, replacing the bromide ion in a nucleophilic substitution reaction and increasing the carbon chain length by one carbon atom.
The reagent used is potassium cyanide (or sodium cyanide), which provides \( \mathrm{CN^-} \) ions.
The reaction is carried out by heating under reflux in ethanol. Ethanol is used as the solvent because the haloalkane is much more soluble in ethanol than in water.
Hydrogen cyanide is not used for this reaction because it does not provide a high concentration of free \( \mathrm{CN^-} \) ions.
Therefore, the correct answer is (B).
Question 35
Which row shows the products formed when cyclohexene reacts with acidified \( \mathrm{KMnO_4} \) under different conditions?
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
The products depend on the reaction conditions:
- Cold, dilute acidified \( \mathrm{KMnO_4} \) adds two hydroxyl groups across the \( \mathrm{C=C} \) bond to form a vicinal diol (cyclohexane-1,2-diol).
- Hot, concentrated acidified \( \mathrm{KMnO_4} \) oxidatively cleaves the double bond. Since each alkene carbon in cyclohexene bears a hydrogen atom, both ends are oxidised to carboxylic acids, producing hexanedioic acid (adipic acid).
Only row A shows these two products correctly.
Therefore, the correct answer is (A).
Question 36
Two steps in the free-radical substitution reaction between methane and chlorine are shown.
Step 1 \( \mathrm{CH_3^{\bullet} + Cl_2 \rightarrow CH_3Cl + Cl^{\bullet}} \)
Step 2 \( \mathrm{CH_3Cl + Cl^{\bullet} \rightarrow CH_2Cl^{\bullet} + HCl} \)
Which statement is correct?
(B) Step 1 is propagation and step 2 is termination.
(C) Step 1 is initiation and step 2 is termination.
(D) Both steps are propagation.
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
In a free-radical substitution mechanism:
- Initiation produces radicals, e.g. \( \mathrm{Cl_2 \rightarrow 2Cl^{\bullet}} \) under UV light.
- Propagation involves a radical reacting to form a product and another radical, allowing the chain reaction to continue.
- Termination occurs when two radicals combine to form a molecule with no radicals remaining.
In Step 1, the methyl radical reacts with chlorine to produce chloromethane and a chlorine radical, so the radical is regenerated.
In Step 2, the chlorine radical reacts with chloromethane to form another radical, \( \mathrm{CH_2Cl^{\bullet}} \), so the chain continues.
Therefore, both steps are propagation, so the correct answer is (D).
Question 37
Compound P displays cis/trans isomerism and gives a red-brown precipitate with Fehling’s solution.
What is P?
(B) B
(C) C
(D) D
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{D}} \)
Fehling’s solution gives a red-brown (brick-red) precipitate with aldehydes. Therefore, compound P must contain a \( \mathrm{-CHO} \) group.
To exhibit cis/trans isomerism, each carbon atom of the \( \mathrm{C=C} \) bond must be attached to two different groups.
- A: One carbon has two hydrogen atoms, so no cis/trans isomerism.
- B: Contains a ketone, so it does not react with Fehling’s solution.
- C: One carbon has two hydrogen atoms, so no cis/trans isomerism.
- D: Contains an aldehyde group and each carbon of the double bond has two different substituents, so it shows cis/trans isomerism.
Therefore, the correct answer is (D).
Question 38
Fructose is a sugar with more than one chiral centre. The fructose molecule is shown with X, Y and Z indicating three carbon atoms.
Which carbon atoms are chiral centres?
(B) X and Y only
(C) X only
(D) Y only
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
A chiral centre is a carbon atom bonded to four different groups.
- X is attached to \( \mathrm{H} \), \( \mathrm{OH} \), the carbonyl-containing chain above, and a different carbon chain below, so it is chiral.
- Y is also bonded to four different groups, so it is chiral.
- Z is a \( \mathrm{CH_2OH} \) carbon and is bonded to two hydrogen atoms, so it is not chiral.
Therefore, the chiral centres are X and Y only.
Therefore, the correct answer is (B).
Question 39
An organic compound, T, contains only one functional group.
Compound T has the empirical formula \( \mathrm{C_2H_4O} \).
Some functional groups are listed.
1. Alcohol
2. Aldehyde
3. Ester
4. Ketone
Which functional groups are possible for compound T?
(B) 1 and 4
(C) 2 and 3
(D) 2 and 4
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{C}} \)
The empirical formula is \( \mathrm{C_2H_4O} \), so any possible molecular formula must be a whole-number multiple of this ratio.
- Aldehydes can have the formula \( \mathrm{C_nH_{2n}O} \), for example ethanal, \( \mathrm{C_2H_4O} \).
- Esters can also have molecular formulas that simplify to \( \mathrm{C_2H_4O} \), for example compounds with formula \( \mathrm{C_4H_8O_2} \).
- Alcohols have the general formula \( \mathrm{C_nH_{2n+2}O} \), which does not simplify to \( \mathrm{C_2H_4O} \).
- Ketones require at least three carbon atoms, so no ketone has the empirical formula \( \mathrm{C_2H_4O} \).
Therefore, the possible functional groups are 2 and 3, so the correct answer is (C).
Question 40
The mass spectrum of an organic compound has the following features.
- The molecular ion peak is at \( m/e = 60 \). This peak has a relative intensity of \(100\).
- There is a peak at \( m/e = 61 \), which has a relative intensity of \(2.2\).
- There is no fragment peak at \( m/e = 17 \).
- There is a fragment peak at \( m/e = 31 \).
What could be the identity of the organic compound?
(B) Methoxyethane, \( \mathrm{CH_3OCH_2CH_3} \)
(C) Ethanoic acid, \( \mathrm{CH_3COOH} \)
(D) Propan-2-ol, \( \mathrm{CH_3CH(OH)CH_3} \)
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
The molecular ion peak at \( m/e = 60 \) shows that the compound has a relative molecular mass of \(60\).
The peak at \( m/e = 61 \) has an intensity of about \(2.2\%\), indicating the presence of two carbon atoms, since the natural abundance of \(^{13}\mathrm{C}\) is approximately \(1.1\%\) per carbon atom.
There is no fragment peak at \( m/e = 17 \), so a hydroxyl group (\( \mathrm{-OH} \)) is unlikely. This rules out ethanoic acid and propan-2-ol.
A fragment at \( m/e = 31 \) is consistent with the \( \mathrm{CH_3O^+} \) ion, which is commonly produced by esters containing a methoxy group.
Methyl methanoate, \( \mathrm{HCOOCH_3} \), satisfies all of these observations.
Therefore, the correct answer is (A).