All solutions here are SUGGESTED. Mr. Lee will hold no liability for any errors. Comments are entirely personal opinions.
Question 1: B
The element must be from Group IV so the electronic configuration must be either 1s22s22p2 or 1s22s22p63s23p2. The four electrons of highest energy refer to the electrons at shells with higher principal quantum number. The best answer is B.
Question 2: B
Sigma bonds can be formed via head-on overlap between two s orbitals, two p orbitals and between one s and one p orbitals. In hydrocarbons, the carbon atoms have hybrid orbitals consisting of s and p orbitals. Hence, sigma bonds can be formed by either s or p orbitals while the pi bonds can only be formed by side-way overlap between two p orbitals.
Question 3: C
Oxygen atom has 8 electrons. Hydrogen atom has 1 electron. Since hydroxide anion has an extra electron gained, the total number of electrons is 10. Oxygen atom has 8 neutrons while hydrogen atom has no neutrons. Hence, the answer is C.
Question 4: C
A higher charge on Ca2+ would mean that the electrostatic forces of attraction between the cation and the sea of mobile electrons are stronger as compared to Na+. Option D is wrong because the electrons in the calcium ion are not mobile and even if calcium ion has more electrons than sodium ion, it does not explain the stronger metallic bonding. The correct explanation should be calcium metal has more mobile electrons than sodium metal, instead of ion.
Question 5: B
In zwitterion, –NH2 (trigonal pyramidal, 107°) group becomes –NH3+ (tetrahedral, 109.5°).
Question 6: D
Lattice energy is the enthalpy changed when 1 mole of solid ionic compound is formed from its constituent gaseous ions.
Question 7: B
Number of moles of NaOH = 12.5 / 1000 x 0.0500 = 6.25 x 10-4 mol (limiting reagent)
Number of moles of HCl = 25.0 / 1000 x 0.100 = 2.50 x 10-3 mol (in excess)
Number of moles of HCl remaining = 1.875 x 10-3 mol
Concentration of HCl remaining = 1.875 x 10-3 / (37.5 / 1000) = 0.0500 mol dm-3
Question 8: C
CH4 + 2O2 → CO2 + 2H2O
CH4 + 3/2 O2 → CO + 2H2O
9CH4 + 18O2 → 9CO2 + 18H2O
CH4 + 3/2 O2 → CO + 2H2O
Therefore, 10 moles of CH4 react with 19.5 moles of oxygen to produce 9 moles of CO2 and 1 mole of CO (9 : 1 ratio)
Hence, 1 dm3 of CH4 reacts with 1.95 dm3 of oxygen.
Question 9: A
Formation of hydrogen bonds indicates that the reaction is exothermic. ΔH is negative.
The product (1 mole) is more ordered as compared to the reactants (1 mole of CO2 and n moles of H2O). Hence, ΔS is negative.
Question 10: B
Standard conditions means that all concentrations must be 1 mol dm-3.
Question 11: D
Equilibrium concentration for CH3CO2H = C(1 – α)
Equilibrium concentration for CH3CO2– = Cα
Equilibrium concentration for H+ = Cα
Kc = (Cα)2 / C(1 – α) = α2C / (1 – α)
Question 12: C
Increasing the pressure will shift the equilibrium to the right because the forward reaction produces fewer number of moles of gaseous substances. Decreasing the temperature for the exothermic reaction will shift the equilibrium to the right because the forward reaction will release heat energy to counteract the decrease in temperature.
Question 13: D
Total number of moles = 2.0 mol
Number of moles of H2 = 33.3 / 100 x 2.0 = 0.666
Number of moles of CO2 = 33.3 / 100 x 2.0 = 0.666
Percentage of H2O = (100% – 33.3% – 33.3%) / 2 = 16.7%
Number of moles of H2O = 16.7 / 100 x 2.0 = 0.334
Number of moles of CO = 16.7 / 100 x 2.0 = 0.334
Kc = (0.666)2 / (0.334)2 = 3.98 ≈ 4.0
Question 14: B
The reaction is second order with respect to NO. Hence, when concentration decreases by 2, the initial rate will decrease by 4. Hence, x = 6.0 / 4 = 1.5
The reaction is first order with respect to H2. Hence when concentration increases by 2, the initial rate will increase by 2. Hence, y = 1.5 x 2 = 3.0
The reaction is second order with respect to NO. Hence, when the initial rate decreases by (3.0 / 0.75) = 4, the concentration of NO will decrease by 2. Hence, z = 0.5.
Question 15: D
Thermal stability of nitrates increases down Group II due to lower charge density of Group II metal cations as the size of the cations increases.
Question 16: D
Phosphorus exists as P4. Option A is wrong because sulfur can also form two acidic oxides, SO2 and SO3. Option B is wrong because argon should have the highest ionisation energy. Option C is wrong because chlories of silicon (SiCl4) and sulfur (S2Cl2) can react with water to form HCl solution which is acidic.
Question 17: A
Only Option A is correct. Option B is wrong because it is not a precipitate. Option C is wrong because it is a blue precipitate instead of a pale blue solution. Option D is wrong because Cu+ complex should not have any colour because it has completely filled 3d orbitals so no d-d transition is possible.
Question 18: C
Atomic radius should decrease across the period from Mg to P while for melting point, Si has the highest, followed by Al, Mg and P.
Question 19: B
Only dilute sulfuric acid is used to acidify aqueous potassium dichromate(VI). Ethanol can be oxidised to form ethanoic acid.
Question 20: A
PCl5 will react with hydroxyl groups and carboxylic acid groups. The Br atom does not react with PCl5
Question 21: B
The 3 structural isomers are:
Question 22: D
Question 23: B
NaBH4 will reduce aldehyde to alcohol but it does not reduce alkene to alkane.
Question 24: A
Molecular formula of citric acid = C6H8O7
Condensation reaction will lose one molecule of water.
Hence, C6H8O7 + C3H6O2 → C9H12O8 + H2O
Hence, option A is the best answer as the structural formula of C2H5CO2H fits the molecular formula of C3H6O2
Question 25: D
Ethanoyl chloride is an acid chloride. It will react with water to produce ethanoic acid and Cl– and the Cl– ion will form a white precipitate (AgCl) with Ag+
Question 26: D
pH 11 is alkaline so the functional groups will be deprotonated.
Question 27: D
Addition of molecular mass of 14 suggests that one oxygen atom is added but two hydrogen atoms are removed. This indicates that primary alcohol (–CH2OH) is converted to carboxylic acid group (–COOH).
Question 28: C
Cold NaOH(aq) is able to react with the acidic phenol group but not with the alcohol group.
Question 29: B
The phenol group is an activating group and is 2,4,6-directing.
Question 30: A
The carbon-carbon double bond will attack the Br–Br first to form a carbocation containing one Br atom.
Question 31: D
Mn2+: [Ar] 3d5 (no paired 3d electrons)
Fe2+: [Ar] 3d6 (one paired 3d electrons)
Co3+: [Ar] 3d6 (one paired 3d electrons)
Question 32: A
Simple covalent compounds have low boiling point. AlBr3 has a simple molecular structure because it has a low boiling point even though it has a metal and non-metal. If AlBr3 is ionic, the boiling point should be 4 digit. Hence, all 3 compounds are covalent.
Question 33: C
The molecules of the gas have mass, so option 1 is wrong.
Question 34: A
Zinc will be oxidised while NH4+ will be reduced. So the overall potential will be +1.50V (energetically feasible). Option 2 and 3 are correct also.
Question 35: B
Rate constants are affected by activation energy and temperature based on Arrhenius equation k = Ae–Ea / RT. Introducing a catalyst will increase both kf and kb equally while increasing the temperature for an exothermic reaction will increase BOTH kb and kf but will increase kb more than kf so equilibrium shifts to the left. Rate constant does not depend on concentration of the reactants.
Question 36: A
Methanal is trigonal planar with respect to carbon and the carbon has an oxidation number of 0 (hydrogen has an oxidation number of +1 while oxygen has an oxidation number of –2). The equation for complete combustion of methanal is:
CH2O + O2 → CO2 + H2O
Hence, 1 mole of oxygen is required for 1 mole of methanal.
Question 37: C
Radicals do not have a lone pair of electrons. They have unpaired electrons and can be formed by homolytic fission of a covalent bond.
Question 38: C
Geometric isomerism is possible if there are two different groups attached to the sp2 carbon in the carbon-carbon double bond. Hence, Y cannot be H and can be Br so that the carbon-carbon double bond with Y can exist as geometric isomers. The carbon-carbon double bonds with Z and X cannot exist as geometric isomers since the groups attached to the sp2 carbon in the carbon-carbon double bond are the same.
Question 39: A
Thymol has a phenol group so the acidic phenol group can react with the alkali present in alkaline KMnO4 via neutralisation. The phenol group can also react with ethanoyl chloride to form an ester via condensation and it can also react with sodium metal to form phenoxide ion.
Question 40: B
Option 1 and 2 have no line of symmetry while option 3 has a line of symmetry. Hence, option 3 is not optically active.
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