2024 Prelims VJC H2 Chem P1 (Ans)

1 VICTORIA JUNIOR COLLEGE 2024 JC2 H2 CHEMISTRY PRELIM EXAM PAPER 1 ANSWERS 1 A 7 C 13 B 19 A 25 D 2 B 8 A 14 A 20 B 26 A 3 B 9 C 15 C 21 D 27 D 4 D 10 D 16 D 22 C 28 D 5 C 11 A 17 D 23 A 29 D 6 B 12 B 18 B 24 C 30 C 1 A Since there are 4 valence electrons in the Group 14 elements, the electronic configuration of the valence shell at its ground state should be ns2np2. This corresponds to the electrons with four highest energy. 2 B Element X has the highest 6th ionisation energy, it means that the sixth electron removed from X is from its next inner shell. Hence, X has five valence electrons and X is in Group 15. As W, X, Y and Z are four consecutive elements in the Periodic Table, W is in group 14. Hence, the formula of the fluoride of W is WF4. 3 B (1 and 2 only) Option 1: Correct Lone pair electrons on oxygen in CH3OCH3 is donated to the vacant orbital of boron in BF3 to form the dative bond. Option 2: Correct The bond angle around B changes from 120 ° in BF3 (3 bond pairs, trigonal planar) to 109.5 ° in J (4 bond pairs, tetrahedral). Option 3: Incorrect In CH3OCCH3, there are 2 bond pairs 2 lone pairs around O (bent). In Q, there are 3 b ond pairs and 1 lone pair around O (trigonal pyramidal). 4 D Option A: correct statement Since O is more electronegative than N, the H2O molecule has a greater dipole moment than the NH 3 molecule, which results in stronger permanent dipole- permanent dipole (pd-pd) interactions between H 2O molecules. Hence, more energy is required to overcome the stronger pd-pd between H2O molecules. Option B: correct statement Since O is more electronegative than N, th e hydrogen bonds formed between H2O molecules are stronger than that between NH3 molecules. More energy is required to overcome the stronger hydrogen bonding between H 2O molecules. Option C: correct statement The extent of hydrogen bonds is greater for H 2O (2 hydrogen bonds per molecule) than NH3 (1 hydrogen bond per molecule). Hence, more energy is required to overcome the more extensive hydrogen bonds between H2O molecules. Option D: Incorrect statement Energy is required to overcome the intermolecular force of attractions (hydrogen bonds) between H2O molecules and that between NH3 molecules. Hence, the strength of O–H bonds in water and N –H bonds in NH 3 does not affect the boiling points of H2O and NH3. 5 C pV = nRT Since V is the same for both gases, V R = nT p = constant nATA pA = nD2TD2 pD2 where D2 is deuterium gas, 2H2 Since, n = m Mr , mA pAMrA TA = mD2 pD2Mr D2 TD2 18.0 2 × pD2 × MrA (40 + 273) = 0.6 pB× 2.0 × 2 (20 + 273) MrA = 64.1 6 B Ion of V is positively charged as its ionic radius is smaller than its atomic radius. Hence, V could be Na, Mg, Al or Si as these elements form cations with 10 electrons. Ion of Z is negatively charged as its ionic radius is bigger than its atomic radius. Hence, Z could be P, S or C l as these elements form anions with 18 electrons. Option A: Incorrect Ions of V and Z have different number of full electronic shells, i.e. 2 and 3 respectively. Option B: Correct Electronegativity increases across the period. Z has a greater electronegativity than V as Z has a greater atomic number than V in Period 3. Option C: Incorrect Ions of Z are negatively charged. Option D: Incorrect Number of outer electrons increases across the period. V has fewer outer electrons than Z as V has a smaller atomic number than Z in Period 3. 7 C (1 and 3 only) Option 1: Correct Down the group, the ionic radius increases, hence, charge density of the metal ion (∝ q+/r+) decreases. The polarising power of the metal ion decreases. There is less distortion of the anionic charge cloud by the metal ion and the stability of the carbonate to heat increases. Option 2: Incorrect Down the group, the ionic radius increases, hence, valence electrons are less strongly attracted by the nucleus and more easily lost . Reducing power of Group 2 elements increases. Hence, reaction between Group 2 elements and water becomes more vigorous down the group. Option 3: Correct Down the group, since number of electronic shells increases, valence electrons are further away from the nucleus. Hence, there is weaker electrostatic forces of attraction between the nucleus and the valence electrons. Ability to attract shared electrons i n a covalent bond towards itself decreases, so electronegativity decreases. 8 A For Group 17, • strength of covalent bonds: C l–Cl > Br–Br > I–I. As atomic radius increases from C l to I, the extent of orbital overlap between the halogen atoms becomes less effective, leading to weaker covalent bond between the halogen atoms.

2 • strength of instantaneous dipole –induced dipole interactions: I2 > Br 2 > C l2. As number of electrons increases from Cl2 to I2, the larger electron cloud size will lead to greater ease of electron cloud distortion and stronger instantaneous dipole –induced dipole interactions. • strength of oxidising agent: C l2 > Br 2 > I2. Down the group, the distance between the nucleus and valence electrons increases due to increasing number of electronic shells. There is weaker electrostatic forces of attraction between the nucleus and the incoming electron and the ease of gaining electrons decreases, thus oxidising power of halogens decreases down the group. Hence, X2 is I2 which has weaker covalent bonds than Y2 which is C l2. X2 (I2) has stronger instantaneous dipole – induced dipole forces than Z 2 which is Br 2. Y2 (Cl2) is a stronger oxidising agent than Z2 (Br2). 9 C Amount of Au = 10000 ÷ 197 = 50.76 mol Amount of O2 required = 50.76 ÷ 4 = 12.69 mol Volume of O2 required = 12.69 × 24 = 304.6 dm3 Volume of air required = 304.6 ÷ 20% = 1520 dm3 10 D Standard enthalpy change of formation is defined as the enthalpy change when one mole of the substance in their specific state is formed from its constituent elements in their standard states under standard conditions of 298 K and 1 bar. Option A: Incorrect For the standard enthalpy change of formation LiC l(s), 1 mol of LiC l(s) should be formed from 1 mol of Li(s) and ½ mol of Cl2(g). Option B: Incorrect 2 mol of H 2O is formed. Hence, t he option represents 2×ΔHfo(H2O). Option C: Incorrect For the standard enthalpy change of formation of C2H6(g), 1 mol of C2H6(g) should be formed from 2 mol of C(s) and 3 mol of H 2(g). Carbon should be in solid state and not gaseous state at 298 K. Option D: Correct 1 mol of NaC l(s) is formed from their constituent elements. 11 A Vigorous r eaction occurred shows that the reaction is spontaneous so G < 0. Temperature of the solution falls means the reaction is endothermic, H > 0. Since G = H – TS, S = (H – G)/T, since T > 0, G < 0 and H > 0, S must be positive. 12 B The question states that time is measured when a fixed volume of O 2 is produced. Hence, rate  1/time. Thus, the order of the reaction with respect to H 2O2 can be determined from the [H 2O2] against 1/time graph , which is equivalent to [reactant] against rate of the reaction graph. Thus, Option B is correct. [H2O2] and volume of O 2 are not measured at different time of the experiment to show the progress of reaction. Thus, options A and C are incorrect. The fixed volume of O 2 does not show the progress of reaction. Option D is incorrect. 13 B With a decrease in the temperature, t he Boltzmann distribution curve will shift to the left (as seen by the dotted line) . There is a smaller number of molecules with higher energy so n at Y will be lower, while there are a larger number of molecules with lower energy, so n at X will be higher. 14 A SO2 + ½O2 → SO3 H < 0 (exothermic) Increasing the temperature shifts the position of equilibrium to the left to absorb excess heat