2024 Prelims RVHS H2 Chem P2 (Ans)

1 River Valley High School 2024 JC2 H2 Chemistry 9729 Prelim Exam Paper 2 Suggested Solutions 1 (a) (i) Proton and deuterons are deflected to the negative plate , but hydrogen atoms is not deflected . This is because protons and deuterons are positively charged , whereas hydrogen atoms are electrically neutral. The mass of a proton is half of the mass of a deuteron (1 neutron and 1 proton), thus angle of deflection for proton doubled to that of deuterons. [2] (ii) +1310 kJ mol−1 (Same value as first IE for hydrogen.) The electron to be removed from deuterium atom and hydrogen atom experience the same nuclear charge and no shielding effect. They are of similar/same distance from the nucleus . Thus, the attraction of positive nucleus for the (1s) electrons to be removed is the same. Or Hydrogen and deuterium have the same electronic configuration . Although deuterium has an additional neutron , it has no charge , therefore does not affect the attraction of nucleus for the electrons to be removed. [2] (b) (i) 1s2 2s2 2p6 3s2 3p6 3d8 4s2

2 (ii) For all atoms, there is an increase in successive ionisation energies as the electron is removed from an ion of increasing positive charge. The significant jump from 2 nd to the 3rd ionisation energy for element A is due the removal of the 3 rd/3p electron from an inner principal quantum shell which is more strongly attracted to the positive nucleus compared to a 3s electron. 3d and 4s electrons are very close in energy. Similar amount of energy is required to remove the 4s/2nd and 3d/3rd electron from Ni. (iii) Element B is Mg. [1] (c) (i) 2NO + 2CO → N2 + 2CO2 Or NO2 + 2CO → ½ N2 + 2CO2 [1] (ii) During adsorption, the formation of weak bonds between the NO x molecules and the active sites on rhodium surface weakens/breaks the bonds within the NOx molecules. Reaction occurs more readily as the NO x particles are held on rhodium surface in close proximity and in the correct orientation. During desorption, the weak bonds between the product particles and the rhodium surface are broken . The products formed diffuse away from the surface of the catalyst and the active sites become available again. [3] (d) (i) First order with respect to CxHy at low concentration, but zeroth order at high concentration. At high [C xHy], adding more substrates cannot accelerate the reaction as all the active sites on the catalyst surface are saturated/ occupied. [2] (ii) [2]

3 A catalyst provides an alternative reaction pathway with a lower activation energy . As a result, more reactant particles possess energy  activation energy required for an effective collision and the frequency of effective collisions increases . Thus, rate constant is increased. (iii) (e) NO2 forms photochemical smog which causes respiratory problems. [1] 2 (a) (i) A B [2] (ii) Condensation/ Addition-elimination [1] (b) E F [2]

4 (c) H I [2] (d) The tertiary carbocation intermediate in the formation of G is more stable as there are 3 electron-donating alkyl groups to spread out the positive charge. [1] (e) Test: add acidified KMnO4 (or K2Cr2O7), heat. Observations: KMnO4 remains purple (or K 2Cr2O7 remains orange) for C, while purple KMnO4 decolourises (or orange K2Cr2O7 turns green) for J. or Test: add NaOH, heat; followed by add I2(aq), warm. Observations: no pale yellow ppt formed for C, while pale yellow ppt formed for J. [2]

5 3 (a) An Arrhenius acid is a compound that dissolves in water to yield hydrogen ions. [1] (b) (i) [2] (ii) When 30.0 cm3 of NaOH has been added, the solution contains excess NaOH and conjugate base of the amino acid. Amount of excess NaOH = 10.0 1000 × 0.100 = 1.000  103 mol Concentration of OH = 1.000  103  40.0 1000 = 2.500  102 mol dm3 pOH = 1.602 pH = 14  1.602 = 12.4 [2] (iii) [1] (iv) -CO2H is a stronger acid than H2CO3 so -CO2 will not be protonated by H2CO3. -NH3+ is weaker acid than H 2CO3 so -NH2 will be protonated by H2CO3. [2]

6 - NH2 is a stronger base than CO2, so -NH2 is protonated. (c) [2] 4 (a) Lactide has more electrons than lactic acid. More energy is needed to overcome the stronger instantaneous dipole -induced dipole interactions between lactide molecules than the hydrogen bonds between lactic acid molecules. Since O is more electronegative than N, O−H bond is more polar than N−H bond. More energy is needed to overcome the stronger hydrogen bonds between lactic acid molecules than that between 1,4-butanediamine molecules. [3] (b) (i) [1] p (ii) G = H −TS Since H < 0, S < 0 , −TS > 0 , so as temperature increases, |−TS| > |H|, G becomes positive. Therefore, the polymerisation becomes less spontaneous. [2] (c) (i) Increasingly positive/ decreasingly negative due to the increase in disorder in the system as the number of ways to arrange the polymer [1]

7 chain/ greater flexibility in the rotation of bonds in the open chain increases.

8 (ii) H = 2(360) + 2(390) − 2(305) − 2(460) = −30.0 kJ mol−1 [2] (d) The sp2 C1 in norbornene becomes sp 3 hybridised in B. The sp 2 orbital is closer to the nucleus/ smaller in size , so C 1 – H bond is shorter in norbornene. [1] (e) (i) [2] (ii) The sp2 carbocation is trigonal planar. The bromide ion/ nucleophile attack from both sides of the plane, which give rise to a pair of stereoisomers. [1] (iii) There is an internal plane of symmetry the mirror images of stereoisomer are superimposable , thus this structure is optically inactive. [2] 5 (a) (i) A transition element is a d -block element which forms at least one stable ion with a partially filled d subshell. [1] (ii) In the presence of (H 2O) ligands, the degenerate partially filled 3 d orbitals are split into 2 sets of orbitals with a (small) energy difference (E). This E is different for V 2+ and V3+ because these ions have different oxidation states/ charges on ions/ number of electron/ electronic configuration . Radiation from visible light spectrum is absorbed to promote an electron from a lower energy d -orbital to another d -orbital of higher energy. The green colour observed for V 3+ corresponds to the complement of the red colours absorbed. [3]

9 The violet colour observed for V2+ corresponds to the complement of the yellow colours absorbed. (iii) Type of isomer: trans Type of isomer: Cis [2]

10 (b) (i) Electrode A (through external circuit) to electrode B. [1] (ii) Ecell  = E(VO2+| VO2+) – E(V3+| V2+) = (+1.00) – (– 0.26 ) = +1.26V [1] (iii) As the pH increased, [H+] decreased . VO2+(aq) + 2H+ + e = VO2+(aq) + H2O(l) ----- (1) The position of equilibrium of (1) shifts to the left and E(VO2+(aq)/VO2+(aq)) becomes less positive. Thus, Ecell becomes less positive. [2] (c) (i) Yellow to green/greenish-blue/blue-green [1] When the cell becomes flat , 76.9% of yellow VO2+ is converted to blue VO 2+. The colour is green as the solution contains a mixture of yellow VO2+ and blue VO2+. (ii) VO2+  V2+  e− Amount of electrons, n = 76.95.00 2.00 100 = 7.69 mol Capacity = Q = nF = 7.69  96500 = 7.421 105 C [2]