2022 RVHS JC2 H2 CM Prelim P1 (Worked Solutions)

1 2022 JC 2 H2 Chemistry Prelim Paper 1 Worked Solutions Qn Ans Solutions 1 D Number of proton > electron => cation Number of proton > neutron => likely contain D (highlighted by question) No. of proton No. of neutron No. of electron A. NO2+ 7 + 2(8) 7 + 2(8) 7 +2(8) -1* B. ND2H 7 + 2(1)+ 1 7 + 2(1) + 0 7 + 2(1) + 1 C. NDH− 7 + 1 + 1 7 + 1 + 0 7 + 1 + 1 +1* D. ND3H+ 7 + 3(1) + 1 7 +3(1) + 0 7 + 3(1) + 1 -1* 2 A A. V− → V2− [Ar] 4s23d5 B. N → N− 1s2 2s2 2p4 C. Se+ → Se [Ar] 4s23d104p4 D. Ti → Ti− [Ar] 4s23d3 3 B Significant increment from 5th to 6th I.E. Element M from period 3 and has 5 valence electrons. => Phosphorous Equation 2 is formation of PCl3 ,Equation 3 is formation of PCl5 4 D Option A: The dipole moment on the HF molecule should be larger (HF dipole moment is larger and factoring in partial cancellation of 2 smaller OH dipole moments). Option B: As fluorine is more electronegative, the hydrogen bond formed between HF molecules will be stronger than the hydrogen bond formed between H 2O molecules. Option C: The change in id -id interactions is negligible. id-id interactions are not the predominant IMF for HF and H2O molecules. Option D: H2O forms more extensive hydrogen bonds than HF. 5 C Option A: In BF3, B shared its 3 valence electrons to form 3 covalent bonds with 3 F atoms. B is electron deficient. Option B: In CO, C shares its 2 valence electrons with 2 valence electrons on O (to maintain an octet electronic configuration for O). To achieve an octet electronic configuration for C, O will donate a lone pair of electrons to C and form a dative bond. Option C: In NO, N shares its 2 valence electrons with the 2 valence electrons on O (to maintain an octet electronic configuration for O). N is electron deficient. Option D: In SiO2, which has a giant molecular structure, Si shares its 4 valence electrons with 4 oxygen atoms (which will further extends to obtain a macromolecule).

2 6 B Deviation depends on IMF and size of particles. All 4 compounds has id-id, SO3 has the highest Mr. 7 C When an ideal gas is compressed from 20 atm to 80 atm with no further reaction, the volume is expected to change from 67.0 cm3 to 16.75 cm3 (instead of 15.5 cm3 as stated in the question). Option C is not a valid explanation since the dissociation of a gas would result in an observed gas volume of larger than 16.75 cm 3 (instead of 15.5 cm3 as stated in the question). 8 A From the information given, 20.0 cm3 of 0.20 mol dm −3 acid reacts exactly with 40 cm3 0.10 mol dm−3 of base, indicating that reacting mole ratio of acid and base is 1:1. This would be inconsistent with option A since CH3COOH and Ba(OH) 2 would react in a 2:1 ratio. Hence, option A is false, making A the answer. 9 A H must be negative for the overall process must be exothermic since heat must be generated by the reaction to enable the heating process. S must be negative as indicated by the reacti