2025 JC2 Prelims H1 Chem Paper 1 Solutions_TJC

2025 TJC H1 Chemistry Prelim Paper 1 Worked Solutions 1 2 3 4 5 6 7 8 9 10 B A C B A B B C A D 11 12 13 14 15 16 17 18 19 20 B B C D D D D A A B 21 22 23 24 25 26 27 28 29 30 C A D C B A D C D C 1 Answer: B Al : 1s22s22p63s23p1 Al3+ : 1s22s22p6 Ca : 1s22s22p63s23p64s2 Ca2+ : 1s22s22p63s23p6 Cu : 1s22s22p63s23p63d104s1 Cu2+ : 1s22s22p63s23p63d9 S : 1s22s22p63s23p4 S2– : 1s22s22p63s23p6 Only options A, B and D have noble gas configuration. Only options A and B have more protons than electrons Option A involves removing electron from 3p and 3s orbitals Option C : 2 -nitrophenol can form intramolecular hydrogen bonding giving rise to less extensive H bonding between molecules and lower melting point. Option D : Ethanoic acid can dimerise in benzene by forming two hydrogen bonds between two acid molecules. The apparent M r (120.0) is double than of its actual Mr (60.0). 5 Answer: A A Sr: giant metallic structure Al2O3: giant ionic lattice structure I2: simple molecular structure B Be: giant metallic structure Al2Cl6: simple molecular structure SO3: simple molecular structure C Al: giant metallic structure C: giant molecular structure SiO2: giant molecular structure D Ni : giant metallic structure PCl3: simple molecular structure SiCl4: simple molecular structure 2 Answer: A Angle of deflection  charge/mass Option A : 2/4 = 0.5 Option B : 1/19 = 0.05 Option C : 1/28 = 0.036 Option D : 2/32 = 0.06 3 Answer: C Cu: 1s2 2s2 2p6 3s2 3p6 3d104s1 Mn: 1s2 2s2 2p6 3s2 3p6 3d5 4s2 Fe: 1s2 2s2 2p6 3s2 3p6 3d6 4s2 Fe2+: 1s2 2s2 2p6 3s2 3p6 3d6 4 Answer: B Option A : Water molecules are attracted to other water molecules via hydrogen bonding resulting in a regular three dimensional tetrahedral structure. The arrangement of water molecules in ice create an ‘open cage’ structure resulting in ice to be less dense than water. Option B : Both compounds have hydrogen bonding between molecules. However, volatility depends on the electron cloud size (inferred from M r) which affects the strength of id-id. 6 Answer: B A simple molecular solid should have a lower melting point (due to weak intermolecular forces) and does not conduct electricity (due to lack of mobile charge carriers – ions or free electrons). It may be soluble or insoluble in water depending on the type of intermolecular forces that it can form with water. Since the unknown compound is a solid at room temperature, the answer cannot be A due to the melting point.

7 Answer: B Option 1 has a dative bond from O to C while option 3 has a dative bond from Cl to Al. Option 2 has no dative bond present. 12 8 Answer: C At standard temperature and pressure, graphite is thermodynamically more stable than diamond. However, the conversion from diamond to graphite, has a very high activation energy and therefore, takes place at an extremely slow rate. 13 Answer: C Y: standard enthalpy change of com