EJC Nov 2019 H1 Chemistry 8873 Paper 1 Worked Solutions

Eunoia Junior College 8873 H1 Chemistry 2019 Paper 1 Worked Solutions 1 As Z is not deflected, it is uncharged. Hence, Z is a neutron. Based on the masses of X and Y, X is an electron and Y is a proton. Note: Masses of sub-atomic particles are given in the Data Booklet.  B 2 Arsenic is in Group 15 and Selenium is in Group 16. ✓ 1 Atomic radius of selenium is smaller than that of arsenic as atomic radii decreases across period. ✓ 2 1st I.E. of selenium is smaller than that of arsenic as 1st I.E. generally decreases across period.  3 Both selenium (4s2 4p4) and arsenic (4s2 4p3) have three 4p orbitals containing one or more electrons.  A 3 Electronic configuration of atom with 21 electrons: 1s2 2s2 2p6 3s2 3p6 3d1 4s2 Total number of electrons in the p orbitals = 6 + 6 = 12  B 4  1 only has 1  bond. ✓ 2 has 1  bond and 2  bonds in the triple bond. ✓ 3 has 1  bond and 2  bonds in the triple bond.  C 5 There are 4 bond pairs and no lone pair around S. Hence, the shape about S is tetrahedral, and the bond angle is 109.5o.  B 6  1 Tetrahedral See-saw ✓ 2 Tetrahedral Tetrahedral ✓ 3 Tetrahedral Tetrahedral  D 7 Species Shape Polarity BF3 Trigonal planar Non- polar CO2 Linear Non- polar CCl4 Tetrahedral Non- polar PCl3 Trigonal pyramidal Polar (due to lone pair)  D 8 Propane is more easily liquefied as it has stronger intermolecular instantaneous dipole-induced dipole (id-id) interactions. The stronger id-id interactions arise due to propane having more electrons, as compared to methane. Option A is a true statement, but does not explain the phenomenon, although molecular size gives an indication of the strength of the id-id interactions. It is the number of electrons in the molecule that determines the strength of the id-id interactions, not molecular mass/size.  C 9 Intermolecular forces between gaseous oxygen molecules will be greatest when the molecules are closest together. This occurs at high pressure and low temperature, when the given amount of oxygen molecules occupy the smallest volume.  A 10 R cannot conduct electricity when solid or liquid  cannot be metallic or ionic R is a liquid at room temperature  melting point is below room temperature R is soluble in water  cannot be giant molecular  D 11  A H2O is acting as an acid as it loses H+ to form OH−.  B This is not an acid-base reaction as there is no transfer of H+.  C HCl is acting as an acid as it loses H+ to OH to form H2O. ✓ D NH3 is acting as a base as it gains H+ to form NH4+.  D 12 Alkene X undergoes reduction to form Y. Hence, Y is an alkane. With an Mr of 58, Y is C4H10, and X is C4H8. The product formed when X reacts with liquid bromine (in CCl4) in the dark is C4H8Br2. Mr of C4H8Br2 = 4(12.0) + 8(1.0) + 2(79.9) = 215.8  C 13 Overall equation: 2MnO4 + 5SO2 + 2H2O → 2Mn2+ + 5SO42