EJC Nov 2018 H1 Chemistry 8873 Paper 1 Worked Solutions

Eunoia Junior College 8873 H1 Chemistry 2018 Paper 1 Worked Solutions 1 Species No. of electrons No. of protons 1 1 H 1 1 1 1 H − 2 1 1 1 H + 0 1 2 1 H 1 1 2 1 H − 2 1 2 1 H + 0 1 All the particles have the same number of protons as isotopes only differ in the number of neutrons. The smallest particle will be the one with the most number of electrons. Hence, the order must be cation < atom < anion.  C 2 The four electrons of highest energy for an atom of one of the Group 14 elements are the valence electrons. Hence, the electronic configuration will be in the form ns2 np2.  B 3 If shielding increases as electrons are removed, successive ionisation energies should decrease, not increase. From the data illustrated, successive ionisation energies increase. This means that less energy is required to remove the electrons in the outer shells (which are removed earlier) as compared to those in the inner shells (which are removed later).  C 4 The structure of C2H2 is Hence, there are 3  bonds (2 single bonds and 1 in the triple bond) and 2  bonds (in the triple bond).  D 5 1 ✓ The electron-deficient Al atom accepts a pair of electrons from a chlorine atom to form the dimer. 2 ✓ The shape around the aluminium atom changes from trigonal planar to tetrahedral – bond angles decreased from 120o to 109.5o. 3  The shape around the aluminium atom changes from trigonal planar to tetrahedral  B 6 Species Shape Polarity BF3 Trigonal planar Non- polar PCl3 Trigonal pyramidal Polar (due to lone pair) CCl4 Tetrahedral Non- polar SF6 Octahedral Non- polar  B 7 HBr has a higher boiling point than HCl. HBr has more electrons than HCl, and hence a larger and more polarisable electron cloud. This leads to stronger instantaneous dipole-induced dipole forces between HBr. Note: HCl is a more polar molecule as compared to HBr. Hence, permanent dipole-permanent dipole forces between HCl should be stronger i.e. option D is wrong.  C 8 1.00 g of X = 0.009434 mol of X Amount of Y formed = 0.009434 mol Mass of Y formed = 1.57 g  D 9 A buffer comprises a mixture of a weak acid/base and its conjugate base/acid. Option D involves the addition of an excess of CH3CO2H, which is a weak acid. This will produce a mixture of CH3CO2 Na+ and CH3CO2H, which forms a buffer.  D 10 In the same period, anions will have larger radii as compared to cations as anions have one more filled quantum shell. Anionic radii decrease across a period. Hence, Sb3 will have the largest radius.  C 11 Option A is false as the H3PO4 formed from the reaction of PCl5 with water will not form a white precipitate with NaOH(aq).  A 12 1  Bond energy decreases down the group as the size of the halogen atom increases. With a larger halogen atom, the orbital overlap between hydrogen and the halogen atoms is less effective, leading to a we