RI 2021 Prelim P1 Answers

1 2021 H2 Chemistry Y6 Prelim Paper 1 Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer B B D C A A C D C D B A C B D Question 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer C C A D A C C D A B A B D A B Question 1 (B) Electronic configuration of Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (not 1s2 2s2 2p6 3s2 3p6 3d9 4s2 – see Atomic Structure notes if you are unsure why) Number of electrons in d orbitals with 4 lobes = 8 (dz2 does not have 4 lobes) Question 2 (B) In Period 2, there are two irregularities in the first ionisation energies of the elements. One irregularity occurs at Group 13 (i.e. B) and the other occurs at Group 16 (i.e. O). Two irregularities are also observed in the second ionisation energies at G roup 14 (removing e – from C+) and at Group 17 (removing e– from F+). A quick way to realise is to check the Data Booklet. Therefore, K, the element with the lower 2nd IE can be either from either Group 14 or Group 17, corresponding to carbon and fluorine respectively. Question 3 (D) Q is non -volatile (does not vapourise easily ) eliminates nitrogen dioxide which has a simple covalent structure with weak instantaneous dipole- induced dipole interactions and is a gas at rtp. Q does not conduct electricity in its standard state eliminates sodium as metals can conduct electricity. Q dissolves in water eliminates silicon dioxide is insoluble in water due to its giant covalent structure. Sodium oxide is the only option that • is non -volatile (due to strong ionic bonds holding the giant ionic lattice), • does not conduct electricity in its standard state (no mobile ions as charge carriers in solid state), and • dissolves in water (by reacting with water to form NaOH(aq)). Question 4 (C) BeF2 is the simplest compound of beryllium and fluorine. F–Be–F In BeF 2, Be is sp hybridised and contain two unhybridised p-orbitals which, in this molecule, are empty. Hence Be in BeF 2 can accept 2 pairs of electrons into its two unhybridised p-orbitals. 1 Incorrect. • F donates a lone pair to Be which accepts the pair of electrons from F, thus the arrow representing the dative bond should point from F to Be. 2 Correct. • With 2 empty unhybridised p -orbitals, Be can accept 2 pairs of electrons from 2 F i.e. Be in BeF 2 forms 2 dative bonds with 2 fluorines from other BeF 2, resulting in the polymeric structure. 3 Correct. • Similarly, BeF 2 can form 2 dative bonds with two F– to give BeF42–. Question 5 (A) X, Y, and Z are Period 3 elements. Oxide of X is amphoteric ⇒ X is Al Oxide of Y is basic ⇒ Y is Na or Mg Oxide of Z is acidic ⇒ Z is non-metal (P, S, Cl)

2 Since Z is a non-metal, it forms anions. Since X and Y are metals, they form cations. Since ionic radii of cations are smaller than that of anions (formed from non -metals) and ionic radii decreases from Na+ to Al3+, ionic radii increases in this order X < Y