JPJC 2022 JC1 promos P1_Worked Solution

©Jurong Pioneer Junior College 1 2022 JC1 YEE Paper 1 Worked Solutions Suggested Worked Solution for 2022 JC1 YEE Paper 1 (9729/01) 1 C 6 D 11 A 16 C 21 C 26 C 2 B 7 A 12 D 17 B 22 A 27 C 3 D 8 C 13 A 18 D 23 C 28 B 4 B 9 A 14 B 19 D 24 C 29 B 5 D 10 A 15 A 20 A 25 B 30 B 1 Answer: C The shapes of the five d−orbitals are shown below. Of the five, the first 4 have four lobes. 2 Answer: B For W, since there is a sharp increase from the 7 th to 8th IE, it implies that the 8 th electron of W is removed from the next inner quantum shell and hence, W has 7 valence ele ctrons. Thus, W is from Group 17. Since W, X, Y and Z are co nsecutive elements, i t implies th at X is from Group 18 , Y is from Group 1 of the next Period and Z is from Group 2 of the next Period. Hence, for the same nth IE, since the IE generally increases across the Period and decreases down the Group, X has the highest 1st IE. 3 Answer: D Interpret the two information as follow: • X and Y form ionic compounds Na2X and Na2Y respectively.  X and Y form anion of charge 2−. Hence, they are likely from Group 16 and has a valence electronic configuration of ns2 np4. (Reject Option A and B) • Element Y forms YF6 molecules where X is not able to do so.  Y can expand octet but not X. Hence, X is from Period 2 while Y is from Period 3 onwards. (Reject Option C) 4 Answer: B •  bond is formed from head−on overlap between orbitals (atomic orbitals and/or hybrid orbitals). (Reject Option C and D) s s s p p p •  bond is formed from sideway overlap of p−orbitals. (Reject Option A)

©Jurong Pioneer Junior College 2 2022 JC1 YEE Paper 1 Worked Solutions 5 Answer: D Molecular shape Polarity A B is from Grp 13 (3 valence e−). Around central B, there are 3 bp & 0 lp.  Shape is trigonal planar. Since there is no lp and the surrounding atoms are identical (all C l), the dipole moment from each B−Cl bond cancelled out ( i.e. BCl3 has no net dipole moment).  BCl3 is non−polar. B N is from Grp 15 (5 valence e−). Around central B, there are 3 bp & 1 lp.  Shape is trigonal pyramidal. Since there is 1 lp, the dipole moment from each N−Cl bond are not cancelled out even though the surrounding atoms are th e same (i.e. NCl3 has net dipole moment).  NCl3 is polar. C S is from Grp 16 (6 valence e−). Around central S, there are 2 bp & 1 lp.  Shape is bent. Since there is 1 lp, the dipole moment from each S=O bond are not cancelled out even though the surrounding atoms are the same (i.e. SO2 has net dipole moment).  SO2 is polar. ✓D C is from Grp 14 (4 valence e−). Around central C, there are 4 bp & 0 lp.  Shape is tetrahedral. Since there is 1 lp and the surrounding atoms are different, the dipo le moments do not cancelled out ( i.e. CHCl3 has net d ipole moment).  CHCl3 is polar. 6 Answer: D Both Al2Cl6 and AlCl3 have simple molecular/covalent structures. (Reject Option A and B). Al2Cl6 is form