JPJC 2026 Chemical Bonding Revision Answer

© Jurong Pioneer Junior College 1 Topic 3: Chemical Bonding JURONG PIONEER JUNIOR COLLEGE Topical Revision for Physical Chemistry (Topic 3: Chemical Bonding) FOUNDATION KIT Question 1 2 3 4 5 6 7 8 9 10 Answer A C C C A B D A D A Question 11 12 13 14 15 Answer D C D C C 1 ANS: (A) A AlCl3 and SO3 ✓ B AlCl3 and PCl3  C BF3 and NH3  D CO2 and SO2  2 ANS: (C) C 3 ANS: (C) Al Cl Cl Cl B F F F Al Cl Cl Cl S O O O Cl Cl P Cl H H N H C O O S O O  3bp, 1 lp around N  trigonal pyramidal 3bp, 0 lp around B  trigonal planar 4bp, 0 lp around N  tetrahedral 4bp, 0 lp around B  tetrahedral        2 Total 6, 3

© Jurong Pioneer Junior College 2 Topic 3: Chemical Bonding 4 ANS: (C) 5 ANS: (A) C C CH2 H H H S O CH2 C NH2 H CO2H y z x 6 ANS: (B) H−N=N−H 7 ANS: (D) 1✓ Each aluminium atom is surrounded by four chlorine atoms. 2 There are twelve non-bonded electron pairs in the molecule. 3 Each aluminium atom contributes electrons to four covalent bonds. Each Al only contributes 3 valence electron to form 3 covalent bonds with Cl. The last covalent bond is a dative bond where both electrons are contributed from Cl. H H H O H C O C H α β γ N  3bp, 1lp around N  trigonal pyramidal  α  107 3bp, 0lp around C  trigonal planar  β = 120 2bp, 2lp around O  bent  γ  104.5 Bond angle β > α > γ  3bp, 0lp around C  trigonal planar  x = 120 3bp, 1lp around S  trigonal pyramidal  y  107 110 4bp, 0lp around C  tetrahedral  x = 109.5  110   2bp, 1lp around each N  bent  angle < 120  angle > 109.5 (tetrahedral angle)                 16 lone pairs (non-bonded e pairs)

© Jurong Pioneer Junior College 3 Topic 3: Chemical Bonding 8 ANS: (A) A Al2Cl6 is a planar molecule. B✓ N2O4 is formed through the dimerisation of two NO2 molecules via the formation of a N−N covalent bond. C✓ Each aluminium atom is surrounded by four chlorine atoms in Al2Cl6. D✓ Hydrogen bonds hold the CH3CO2H molecules together in the dimer. 9 ANS: (D) 1✓ The C−C−C bond angle between nearest neighbours is smaller in diamond than in graphite. In diamond, every C atom is tetrahedral  bond angle = 109.5 In graphite, every C atom is trigonal planar  bond angle = 120 2 The shortest carbon−carbon bond occurs in diamond. C atom in graphite is sp2 hydridised. C atom in diamond is sp3 hybridised. C-C bond in graphite involves sp2-sp2 overlap which is stronger than the sp3-sp3 overlap as there is greater s character in sp2 which leads to a shorter and stronger C-C bond. 3 All covalent bonds in diamond are of the same strength but those in graphite are not. C-C bonds in graphite are of the same length and strength A 1, 2 and 3 are correct B 1 and 2 are correct C 2 and 3 are correct D 1 only is correct N O O N O O N O O 2 4bp, 0lp around each Al  tetrahedral around each Al (not planar) 4bp, 0lp around each Al  tetrahedral around each Al (not planar)

© Jurong Pioneer Junior College 4 Topic 3: Chemical Bonding 10 ANS: (A) Silicon carbide is a covalent compound. Since it has a high m.p., the structure should be a giant covalent structure. A✓ a giant structure with covalent bonds between silicon and carbon atoms. B a giant structure containing metallic bonding. Silicon carbide has a giant covalent structure C a giant layer structure with covalent bonds between atoms and instantaneous dipole -induced dipole forces between the layers. This description is about graphite D a simple molecular structure with covalent bonds between atoms of silicon and carbon. Silicon carbide has a giant structure 11 ANS: (D) The millions of microscopic hairs with thousands of pads at its tip ensures a large surface area of contact where extensive instantaneous dipole -induced dipole attractions between the gecko’s feet and the glass surface enabl es the gecko to hang on to the glass surface while climbing. The attraction should not be too strong to prevent the gecko’s feet to be permanently bonded to the glass surface. A co−ordinate bonds (bond too strong) B ionic bonds (bond too strong) C covalent bonds (bond too strong) D✓ instantaneous dipole−induced dipole force 12 ANS: (C) CH3CH2CO2H has hydrogen bonding between its molecules. CH3CH2CH2Cl has permanent dipole-permanent dipole (pd-pd) attraction between its molecules. (CH3)3CH and CH 3CH2CH2CH3 have instantaneous dipole -induced dipole (id -id) attraction between its respective molecules. (CH3)3CH is more spherical and has a smaller surface area of contact. Less energy is required to overcome the less extensive id -id attraction between (CH3)3CH molecules than CH3CH2CH2CH3 molecules. Since strength of hydrogen bonding > pd-pd attraction > id-id attraction Boiling point of CH3CH2CO2H > CH3CH2CH2Cl > CH3CH2CH2CH3 > (CH3)3CH 3 4 2 1

© Jurong Pioneer Junior College 5 Topic 3: Chemical Bonding 13 ANS: (D) CO2 has a simple molecular structure. SiO2 has a giant covalent structure. CO2 is a gas as it has low b.p. only a small amount of energy is required to overcome the weak id-id attraction between CO2 molecules. SiO2 is a solid as it has high m.p. A lot of energy is required to overcome the strong covalent bonds between Si and O atoms. A carbon-oxygen bonds are less polar than silicon-oxygen bonds. For CO2, id-id attraction is broken not C=O covalent bond. B carbon dioxide contains double covalent bonds while silicon dioxide contains single covalent bonds. For CO2, id-id attraction is broken not C=O covalent bond. C the relative molecular mass of carbon dioxide is less than that of silicon dioxide. Both compounds have different structures. Mr cannot explain the difference in m.p./b.p. D✓ instantaneous dipole-induced dipole attractions are much weaker than covalent bonds. 14 ANS: (C) Both H2O and HF have simple molecular structures. H-F bond is more polar than H-O bond. So, the hydrogen bonding between HF molecules should be stronger than the hydrogen bonding between H2O molecules. But H2O has a higher b.p. as it can form more extensive hydrogen bonding between H2O molecules then the hydrogen bonding between HF molecules. H-F H2O 1 H atom 2 H atoms 3 lp on F 2 lp on O A Each hydrogen bond formed between water molecules is stronger than that formed between HF molecules. Hydrogen bonding between HF molecules should be stronger B There are more atoms in a water molecule than there are in an HF molecule, so instantaneous dipole induced dipole attractions are stronger in water. Strength of id-id attraction is dependent on number of electrons in the molecule. Both HF and H2O have the same number of electrons so id -id attraction strength should be similar. The difference in b.p. between HF and H 2O is not attributed to id-id attraction but the different extensiveness of hydrogen bonding between the respective molecules. C✓ There are, on average, more hydrogen bonds between water molecules than there are between HF molecules. H2O can form on average 2 hydrogen bonds per molecule and HF can form on average 1 hydrogen bond per molecule. D The water molecules have a greater permanent dipole than the HF molecule. H-F bond is more polar than H -O bond. The permanent dipole on HF should be greater than that in H2O.

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