RI 2024 H2 Chemical Bonding I Tutorial

1 Answers to Practice Questions (Long) 10. (a) (b) H3O+ and Cl– ions formed Species Bond angle No. of e– pairs No of bond pairs No. of lone pairs Shape H2O 105 4 2 2 bent H3O+ 107 4 3 1 trigonal pyramidal Lone pair – lone pair repulsion > l one pair – bond pair repulsion > b ond pair – bond pair repulsion H2O → H3O+ One lone pair becomes a bond pair through co-ordinate bonding  No lone pair – lone pair repulsion in H3O+. this results in a greater bond angle. Dot-and-cross Structural formula H2O H3O+ (c) BeF2 ∙ 2NH3 adduct tetrahedral wrt to both Be and N in the product 11. (a) (b) Both molecules have three regions of electron density and thus their electron -pair geometry is trigonal planar. • Electron pairs exert greater repulsion than an unpaired electron. As a result, the bond pair – bond pair repulsion > bond pair – unpaired electron repulsion in NO 2. Thus, the bond angle in NO 2 would be greater than 120 o, e.g. 130 o (The actual bond angle is 134 o. Acceptable answers are any value from 120o to 170o). • Lone pair – bond pair repulsion > b ond pair – bond pair repulsion. For O3, the lone pair of electrons on the central O exert greater repulsion than the bond pairs of electrons in O-O •• x O OO xxx x• • •• •••• •• x •• x N OO xxx x• • •• •••• ••

2 bond. Thus, the bond angle in O3 would be less than 120o, e.g. 118o (The actual bond angle is 117o. Acceptable answers are from 110o to 120o). Teachers, pls use this question as a teaching point: A lone pair or a bond pair exerts greater repulsion than a single electron as the repulsion exerted by 2 electrons is greater than 1 electron. (c) NO2 has an odd number of electrons i.e . there is an unpaired electron in N. Dimerisation only involves the formation of N –N covalent bond which releases energy to the surroundings. The energy of the products is less than that of the reactants, hence the reaction is feasible. (Please note t hat there are other considerations that will determine the feasibility of reactions. They will be discussed in Lecture 5 – Energetics.) (d) NO2+ NO2– Dot-and-cross diagram No. of electron densities around N: 2 3 To minimise electrostatic repulsion, the electron–pair geometries are: linear trigonal planar Number of lone pairs 0 1 Molecular shape is linear bent 12 (a) (i) Phosgene, Cl2C=O, has 3 bond pairs and 0 lone pairs of electrons around the central C atom. To minimize electronic repulsion between the bond pairs, the shape of the phosgene molecule is trigonal planar. (ii)  (sigma) bond head-on overlap of p orbitals (show head-on overlap of either s/p orbitals)  (pi) bond side-to-side overlap of p-orbitals Note: it is necessary to label the orbitals and state the type of overlap (b)(i) Electronegativity is the relative ability of an atom in a molecule to attract bonding/shared electrons. N N O O O O O N O− O N O +

3 (ii) Phosgene molecules have intermolecular instantaneous dipole-induced dipole (id-id) and permanent dipole-permanent dipole (pd-pd) interactions. Electrons ar e constantly moving and a t any given moment, the electron density of a phosgene molecule can be unsymmetrical, resulting in an instantaneous dipole, which induces a short-lived dipole in a neighbouring phosgene molecule, hence resulting in id- id interactions. Phosgene molecules are polar with permanent dip oles in their structures. Pd-pd interactions arise due to the electrostatic attraction between the + end of one phosgene molecule and the – end of the other phosgene molecule. (c) 13 Modified from RI 2017 Year 5 Common Test Section C Q1 (a) The  bond between Cl and F is formed by the head-on (collinear/head-to-head) overlap of a 2p orbital of F and a 3p orbital of Cl. [Note that the size of the 2p orbital has to be drawn smaller than the 3p orbital.] (b) There are 3 bond pairs and 2 lone pairs around Br. To minimise repulsion, the 2 lone pairs are positioned at the equatorial positions of the trigonal bipyramid.  BrF3 is T-shaped. (c) . . Br. . x . x x x x x x F x . x x x xF x x + Br . . x x x x F x . F x x x . x . FF . . x x x x x x x x x x x x x x x x x x - x. Bond angle in BrF2+: 105 (values between 90o and 107 are accepted) (d) (e) Chlorine is a smaller atom compared to iodine, so it cannot ‘pack’ as many fluorine atoms around itself, due to repulsion of electron clouds between the F atoms / overcrowding of the F

4 atoms/ steric factors between the F atoms. 14. (a) H2O, NH3 and HF all have intermolecular hydrogen bonding . However, H2O can form on average 2 hydrogen bonds per mo lecule whereas NH 3 and HF can only form 1 hydrogen bond per mole cule. Therefore , more energy is needed to overcome the more extensiv e hydrogen bonding between water molecules, resulting in its highest boiling point. As F is more electronegative than N , the hydrogen bond between HF molecules is stronger than that betw een NH 3 due to the greater - formed on F and + on H atoms on HF molecule. More energy is therefore needed to overcome the hydrogen bonds between HF molecules, resulting in HF having a higher boiling point than NH3. (b) Due to the close proximity of the –OH to the –COOH groups, 2-hydroxybenzoic acid form s intramolecular hydrogen bonding as shown i n the diagram on the right. Thus , it has less sites available for the formation of intermolecular hydrogen bonding with water molecules. Hence 2-hydroxybenzoic acid forms less extensive intermolecular hydrogen bondi ng with water molecules compared to 4 -hydroxybenzoic acid , resulting in lower solubility in water. 15. (a) Both compounds exist in giant ionic lattices . M elting i nvolves overcoming the strong electrostatic forces of attraction between the cations an d a nions. This strength of ionic attraction is approximated by the magnitude of lattice energy, |L.E.|  |(q+q−) / (r+ + r−)|. Since Ca2+ and O2− are both doubly-charged compared to the singly -charged Na+ and Cl−, the magnitude of L .E. for CaO is greater than that of NaCl (OR L.E. of CaO is more exothermic than that of NaC l) and the ionic bond in C aO is stronger than that in NaC l. Hence CaO has a higher melting point. (b) They are covalent substances with simple molecular structures. SiCl4 is a non-polar molecule which forms weak instantaneous dipole–induced dipole interactions only. PCl3 is a polar molecule which forms slightly stronger permanent dipole–permanent dipole and instantaneous dipole–induced dipole interactions. Hence the boiling point of PCl3 is higher than SiCl4 that of since more energy is required to overcome the intermolecular forces of attraction between PCl3 molecules during boiling. SiBr4 is non -polar but it has a much larger electro n cl oud compared to PC l3. A larger electron cloud is mo re easily polarised, hence, the intermolecular instantane ous dipole - induced dipole interactions found in SiBr 4 is significantly stronger than those found in PCl3 which has a much smaller electron clou d. This results in the h igher boiling point of SiBr 4, even though SiBr4 does not have pd-pd interactions. + − − OH O C H O + −− HH OO C O or - - - - represents intramolecular H bond.