RI 2025 H2 Chemical Bonding I Tutorial Answers

10 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 > lone pair – bond pair repulsion > bond 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+ O H H H 1070 (c) BeF2 ∙ 2NH3 adduct N Be N H H H H H H F F 109o 109o 109o 109o 109o 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 – unpai red electron repulsion in NO 2. Thus, the bond angle in NO2 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 > bond pair – bond pair repulsion. For O 3, the lone pair of electrons on the central O exert greater r epulsion than the bond pairs of electrons in O-O  x O OO xxx x     x  x N OO xxx x    

11 bond. Thus, the bond angle in O 3 would be less than 120 o, 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) NO 2 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 surrou ndings. The energy of the products is less than that of the reactants, hence the reaction is feasible. (Please note that there are other considerations that will determine the feasibility of reactions. They will be discussed in Lecture 5 – Energetics.) (d) NO2+ NO 2– 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 p p head-on overlap of p orbitals (show head-on overlap of either s/p orbitals)  (pi) bond p p side-to-side overlap of p-orbitals Note: it is necessary to label the orbitals and state the type of overlap N N O O O O O N O O N O +

12 (b)(i) Electronegativity is the relative ability of an atom in a molecule to attract bonding/shared electrons. (ii) Phosgene molecules have intermolecular instantaneous dipole-induced dipole (id-id) and permanent dipole-permanent dipole (pd-pd) interactions. Electrons are constantly moving and at 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 dipoles in their structures. Pd-pd interactions arise due to the electrostatic attraction between the partial positive end of one phosgene molecule and the partial negative end of the other phosgene molecule. (c) 13 (a) . . 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) (b) (c) 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 atoms/ steric factors between the F atoms. 14. (a) H 2O, NH 3 and HF all have intermolecul ar hydrogen bonding. However, H 2O can form on average 2 hydrogen bonds per molecule whereas NH 3 and HF can only form 1 hydrogen bond per molecule. Therefore, more energy is needed to overcome the more extensive 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 between 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.

13 (b) Due to the close proximity of the –OH to the –COOH groups, 2-hydroxybenzoic acid forms intramolecular hydrogen bonding as shown in 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 bonding with water molecules compared to 4-hydroxybenzoic acid, resulting in lower solubility in water. 15. (a) Both compounds exist in giant ionic lattices. Melting involves overcoming the strong electrostatic forces of attraction between t he cations and anions. This strength of ionic attraction is approximated by the magnitude of lattice energy, |L.E.|  |(q+q) / (r+ + r)|. Since Ca 2+ and O 2 are both doubly-charged compared to the singly-charged Na + and C l, the magnitude of L.E. for CaO is greater than that of NaC l (OR L.E. of CaO is more exothermic than that of NaC l) and the ionic bond in CaO 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. PC l3 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 la rger electron cloud compared to PC l3. A larger electron cloud is more easily polarised, henc e, the intermolecular instantaneous dipole- induced dipole interactions found in SiBr 4 is significantly stronger than those found in PC l3 which has a much smaller electron cloud. This results in the higher 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.