Exercise practice: Chemical Bonding and Molecular Structure

5.1 How the exercises are grouped

The chapter prints exercises 4.1 to 4.40. The exercise practice page groups them by skill and works a representative question in each group.

5.2 Group A: Lewis structures and formal charge

Exercise 4.2 asks for Lewis symbols. Magnesium has two valence electrons, sodium one, fluorine seven, oxygen six, nitrogen five and bromine seven. The dots are arranged around the symbol, and the number of dots equals the group number for the main group elements.

Exercise 4.4 asks for Lewis structures of H2S, SiCl4, BeF2, the carbonate ion and formic acid. In each case count the valence electrons, connect the atoms with single bonds, and place the remaining electrons as lone pairs. The carbonate ion has 24 valence electrons. With three C-O bonds and three lone pairs on each oxygen, the central carbon is one octet short, so one C-O bond becomes a double bond and the other two remain single. The structure with the double bond on any one oxygen is equally valid, which is the resonance of exercise 4.10.

Exercise 4.10 asks why the carbonate ion is best represented by resonance. The three C-O bonds are experimentally identical, so no single structure with one double bond and two single bonds can be correct. The real ion is a hybrid of three equivalent structures, and each bond has a bond order of 1.33.

5.3 Group B: octet rule and bond parameters

Exercise 4.5 defines the octet rule and lists its limitations. The rule works for the s- and p-block compounds that reach eight electrons, but it fails for molecules with an odd electron count such as nitric oxide, and it cannot explain the stability of molecules such as sulfur hexafluoride where the central atom carries twelve electrons.

Exercise 4.6 lists the factors that favour ionic bond formation: low ionization enthalpy of the metal, high negative electron gain enthalpy of the non-metal, and a large lattice enthalpy for the resulting solid. A large lattice enthalpy is what makes the solid stable even though ion formation costs energy.

Exercise 4.8 asks how bond strength is expressed in terms of bond order. A higher bond order means more electron pairs between the atoms, which pulls the nuclei closer and needs more energy to break. Bond order and bond dissociation enthalpy therefore rise together while bond length falls.

Exercise 4.9 asks what bond length means. It is the equilibrium separation of the two nuclei in a bond. The value can be measured from microwave spectroscopy or from X-ray and electron diffraction, and it is approximately the sum of the two covalent radii.

5.4 Group C: shapes, hybridisation and related bond properties

Exercise 4.15 asks about the shapes of BeCl2, BCl3, SiCl4, AsF5, H2S and PH3. Beryllium chloride has two domains and is linear. Boron trichloride is trigonal planar. Silicon tetrachloride is tetrahedral. Arsenic pentafluoride is trigonal bipyramidal. Hydrogen sulphide and phosphine each have one or two lone pairs on the central atom, giving bent and trigonal pyramidal shapes respectively.

Exercise 4.16 discusses why the bond angle in water is smaller than in ammonia. Both have a tetrahedral electron geometry, but oxygen carries two lone pairs while nitrogen carries one. Lone pairs repel more strongly than bond pairs, so water's angle is squeezed to about 104.5 degrees while ammonia's is about 107 degrees.

Exercise 4.20 asks why methane is not square planar. The four bond pairs repel each other equally, and the arrangement that keeps them farthest apart is the tetrahedron, with angles of 109.5 degrees. A square planar arrangement would put neighbouring bond pairs at 90 degrees, which is a higher energy arrangement.

Exercise 4.21 explains why BeH2 has zero dipole moment even though each Be-H bond is polar. The molecule is linear, so the two identical bond moments point in opposite directions and cancel exactly.

Exercise 4.22 compares the dipole moments of NH3 and NF3. The N-H bonds point with the hydrogen ends positive, and the lone pair adds a moment on the same side as the hydrogens, so the moments reinforce. In NF3 the N-F bonds point with the fluorine ends negative, and the lone pair moment opposes them, so the net moment is much smaller. Ammonia has the larger dipole moment even though fluorine is more electronegative than hydrogen.

Exercise 4.26 discusses the change in hybridisation when AlCl3 dimerises. In the monomer each aluminium has three domains and is sp2. In the dimer each aluminium gains a fourth bond from a chlorine lone pair, giving four domains and sp3 hybridisation.

Exercise 4.29 asks for the number of sigma and pi bonds. Ethyne, C2H2, has three sigma bonds and two pi bonds. Ethene, C2H4, has five sigma bonds and one pi bond, counting the four C-H bonds and the C-C sigma bond.

Exercise 4.31 asks which hybrid orbitals the central atoms use. The carbon in HCN is sp, the carbon in H2CO is sp2, the carbon in CH3Br is sp3, the carbon in CH3OH is sp3, and the carbon in CH3COOH is sp3 for the methyl carbon and sp2 for the carboxyl carbon.

Exercise 4.39 asks why the axial bonds in PCl5 are longer than the equatorial bonds. The three equatorial positions are at 90 degrees to the two axial positions, so each axial bond pair feels more repulsion than an equatorial one and is stretched slightly.

5.5 Group D: molecular orbitals, bond order and magnetism

Exercise 4.36 asks why Be2 does not exist. Beryllium has two valence electrons, giving Be2 a configuration with both electrons in the bonding sigma 2s orbital and two in the antibonding sigma* 2s orbital. The bond order is (2 - 2)/2 = 0, so the molecule is not stable.

Exercise 4.37 compares the stability and magnetism of superoxide, peroxide and oxygen. Bond orders are 1.5 for O2-, 1.0 for O2^2- and 2.0 for O2, so stability follows O2 greater than O2- greater than O2^2-. O2 has two unpaired electrons and O2- has one, so both are paramagnetic; O2^2- has none and is diamagnetic.

Exercise 4.40 asks for bond orders. Nitrogen has bond order 3, oxygen 2, O2+ 2.5 and O2- 1.5. This ordering matches the observed bond lengths, which shorten as the bond order rises.