Predict and explain the bond lengths and bond strengths of the carbon-oxygen bonds in \({\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CO}}{{\text{O}}^ - }\).

(i)     State the meaning of the term hybridization.

(ii)     Describe the hybridization of the carbon atom in methane and explain how the concept of hybridization can be used to explain the shape of the methane molecule.

(iii)     Identify the hybridization of the carbon atoms in diamond and graphite and explain why graphite is an electrical conductor.

Aluminium chloride, \({\text{A}}{{\text{l}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{6}}}\), does not conduct electricity when molten but aluminium oxide, \({\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\), does. Explain this in terms of the structure and bonding of the two compounds.

\({\text{A}}{{\text{l}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{6}}}\):

\({\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\):

bond length and bond strength identical for both carbon to oxygen bonds;

intermediate between single and double bond length and strength;

due to delocalization of the electrons (in the p orbitals);

Accept use of Data Booklet values of bond lengths and bond enthalpies.

Accept diagram of delocalization or the two resonance structures for M3.

(i)     mixing/joining together/combining/merging of (atomic) orbitals to form molecular/new orbitals (of equal energy);

(ii)     \({\text{s}}{{\text{p}}^3}\);

isolated C atom electron configuration \({\text{1}}{{\text{s}}^2}{\text{2}}{{\text{s}}^2}{\text{2}}{{\text{p}}^2}\) / excited state C electron configuration is \({\text{1}}{{\text{s}}^2}2{{\text{s}}^{\text{1}}}{\text{2}}{{\text{p}}^{\text{3}}}\);

\({\text{2}}{{\text{s}}^{\text{1}}}{\text{2}}{{\text{p}}^{\text{3}}}\) electrons blend to form four identical hybrid orbitals;

hybrid orbitals lower in total energy than atomic orbitals;

repulsion of (identical hybrid) orbitals creates a tetrahedral shape;

Accept suitably annotated diagram for M2, M3 and M4.

(iii)     diamond:

\({\text{s}}{{\text{p}}^{\text{3}}}\);

graphite:

\({\text{s}}{{\text{p}}^2}\);

(p) electrons delocalized (around layer);

Al₂Cl₆:

covalent bonding / dimer/molecular structure;

no free charges when molten so not an electrical conductor;

Al₂O₃:

ionic / lattice structure;

ions free to move/mobile in molten state;

Question 7 was a popular one and well-answered in general. Part (a) required definitions which were not well known but most candidates determined the empirical and molecular formulas and correctly drew the structural formula of the carboxylic acid. Fewer candidates could correctly draw the structural formula of an ester. Identification of the stronger and longer carbon-oxygen bond was answered correctly by nearly all candidates, but explaining the bond lengths in the propanoate ion was only answered correctly by the very best candidates. Even those who realized that the electrons are delocalized did not give a complete explanation and often scored only 2 marks out of 3.

In part (b) many candidates struggled to define hybridization, frequently referring to overlapping of orbitals. Most could state that the carbon atom in methane is \({\text{s}}{{\text{p}}^{\text{3}}}\) hybridized and that the molecule is tetrahedral, but few gave detailed responses about electron configurations or repulsion of electron pairs. However, most candidates correctly identified the hybridization of carbon in diamond and graphite, and explained why graphite conducts electric current.

In (c) few candidates knew that \({\text{A}}{{\text{l}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{6}}}\) is a covalent compound and that \({\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\) is ionic. Some answers mentioned many types of bonding for one compound.