Covalent structures and bonding
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Some exam "Dos and Don'ts": Covalent bonding
Advice on answering questions on topics often highlighted by examiners as being poorly answered in the past. Make sure that the detail covered by the question is included in your syllabus specification.
A covalent bond
Defintion: a covalent bond is formed between two atoms, usually non-metals, by the sharing of a pair of outer shell (valence) electrons.
In orbital terms this amounts to overlap by orbitals from adjacent atoms. You may need to know that 'end-on' overlap of s- or p- orbitals forms sigma bonds. 'Side-ways' overlap of p-orbitals form a pi-bond, part of a double or triple bond - make sure you can give sketches to illustrate your answer.
Note that for the non-metallic elements in Period 2 (C,N,O,F), the outer (valence shell) can contain no more than 8 (4 pairs) of electrons, limiting the number of covalent bonds they can form to 4,3,2 and 1 respectively. Elements from Period 3 onwards can accommodate more electrons in their outer shells, and form more covalent bonds than their Group No. may suggest, e.g. SF6.
A covalent bond can also be formed by one atom donating both electrons - called a dative or coordinate bond, often shown as X->Y. Note that when an atom forms a covalent bond using a lone pair of electrons, it does not expand its outer shell, although an extra bond has been formed. Thus in NH4+ nitrogen forms 4 covalent bonds, whereas N normally forms only 3 bonds.
Some examples of dative bonds you may come across are in CO (containing a C<-O dative bond!), NH4+ ions, H3O+ ions, metal ion->ligand bonding, in H3N->BF3 and, later, between ligands and metal ions in complexes.
Shapes of molecules (VSEPR theory)
Examiners often complain about the quality of answers to questions on this topic. Very poor answers are usually given to questions asking for an explanation of the theory. Remember the basic principles:
1. Pairs of electrons in the outer (valence) shell of a central atom in a molecule or ion repel each other - NOT 'atoms repel' or 'bonds repel'.
2. They take up an orientation in space around a central atom so as to minimise the repulsion between them (the lowest energy arrangement).
3. Lone pairs repel bonding pairs more than bonding pairs repel each other.
[A problem arises in how to treat multiple bonds - generally they can be regarded as one 'electron cloud', like a bonding pair of electrons, but IB (International Baccalaureate) also uses the term 'negative charge centre' for a bonding pair, a lone pair or a multiple bond.]
Remember to distinguish between the basic electron distribution (4 pairs = tetrahedral, 3 pairs = trigonal planar etc) and the final shape of the molecule or ion, taking into account how many of the pairs (neg. charge centres) are lone pairs. Thus the electron distribution in H2O is basically tetrahedral, but the shape of the molecule is angular or V-shaped, with an H-O-H angle of 105 o due to lone pair-bonding pair repulsion.
In explaining your answer always state how many bonding pairs and how many lone pairs are involved. A diagram of the molecule/ion with the important angle labelled is worth adding, even if not requested.
Polarity of molecules
Students seem to have difficulty explaining why a molecule is non- polar or polar (i.e. behaves like a small dipole) in exams. First you need a diagram showing the shape of the molecule and the individual polar bonds (bond dipoles) shown using d+ and d- symbols. Remember, C-H bonds are NOT usually regarded as polar.
Depending on the shape of the molecule, the individual bond dipoles will cancel out (assuming for the moment they are all the same) or add up to a resultant dipole for the molecule as a whole, making it polar. If there are only two bond dipoles at 180 o or three at 120 o in a plane, then it is fairly obvious to see they will cancel out, giving a non-polar molecule, e.g. CO2 & BX3.
The problems arise with tetrahedral or pyramidal shapes. Remember that in any tetrahedral arrangement of bonds, say around a C-atom, any two of the bonds are in one plane (e.g. the paper, if drawn) and the other two are in a plane perpindicular to it. When considering if the bond dipoles cancel out, consider each pair of bonds in turn to give a new dipole. These two 'new' dipoles, which will be in exactly opposite directions, will usually cancel out, giving a non-polar molecule. You really need to do this exercise with models and convince yourself (and later an examiner possibly!) that CHCl3 is polar whereas CCl4 is not.
In an exam it will probably be enough to say the because of the shape (with a clear diagram), the bond dipoles will cancel out, or not, as the case may be.
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updated: 10 May 2007
