CHEM1101 Worksheet 6: Lone Pairs and Molecular Geometry
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1 CHEM1101 Worksheet 6: Lone Pairs and Molecular Geometry Model 1: Oxidation numbers Oxidation numbers are a useful accountancy tool to help keep track of electrons in compounds and reactions. This is particularly important in redox reactions where some atoms lose (are oxidised) and others gain (are reduced) electrons. It is the positive or negative charge the atom would have if the molecule was completely ionic. Rules for working out oxidation numbers The rules should be used in this order the higher the rule, the higher its priority. 1. An atom in its elemental form (e.g. Fe, Cl 2, graphite etc) has oxidation number = 0 2. The sum of the oxidation number of all the atoms in a molecule equals zero. 3. The sum of the oxidation number of all the atoms in an ion equals the charge of the ion. 4. The oxidation number of fluorine is -1 (except in F 2 where it is 0 [rule 1]). 5. The oxidation number of group 1 elements is The oxidation number of group 2 elements is 2 7. The oxidation number of oxygen is -2, except peroxides where it is The oxidation number of halogens is usually The oxidation number of hydrogen is 1 when bonded to non-metals and -1 when bonded to metals. (a) ClF 3 F has oxidation number = -1 (rule 4) Cl must have oxidation number = 3 so that overall charge is zero: (3) (3-1) = 0 (b) ClF 4 F has oxidation number = -1 (rule 4) Ion has charge of 1. Cl must have oxidation number = 5 so that overall charge is 1: (5) (4-1) = 1 (c) ClF 4 - F has oxidation number = -1 (rule 4) Ion has charge of -1. Cl must have oxidation number = 3 so that overall charge is -1: (3) (4-1) = Work out the oxidation number of the N atom in the molecules and ions below: System Oxidation number System Oxidation number HNO 3 N 2 O NO 3 N 2 NO 2 NH 2 OH N 2 O 4 N 2 H 4 HNO 2 NH 3 NO 2 NH 2 NO NH 4
2 Model 2: Lone Pairs Lone pairs are pairs of electrons in a molecule which do not participate in the bonding. As described in Model 3, they play a key role in determining molecular shape and are also the source of the reactivity of many molecules. For molecules and ions of the type n in which an atom is surrounded by n atoms, the oxidation number of tells us how many of its electrons are used in bonding. The number of lone pairs, m, can then be easily worked out by subtracting this from the number of valence electrons it possesses: number of lone pairs = m = ½ (number of valence electrons oxidation number ) * For an s-block element, the number of valence electrons is equal to the group number. For a p-block element, it is equal to the group number The number of valence electrons gives the total number of electrons that an atom has available. The oxidation number reflects the number used to make bonds. The difference between them gives the number of unused electrons: this is halved to give the number of unused or lone pairs. Cl and F are both in group 17 so both have (17 10) = 7 valence electrons. (a) In ClF 3, F has an oxidation number of -1 and Cl has an oxidation number of 3: number of lone pairs on Cl = ½ (7 3) = 2 (b) In ClF 4, F has an oxidation number of -1 and Cl has an oxidation number of 5: number of lone pairs on Cl = ½ (7 5) = 1 (c) In ClF 4, F has an oxidation number of -1 and Cl has an oxidation number of 3: number of lone pairs on Cl = ½ (7 3) = 2 2. How many lone pairs are there on the underlined atom in the molecules and ions below? System Number of Lone Pairs System Number of Lone Pairs SF 2 ef 2 SF 4 ef 4 SF 6 ef 3 SOF 4 eo 2 F 2 ** Note that the absolute value of the oxidation number is used: the or sign is dropped.
3 Model 3: Molecular Shape The Valence Shell Electron Pair Repulsion (VSEPR) model assumes that the because of the repulsion between electrons, the bonds and lone pairs surrounding an atom try to get as far from each other as possible. For molecules and ions of the type n in which the central atom makes n bonds and has m lone pairs, the shape is dictated by the total number of bonds and lone pairs: n m. The 3D-arrangements of bonds and lone pair which maximize the distance between these bonds and lone pairs for common values of (n m) are shown in the table below. n m Arrangement n m Arrangement 2 linear 180 o 5 bipyramid 120 o 90 o 3 planar 6 octahedral 90 o 120 o 4 tetrahedral o 7 pentagonal bipyramid 72 o 90 o (a) (b) There are 4H attached to the C atom in CH 4 : n = 4. The C atom has an oxidation number of -4 so no lone pairs; m = 0. With n m = 4 0 = 4, the arrangement is tetrahedral. There are 2O attached to the C atom in CO 2 : n = 2. The C atom has an oxidation number of 4 no lone pairs: m = 0. With n m = 2 0 = 2, the arrangement is linear. 3. Complete the table below. The central element is underlined. System n m n m Arrangement BeF 2 BCl 3 BCl 4 PF 5 SF 6 IF 7
4 Model 4: Positioning the Lone Pairs The molecular shape or geometry is the arrangement of the bonds. If lone pairs are present, they help to dictate what this arrangement is but are not included when the geometry of the molecule is described. The table below shows the geometries possible when n m = 4-6. n m m = 0 m = 1 m = 2 m = 3 3 planar bent 4 tetrahedral pyramidal bent or V shaped 5 bipyramidal see saw T shaped Linear 6 octahedral square based pyramid square planar T shaped Note that when n = 5, the lone pairs occupy the sites in the equatorial plane. 4. Sketch and describe the molecular geometries of the following systems. (a) H 2 O (b) eo 3 (c) ClF 3 (d) ClF 4 (e) ClF 4 (f) ef 2 (g) ef 4 (h) eo 2 F 2
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