Topic 5: Structure and Shape

Transcription

1 Topic 5: Structure and Shape Lewis structures Lewis structures are a means of determining stable electron arrangements in molecules. It considers the valence electrons of an atom only. A stable arrangement is one in which each atom has achieved a noble gas electron configuration by distribution of the electrons as bond pairs or lone pairs (non- bonded pairs). A noble gas electron configuration is 2 for hydrogen and 8 for, N, O and F. This is sometimes called The Octet Rule. To draw a Lewis Structure you need to know which atoms are bonded to which. Then: 1. Add up the total number of valence electrons present, add or subtract electrons to account for any charge. 2. Join the appropriate atoms using an electron pair for each bond. 3. Distribute the remaining electrons to result in an octet of electrons on each atom (except hydrogen that always has two electrons associated with it). 4. If there are too few electrons to give every atom an octet, move non- bonded pairs between atoms to give multiple bonds. 5. If there are electrons left over after forming octets, place them on the central atom. 6. Indicate the overall charge. Example: Draw the Lewis structure of N 3. Total number of valence electrons: 5 (N) () = 8 N is at the centre, three N- bonds require 3 2 electrons: Electrons remaining = 8 (valence) 6 (bonding) = 2; place lone pair on nitrogen: Nitrogen has octet : 3 2 electrons (in bonds) + 2 electrons (lone pair). Example: Draw the Lewis structure of O Total number of valence electrons: 6 (O) + 1 () + 1 (charge) = 8 One O- bond requires 2 electrons: 6 electrons remaining; three lone pairs: Oxygen has octet : 2 electrons (in bond) + 6 electrons (lone pairs). Indicate charge outside brackets. Example: Draw the Lewis structure of O 2 Total number of valence electrons: 2x6 (O) = 12 One O- O bond requires 2 electrons: 10 electrons remaining; 5 lone pairs: One O atoms has only 6 electrons so share electrons from the other atom to form a double bond. Now both O atoms are associated with an octet of electrons: 40

2 1. Draw the Lewis structures of: 4, N 4 +, 2 O and N 2 The octet rule predicts all stable compounds of, N, O & F (, N, O & are the principle components of organic molecules which represent the vast majority of known compounds). In contrast compounds of Be and B may have less than 8 electrons and are referred to as electron deficient. Example: Draw the Lewis structures of Bel 2 Total number of valence electrons: 2 (Be) + 2x7 (l) = 16 Two Be- l bonds requires 4 electrons: 12 electrons remaining; three lone pairs on each l: While Be does not have an octet further sharing of electrons with l to form double bonds is not found as Be is recognised as electron deficient. 2. Draw the Lewis structure of BF 3 41

3 Elements in the third period and beyond have available d orbitals as well as s & p orbitals and so may form stable compounds with greater than 8 electrons surrounding these atoms. Typically these are observed in compounds of l, Br, I, P and S where stable bonding arrangements containing 8, 10 or 12 electrons around these atoms are observed. Example: Draw the Lewis structure of lf 3 Total number of valence electrons: 7 (l) + 3x7 (F) = 28 Three l- F bonds requires 6 electrons: 22 electrons remaining; three lone pairs on each F and one on l gives each atom an octet: There are still 2 electrons remaining. These form an additional lone pair on the l. While F can only have a maximum of 8 electrons (valence shell is the n = 2 shell which can accommodate only 8 electrons), l may have more than 8 electrons (valence shell is the n = 3 shell which can accommodate 18 electrons though in practice we observe bonding with 8, 10 or 12 electron arrangements). 3. Draw the Lewis structures of Pl 3, Pl 5, SF 6 and Il 4-42

4 Where more than one structure may be drawn, resonance occurs where the actual structure is a weighted combination of all possible structures and the electrons are delocalised in the molecule/ion. The Lewis structures are drawn connected with a double headed arrow. Example: Draw the Lewis structure of O 3 2 Total number of valence electrons: 4 () + 3x6 (O) + 2 (charge) = 24 Three - O bonds requires 6 electrons: 18 electrons remaining; try three lone pairs on each O: does not have an octet so share a pair from one of the O atoms: Equally likely to share a lone pair from any of the three O atoms so there are three resonance structures: When resonance occurs The actual structure is an average of all possible Lewis structures. The actual structure is a single structure. The actual structure is more stable than predicted from the Lewis structure. There is no easy way of drawing the actual structure! 4. Draw the Lewis structure of O 3, O 2 - and NO 3-43

5 Molecular Geometry The valence shell electron pair repulsion model (VSEPR model) assumes that electron pairs repel one another. This produces a set of geometries which depend only on the number of valence shell electron pairs and not on the atoms present. To determine the molecular geometry: 1. Draw the Lewis structure. 2. ount the number of electron areas (that is bond pairs and lone pairs but count each multiple bond as one electron area). 3. Arrange electron areas to minimise repulsion. This is the electron pair distribution. 4. Position the atoms to minimise the lone pair - lone pair repulsion if > 1 lone pair. 5. Name the geometry from the atom positions. This is the molecular geometry and may be different from the electron pair distribution. 6. For large molecules repeat the process for each atom that is attached to at least two other atoms to build up a picture of the overall shape. Blackman Figure 5.13, 5.16,

6 No of Electron Pairs (Lewis Structure) In summary: Arrangement of Electron Pairs No of σ Bond Pairs No of Lone Pairs Molecular geometry Examples Linear Bel 2, O 2, N 3 X 3 X Trigonal planer Angular Bl 3, SO 3, O 3 2- SO 2, O 3, NO Tetrahedral 4, N 4 +, PO X Trigonal pyramid Angular 3 O +, N 3, XeO 3 2 O, N - - 2, lo Trigonal bipyramidal Pl "See- saw" SF 4, PBr 4-5 X 3 2 T- shaped lf 3, XeF Linear Il 2 -, XeF Octahedral SF 6, SiF 6 2-, AsF 6-6 X 5 1 Square pyramidal IF 5, SF 5 -, SbF Square planar Il 4 -, XeF 4 Example: What is the molecular geometry of O 2? Lewis structure is: Two areas of electrons about the central atom, no lone pairs, therefore linear. Example: What is the molecular geometry of O 3 2? Lewis structure shows resonance but examining any one of the resonance structures will give the correct answer: Three areas of electrons about the central atom, no lone pairs, therefore triganol planar. Example: What is the molecular geometry of lf 3? Lewis structure is: Five areas of electrons about the central atom, so electron pair distribution is trigonal bipyramidal. There are two lone pairs which results in a molecular geometry of T- shaped. 45

7 5. Draw the Lewis structures and determine the molecular geometry of the following species: 4 3 N 4 + N 3 N 2 3 O + 2 O 6. Draw the Lewis structures and determine the molecular geometry for the following molecules: a. F 4 b. Asl 5 c. Se 6 7. For the following molecules, draw the Lewis structure(s) and predict the molecular shape. a. PBr 4 b. I 3 46

8 Dipole Moments Any bond between two different atoms will be polar as a result of the electronegativity difference between the atoms. A molecule has a permanent dipole moment if it contains polar bonds and is not a symmetrical shape. Examples of polar molecules: l O O N l l l Examples of non- polar molecules: O O l 8. omplete the following table. Molecule Lewis structure Molecular geometry Is there a dipole moment? l l l F F F S F F F l SO 2 SO 3 SF 4 IF 5 SF 6 47

9 Intra- & Inter-molecular forces We have examined intramolecular forces those that act within a molecule. There are the forces responsible for covalent bonds and, more loosely, the ionic bonds and metallic bonds that hold compounds together. They are STRONG (typically kj mol - 1 ). When you break and form these bonds you are also performing a chemical reaction. There are also non- bonding forces that generally occur between molecules and so are termed inter- molecular forces. These are largely WEAK (typically 0-50 kj mol - 1 ) and when you make and break these you are mainly performing physical changes eg boiling a liquid. There are a range of intermolecular forces: Ion- Dipole - dipoles align (partially or fully) in the electric field of an ion - typical energy kj mol - 1 Dipole- dipole - dipoles align (partially or fully) in the electric field of the neighboring dipole. - typical energy 5-25 kj mol - 1 Ion- induced dipole - a dipole is induced in a polarizable molecule by the presence of an ion. - typical energy: 3-15 kj mol - 1 Dipole induced dipole - a dipole is induced in a polarizable molecule by the presence of another dipole. - typical energy: 2-10 kj mol - 1 Dispersion or London forces - two polarizable molecules induce transient dipole moments in each other. - typical energy: 0.05 upwards kj mol - 1. Dispersion forces exist between ALL molecules. The force increases in strength with molecular mass. Forces associated with permanent dipoles are found only in substances with overall dipole moments (polar molecules). Their existence adds to the dispersion forces. When comparing substances of widely different masses, dispersion forces are usually more significant than dipolar forces. When comparing substances of similar molecular mass, dipole forces can produce significant differences in molecular properties (e.g. boiling point). 9. Match the five forces above to the following situations: 1. Liquid chloroform (l 3 ). 2. Petrol (a mixture of non- polar hydrocarbons). 3. Oxygen dissolved in water. 4. A solution of sodium chloride. 5. Oxygen interacting with the iron of haemoglobin. 48

10 ydrogen bonds ydrogen bonding arises from an unusually strong and directed dipole- dipole force. When is bonded to a very electronegative element (F, O, N) the bond is polar covalent. is unusual because with only one electron, it leaves a partially exposed nucleus ( has no other core electrons to shield the nucleus). The bond can be thought of as forming between the hydrogen atom and the lone pairs of the F, N, or O in F, N 3 and 2 O, respectively. Blackman Figure 6.31 Water is perhaps the most unusual liquid. Each water molecule is - bonded to FOUR other water molecules (donating 2 - atoms and accepting two - atoms to the lone pairs), forming a tetrahedral network (ice) and a loose tetrahedral network (liquid). Measuring Intermolecular Forces The melting transition from solid to liquid is not a good way of measuring intermolecular forces, as the molecules are held together by attractive forces in both phases. In order to get an idea of the strength of the interaction, we need to separate the molecules. This requires a transition to a gas, i.e. vaporisation or sublimation. A measure of this is the molar enthalpy of vaporisation. An indirect measure is the boiling point. 10. Explain the trend in the table in terms of the type and size of IM Forces present. Substance Δ vap (kj mol - 1 ) Molar mass Intermolecular forces present Butane, 3 ( 2 ) Diethyl ether, ( 2 5 )- O- ( 2 5 ) Methanol, 3 O Ethanol, 3 2 O Water, 2 O

11 11. Which of the following molecules can be expected to show hydrogen- bonding with another molecule of the same substance? Draw the - bonding interactions for those that will O 3 F O O N 2 O (glycine an amino acid) 12. ircle the molecule in the following pairs that is likely to have the higher boiling point and identify the intermolecular forces present in that molecule. O 2 and SO 2 l 4 and l 3 3 O 3 and 3 2 O Ar and Xe 2 and F 2 50

12 Phase changes The normal sequence of phases we expect to see on warming is Solid Liquid Gas The phase change (eg solid to liquid) occurs at a sharply defined temperature for most materials. owever, particularly for covalent materials that possess a rod like structure, an intermediate phase displaying some characteristics of both solids and liquids may occur. These are liquid crystals. 13. In the pictures below, each ellipse represents a molecule. Label the pictures that correspond to (a) the solid phase and (b) the liquid phase. 14. In smectic and nematic liquid crystals, the molecules are orientated to be roughly parallel. In smectic liquid crystals, the molecules are also roughly arranged in layers. There are two types of smectic liquid crystal. Labels the pictures that correspond to (a) smectic and (b) nematic liquid crystals. 15. In the smectic A phase, the average orientation of the molecule is normal to the layer. In the smectic phase, the average orientation of the molecule is at an angle to the layer. Labels the pictures that correspond to (a) smectic A and (b) smectic liquid crystals. 16. Summarise the phase transitions by redrawing the pictures in the table below. Solid Smectic Smectic A Nematic Liquid Regular crystal Molecules still Molecules still in lattice in three in layers but layers but dimensions orientate at an orientated on angle to the average normal layer to the layer Examples of liquid crystal materials are: Preferential orientation along the long axis of the molecule but no layers discernible Random orientation of molecules within the liquid N S O O 51