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Chemistry 100 Chapter 9 Molecular Geometry and Bonding Theories.

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1 Chemistry 100 Chapter 9 Molecular Geometry and Bonding Theories

2 Molecular Geometry The three-dimensional arrangement of atoms in a molecule  molecular geometry Lewis structures can’t be used to predict geometry A very simple theory tells us that the repulsion between electron pairs (both bonding and non-bonding) helps account for the arrangement of atoms in molecules

3 The VSEPR Model Electrons are negatively charged, they want to occupy positions such that electron – electron interactions are minimised as much as possible Valence Shell Electron-Pair Repulsion Model  treat double and triple bonds as single bonds  resonance structure - apply VSEPR to any of them Formal charges are usually omitted

4 Molecules With More Than One Central Atom We simply apply VSEPR to each ‘central atom’ in the molecule. Carbon #1 – tetrahedral Carbon #2 – trigonal planar

5 Dipole Moments The HF molecule has a bond dipole – a charge separation due to the electronegativity difference between F and H. The shape of a molecule and the magnitude of the bond dipole(s) can give the molecule an overall degree of polarity  dipole moment.  + H-F 

6 Homonuclear diatomics  no dipole moment (O 2, F 2, Cl 2, etc) Triatomic molecules (and greater). Must look at the net effect of all the bond dipoles. In molecules like CCl 4 (tetrahedral) BF 3 (trigonal planar) all the individual bond dipoles cancel  no resultant dipole moment.

7 Bond Dipoles in Molecules

8 More Bond Dipoles

9 Valence Bond Theory and Hybridisation Valence bond theory description of the covalent bonding and structure in molecules. Electrons in a molecule occupy the atomic orbitals of individual atoms. The covalent bond results from the overlap of the atomic orbitals on the individual atoms

10 The H 2 Molecule In a hydrogen molecule, we observe a single bond indicating the overlap of the 1s orbitals on the individual atoms

11 The Bonding in H 2 Note that bond has cylindrical symmetry with respect to the line joining the atomic centres description. This is known as a  bond Overlap Region 1s (H1) – 1s(H2)  bond

12 The Bonding in H 2

13 The Cl 2 Molecule In the chlorine molecule, we observe a single bond indicating the overlap of the 3p orbitals on the individual atoms. Bonding description 3p z (Cl 1) – 3p z (Cl 2)

14 Is This a  Bond?

15 Hybrid Atomic Orbitals Look at the bonding picture in methane (CH 4 ). Bonding and geometry in polyatomic molecules may be explained in terms of the formation of hybrid atomic orbitals Bonds - overlap of the hybrid atomic orbitals with the atoms. appropriate half- filled atomic orbital on the terminal

16 The CH 4 Molecule

17 The Formation of the sp 3 Hybrids

18 sp 2 Hybridisation What if we try to rationalise the bonding picture in the BH 3 (a trigonal planar molecule)? We mix 2 “pure” p orbitals and a “pure” s orbital to form “hybrid” or mixed sp 2 orbitals. These three sp 2 hybrid orbitals lie in the same plane with an angle of 120  between them.

19 A Trigonal Planar Molecule Overlap regions Overlap region

20 sp Hybridisation What if we try to rationalise the bonding picture in the BeH 2 species (a linear molecule)? We mix a single “pure” p orbital and a “pure” s orbital to form two “hybrid” or mixed sp orbitals These sp hybrid orbitals have an angle of 180  between them.

21 A Linear Molecule The BeH 2 molecule Overlap Regions

22 Double Bonds Look at ethene C 2 H 4. Each central atom is an AB 3 system, the bonding picture must be consistent with VSEPR theory.

23 The  Bond Additional feature  an unhybridized p orbital on adjacent carbon atoms. Overlap the two parallel 2p z orbitals (a  - orbital is formed).

24 The C 2 H 4 Molecule

25 The Bond Angles in C 2 H 4 Bond angles  HCH =  HCC  120 . Note that the  bond is perpendicular to the plane containing the molecule. We can rationalize the presence of any double bond by assuming sp 2 hybridization exists on the central atoms! Any double bond  one  bond and a  bond

26 The Triple Bond Look at acetylene (ethyne) The carbon atoms each have a triple bond and a single bond.

27 The C 2 H 2 Molecule

28 On the carbon central atom, we now have 2 sp hybrid orbitals and two unhybridised p orbitals We can again overlap the 2p y orbitals and the 2p z orbitals on the C central atoms (two p- bonds are formed). The  Bonds

29 The Bond Angles in C 2 H 2 Bond angles  HCH =  HCC = 180 . The  bonds are again perpendicular to the plane containing the molecule. Triple bond  one  bond and two  bonds Rationalise the presence of any triple bond by assuming sp hybridization exists on the central atoms!

30 sp 3 d Hybridisation How can we use the hybridisation concept to explain the bonding picture PCl 5. There are five bonds between P and Cl (all  type bonds). 5 sp 3 d orbitals  these orbitals overlap with the 3p orbitals in Cl to form the 5  bonds with the required VSEPR geometry  trigonal bipyramid. Bond overlaps [sp 3 d (P ) – 3p z (Cl) ] x 5  type

31 sp 3 d 2 Hybridisation Look at the SF 6 molecule. 6 sp 3 d 2 orbitals  these orbitals overlap with the 2p z orbitals in F to form the 6  bonds with the required VSEPR geometry  octahedral. Bond overlaps [sp 3 d 2 (S ) – 2p z (F) ] x 6  type

32 Notes for Understanding Hybridisation Applied to atoms in molecules only Number hybrid orbitals = number of atomic orbitals used to make them Hybrid orbitals have different energies and shapes from the atomic orbitals from which they were made. Hybridisation requires energy for the promotion of the electron and the mixing of the orbitals  energy is offset by bond formation.

33 Delocalised Bonding In almost all the cases where we described the bonding n the molecule, the bonding electrons have been totally associated with the two atoms that form the bond  they are localised. What about the bonding situation in benzene, the nitrate ion, the carbonate ion? In benzene, the C-C  bonds are formed from the sp 2 hybrid orbitals. The unhybridised 2p z orbital on C overlaps with another 2p z orbital on the adjacent C atom.

34 Three  bonds are formed. These  bonds extend over the whole molecule (i.e. the  bonds are delocalised). The  electrons are free to move around the benzene ring. Any species where we had several resonance structures, we would have delocalisation of the  -electrons.

35 Delocalised Electrons in Molecules

36 Molecular Orbital (M.O.) Theory Valence bond and the concept of the hybridisation of atomic orbitals does not account for a number of fundamental observations of chemistry. To reconcile these and other differences, we turn to molecular orbital theory (MO theory). In MO theory, covalent bonding is described in terms of molecular orbitals, i.e., the combination of atomic orbitals that results in an orbital associated with the whole molecule.

37 Recall the wave properties of electrons. constructive interference  the two e - waves interact favourably; loosely analogous to a build-up of e - density between the two atomic centres. destructive interference  unfavourable interaction of e - waves; analogous to the decrease of e - density between two atomic centres.

38 Constructive and Destructive Interference + + Constructive Destructive

39  bonding = C 1  ls (H 1) + C 2  ls (H 2)  anti = C 1  ls (H 1) - C 2  ls (H 2) Bonding Orbital  a centro-symmetric orbital (i.e. symmetric about the line of symmetry of the bonding atoms). Bonding M’s have lower energy and greater stability than the AO’s from which it was formed. Electron density is concentrated in the region immediately between the bonding nuclei.

40 Anti-bonding orbital  a node (0 electron density) between the two nuclei. In an anti-bonding MO, we have higher energy and less stability than the atomic orbitals from which it was formed. As with valance bond theory (hybridisation) 2 AO’s  2 MO’s

41 Bonding and Anti-Bonding M.O.’s from 1s atomic Orbitals

42 The MO’s in the H 2 Atom

43 The situation for two 2s orbitals is the same! The situation for two 3s orbital is the same. Let’s look at the following series of molecules H 2, He 2 +, He 2 bond order = ½ {bonding - anti-bonding e - ‘s}. Higher bond order  greater bond stability.


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