1. Molecular Geometry Chapter 9 AP Chemistry Chapter 9 AP Chemistry

2. VSEPR  Valence Shell Electron Pair Repulsions  Electrons are negatively charged, so each pair will repel other pairs such that they spread out in 3-D space to minimize the repulsions.  Valence Shell Electron Pair Repulsions  Electrons are negatively charged, so each pair will repel other pairs such that they spread out in 3-D space to minimize the repulsions.

3. Electron Domains  Domains are regions about an atom’s shell where electrons are concentrated.  This is easier to see with a Lewis diagram.  For example, the carbon atom above has electrons on two sides (even though they are double bonds). So this carbon atom has 2 domains.

4. How many domains does the central atom have in…

5. C has 4, N has 4 & O has 3

6. Geometry  The shapes that molecules take, and thus the angles between bonds, depends on the number of domains.  2 domains need to be 180 o apart to minimize repulsions.  3 Domains need to be 120 o apart.  2 & 3 domains can remain 2-D. Any more domains and it must be 3-D.

7. # of domainsArrangementDomain GeometryBond Angles 2linear180 3trigonal planar120 4Tetrahedral109.5 5 Trigonal bipyramidal 120 & 90 6octahedral90

8. However,  The shape may not match the domain geometry.  Why?  The shape may not match the domain geometry.  Why?

9. Domain Geometry vs Molecular Geometry  In the Lewis Structure of water, we see 4 domains. Yet when we look at a water molecule, we can only see the bonds, not the nonbonding pairs.  Look back at the angles.  4 domains should have an angle of 109.5.  The water molecule is 104.5.  These angles are too close to be coincidence.

10. Linear Domain Geometry  There are 2 domains  There are Zero nonbonding domains.  The Molecular Geometry is linear  Example:  There are 2 domains  There are Zero nonbonding domains.  The Molecular Geometry is linear  Example:

11. Trigonal Planar Domain Geometry option 1  There are 3 domains  If there is Zero nonbonding domains, then The Molecular Geometry is trigonal planar  Example:  There are 3 domains  If there is Zero nonbonding domains, then The Molecular Geometry is trigonal planar  Example:

12. Trigonal Planar Domain Geometry option 2  There are 3 domains  If there is 1 nonbonding domain, then The Molecular Geometry is bent  Example:  There are 3 domains  If there is 1 nonbonding domain, then The Molecular Geometry is bent  Example:

13. Tetrahedral Domain Geometry option 1  There are 4 domains  If there is Zero nonbonding domains, then The Molecular Geometry is tetrahedral  Example:  There are 4 domains  If there is Zero nonbonding domains, then The Molecular Geometry is tetrahedral  Example:

14. Tetrahedral Domain Geometry option 2  There are 4 domains  If there is 1 nonbonding domain, then The Molecular Geometry is trigonal pyramidal  Example:  There are 4 domains  If there is 1 nonbonding domain, then The Molecular Geometry is trigonal pyramidal  Example:

15. Tetrahedral Domain Geometry option 3  There are 4 domains  If there are 2 nonbonding domains, then The Molecular Geometry is bent  Example:  There are 4 domains  If there are 2 nonbonding domains, then The Molecular Geometry is bent  Example:

16. Trigonal Bipyramidal Domain Geometry option 1  There are 5 domains  If there are zero nonbonding domains, then The Molecular Geometry is trigonal bipyramidal  Example:  There are 5 domains  If there are zero nonbonding domains, then The Molecular Geometry is trigonal bipyramidal  Example:

17. Trigonal Bipyramidal Domain Geometry option 2  There are 5 domains  If there is 1 nonbonding domain, then The Molecular Geometry is SeeSaw  Example:  There are 5 domains  If there is 1 nonbonding domain, then The Molecular Geometry is SeeSaw  Example:

18. Trigonal Bipyramidal Domain Geometry option 3  There are 5 domains  If there are 2 nonbonding domains, then The Molecular Geometry is T-Shaped  Example:  There are 5 domains  If there are 2 nonbonding domains, then The Molecular Geometry is T-Shaped  Example:

19. Trigonal Bipyramidal Domain Geometry option 4  There are 5 domains  If there are 3 nonbonding domains, then The Molecular Geometry is linear  Example:  There are 5 domains  If there are 3 nonbonding domains, then The Molecular Geometry is linear  Example:

20. Octahedral Domain Geometry option 1  There are 6 domains  If there are zero nonbonding domains, then The Molecular Geometry is octahedral  Example:  There are 6 domains  If there are zero nonbonding domains, then The Molecular Geometry is octahedral  Example:

21. Octahedral Domain Geometry option 2  There are 6 domains  If there is 1 nonbonding domain, then The Molecular Geometry is square pyramidal  Example:  There are 6 domains  If there is 1 nonbonding domain, then The Molecular Geometry is square pyramidal  Example:

22. Octahedral Domain Geometry option 3  There are 6 domains  If there are 2 nonbonding domains, then The Molecular Geometry is square planar  Example:  There are 6 domains  If there are 2 nonbonding domains, then The Molecular Geometry is square planar  Example:

23. What is the Domain Geometry and the Molecular Geometry of:  CO 2  CH 4  XeF 4  H 2 CO  CO 2  CH 4  XeF 4  H 2 CO  H 2 O  XeF 2  PCl 5  ICl 5  H 2 O  XeF 2  PCl 5  ICl 5

24. Domain Geometry Molecular Geometry CO 2 linear CH 4 tetrahedral XeF 4 octahedralSquare planar H 2 COTrigonal planar H2OH2Otetrahedralbent XeF 2 Trigonal bipyramidal linear PCl 5 Trigonal bipyramidal ICl 5 octahedralSquare pyramidal

25. A thought Question  The Electron Dot Structure of Carbon shows four unpaired electrons, but the Orbital Notation only shows 2. Why?  *  *C*  *  Will carbon make 2 bonds, or 4?  The Electron Dot Structure of Carbon shows four unpaired electrons, but the Orbital Notation only shows 2. Why?  *  *C*  *  Will carbon make 2 bonds, or 4?

26. Hybridization  Bonding usually involves s-orbitals. For the s-orbital of carbon to bond, one of the electrons has to go somewhere.  That somewhere is the empty p orbital. In order to make 4 bonds, the carbon will combine its s-orbital with its 3 p-orbitals into a new set of 4 orbitals all of equal energy.  This new set is called a hybrid and is referred to as an sp 3 hybrid.  Bonding usually involves s-orbitals. For the s-orbital of carbon to bond, one of the electrons has to go somewhere.  That somewhere is the empty p orbital. In order to make 4 bonds, the carbon will combine its s-orbital with its 3 p-orbitals into a new set of 4 orbitals all of equal energy.  This new set is called a hybrid and is referred to as an sp 3 hybrid.

27. The SP 3 Hybrid  On the left are regular p- orbitals and s-- orbital.  On the right are the 4 hybrized sp 3 -orbitals.

28. More Hybrids  When there are 2 domains, there is an SP hybrid.  When there are 3 domains, there is an SP 2 hybrid.  When there are 4 domains, there is an SP 3 hybrid.  When there are 5 domains, there is an SP 3 D hybrid.  When there are 6 domains, there is an SP 3 D 2 hybrid.  When there are 2 domains, there is an SP hybrid.  When there are 3 domains, there is an SP 2 hybrid.  When there are 4 domains, there is an SP 3 hybrid.  When there are 5 domains, there is an SP 3 D hybrid.  When there are 6 domains, there is an SP 3 D 2 hybrid.

29. What is the hybridization of the central atom in:  CO 2  CH 4  XeF 4  H 2 CO  CO 2  CH 4  XeF 4  H 2 CO  H 2 O  XeF 2  PCl 5  ICl 5

30. the hybridization of the central atoms are:  CO 2 = SP  CH 4 = SP 3  XeF 4 = SP 3 D 2  H 2 CO = SP 2  CO 2 = SP  CH 4 = SP 3  XeF 4 = SP 3 D 2  H 2 CO = SP 2  H 2 O = SP 3  XeF 2 = SP 3 D  PCl 5 = SP 3 D  ICl 5 = SP 3 D 2

31. Bonds  Earlier, we stated that bonding usually involves an s-orbital. How does that happen?  When 2 s-orbitals overlap, the electro- static forces of attraction of the nucleus of one atom will attract the electrons of the other atom and vice versa, forming a bond.  If two s-orbitals directly overlap then the bond formed is linear between the 2 nuclear centers & is called a sigma (  ) bond.  Earlier, we stated that bonding usually involves an s-orbital. How does that happen?  When 2 s-orbitals overlap, the electro- static forces of attraction of the nucleus of one atom will attract the electrons of the other atom and vice versa, forming a bond.  If two s-orbitals directly overlap then the bond formed is linear between the 2 nuclear centers & is called a sigma (  ) bond.

32. Sigma Bond  While this is a depiction of a sigma bond, a sigma bond is not always formed between two s- orbitals.

34. Double Bonds  Let’s examine a C 2 H 4 molecule.  From the Lewis Structure, we expect a double bond. We can also see that carbon has 3 domains, so we expect SP 2 hybridization.

35.  SP 2 hybridized orbitals bond each carbon atom (and hydrogen atoms) along the axis connecting the atoms, forming  bonds.  Since SP 2 uses 3 orbitals, we see that there is an unhybridized P-orbital.  As the  bond forms, the atoms move closer and the p-orbitals of the 2 carbons merge into a 2nd bond called a pi (  ) bond.  The top and bottom portion are both part of the same  bond.  SP 2 hybridized orbitals bond each carbon atom (and hydrogen atoms) along the axis connecting the atoms, forming  bonds.  Since SP 2 uses 3 orbitals, we see that there is an unhybridized P-orbital.  As the  bond forms, the atoms move closer and the p-orbitals of the 2 carbons merge into a 2nd bond called a pi (  ) bond.  The top and bottom portion are both part of the same  bond.

36. Triple Bonds  Let’s examine a C 2 H 2 molecule.  From the Lewis Structure, we expect a triple bond. We can also see that carbon has 2 domains, so we expect SP hybridization.

37.  SP hybridized orbitals bond each carbon atom (and hydrogen atoms) along the axis connecting the atoms, forming  bonds.  Since SP uses 2 orbitals, there must be 2 unhybridized P-orbitals.  As the  bond forms, the atoms move closer and the p-orbitals of the 2 carbons merge into 2 pi (  ) bonds.  SP hybridized orbitals bond each carbon atom (and hydrogen atoms) along the axis connecting the atoms, forming  bonds.  Since SP uses 2 orbitals, there must be 2 unhybridized P-orbitals.  As the  bond forms, the atoms move closer and the p-orbitals of the 2 carbons merge into 2 pi (  ) bonds.

38. Can you figure out…  How many pi bonds and how many sigma bonds are present (in total) in the molecule below?

39. Remember  A single bond consist of 1 sigma bond.  A double bond consist of 1 sigma bond and 1 pi bond.  A triple bond consist of 1 sigma bond and 2 pi bonds.  So the answer to the last question is 11 sigma bonds and 1 pi bond.  A single bond consist of 1 sigma bond.  A double bond consist of 1 sigma bond and 1 pi bond.  A triple bond consist of 1 sigma bond and 2 pi bonds.  So the answer to the last question is 11 sigma bonds and 1 pi bond.