The Relation of Bond Order, Bond Length and Bond Energy Bond Bond Order Average Bond Average Bond Length (pm) Energy (kJ/mol) C O C O C O C C C C C C N N N N N N Table 9.4

VSEPR: Valence Shell Electron Pair Repulsion: A way to predict the shapes of molecules Pairs of valence electrons want to get as far away from each other as possible in 3-dimensional space.

Balloon Analogy for the Mutual Repulsion of Electron Groups TwoThree Four Five Six Number of Electron Groups

AX 2 Geometry - Linear Cl Be BeCl 2 Gaseous beryllium chloride is an example of a molecule in which the central atom - Be does not have an octet of electrons, and is electron deficient. Other alkaline earth elements also have the same valence electron configuration, and the same geometry for molecules of this type. Therefore this geometry is common to group II elements. Molecular Geometry = Linear Arrangement COO Carbon dioxide also has the same geometry, and is a linear molecule, but in this case, the bonds between the carbon and oxygens are double bonds. CO 2

The Two Molecular Shapes of the Trigonal Planar Electron-Group Arrangement

AX 3 Geometry - Trigonal Planar BF 3 B F FF.. N O OO NO 3 - Boron Trifluoride Nitrate Anion All of the boron Family(IIIA) elements have the same geometry. Trigonal Planar ! AX 2 E SO 2.. S O O The AX 2 E molecules have a pair of Electrons where the third atom would appear in the space around the central atom, in the trigonal planar geometry. -

The Three Molecular Shapes of the Tetrahedral Electron-Group Arrangement

AX 4 Geometry - Tetrahedral C H HH H CH 4 Methane C H H H H All molecules or ions with four electron groups around a central atom adopt the tetrahedral arrangement H H N H H H HH+H N H all angles are the same! Ammonia is in a tetrahedral shape, but it has only an electron pair in one location, so the smaller angle! Ammonium Ion

The Four Molecular Shapes of the Trigonal Bipyramidal Electron- Group Arrangement

AX 5 Geometry - Trigonal Bipyramidal Br F F F AX 3 E 2 - BrF3 I I I AX 2 E 3 - I 3 - P Cl AX 5 - PCl 5

The Three Molecular Shapes of the Octahedral Electron-Group Arrangement

AX 6 Geometry - Octahedral S F F F F F F.. AX 6 Sulfur Hexafluoride I F FF F F.. AX 5 E Iodine Pentafluoride Xe F FF F.. Xenon TetrafluorideSquare planar shape

Using VSEPR Theory to Determine Molecular Shape 1) Write the Lewis structure from the molecular formula to see the relative placement of atoms and the number of electron groups. 2) Assign an electron-group arrangement by counting all electron groups around the central atom, bonding plus nonbonding. 3) Predict the ideal bond angle from the electron-group arrangement and the direction of any deviation caused by the lone pairs or double bonds. 4) Draw and name the molecular shape by counting bonding groups and non-bonding groups separately.

Hybrid Orbital Model

The sp Hybrid Orbitals in Gaseous BeCl 2

The sp 3 Hybrid Orbitals in NH 3 and H 2 O

The sp 3 d Hybrid Orbitals in PCl 5

The sp 3 d 2 Hybrid Orbitals in SF 6 Sulfur Hexafluoride -- SF 6

Figure 10.26: Sigma and pi bonds.

Figure 10.27: Bonding in ethylene.

Figure 10.28: Bonding in acetylene.

Restricted Rotation of  -Bonded Molecules A) Cis - 1,2 dichloroethylene B) trans - 1,2 dichloroethylene

Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH 3 NH 2 b) Xenon tetrafluoride, XeF 4 Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms, from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized.

Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH 3 NH 2 b) Xenon tetrafluoride, XeF 4 Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms, from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized. Solution: a) For CH 3 NH 2 : The shape is tetrahedral around the C and N atoms. Therefore, each central atom is sp 3 hybridized. The carbon atom has four half-filled sp 3 orbitals: Isolated Carbon Atom 2s2psp 3 Hybridized Carbon Atom

The N atom has three half-filled sp 3 orbitals and one filled with a lone pair. 2s 2p2p sp 3 C H H H H H N..

b) The Xenon atom has filled 5 s and 5 p orbitals with the 5 d orbitals empty. 5 s5 p 5 d Hybridized Xe atom: 5 d Isolated Xe atom sp 3 d 2

b) continued:For XeF 4. for Xenon, normally it has a full octet of electrons,which would mean an octahedral geometry, so to make the compound, two pairs must be broken up, and bonds made to the four fluorine atoms. If the two lone pairs are on the equatorial positions, they will be at 90 0 to each other, whereas if the two polar positions are chosen, the two electron groups will be from each other. Thereby minimizing the repulsion between the two electron groups. Xe F F F F.. Xe F FF F Square planar 180 0