Chapter 7.2-7.3 – More Hybridization, Multiple Bonds CHM1111 Section 04 Instructor: Dr. Jules Carlson Class Time: M/W/F 1:30-2:20 Wednesday, November 2nd

Organic Chemistry Nomenclature For straight chain hydrocarbons (only have C and H), name with number of carbons 1- Meth, 2- Eth, 3- Prop, 4- But, 5- Pent, etc… If there are all single bonds, end name with –ane If there is a double bond, end name with –ene If there is a triple bond, end name with –yne So first compound with a double bond we can talk about is Ethene (common name ethylene).

I Clicker Question from Monday Which of the following statements are false: Orbital overlap in H2S is between 1 s orbitals for H, and 3 sp3 orbitals in S. For Ammonium ion NH4+ , the N has 4 hybridized 2 sp3 orbitals available for bonding and no lone pairs. For Hydronium ion H3O+, oxygen has 3 hybridized 2 sp3 orbitals available for bonding and one lone pair. Both (a) and (c) are false. No statements are false.

Lone Pairs with Tetrahedral Geometry Carbon has valence electrons 2s22p2 so has 4 valence electrons, and forms hybrid orbitals with one s and three p orbitals to make sp3 orbitals. Carbon in Methane H H H H Nitrogen in Ammonia H H H lone pairs Oxygen in Water H H Even with lone pairs, for elements with a steric number of 4, will form tetrahedral centres with hybridization leading to 4 sp3 orbitals.

Tetrahedral Geometry for Cations Carbon in Methane 2s22p2 CH4 H H H H Carbon has valence electrons 2s22p2 so has 4 valence electrons, and forms hybrid orbitals with one s and three p orbitals to make sp3 orbitals. Since it is positively charged, Nitrogen (normally 2s22p3) in NH4+ has the same number of valence electrons as Carbon (2s22p2). Note unpaired electrons in orbitals can pair by bonding. + Nitrogen in Ammonium 2s22p3 – 1e- 2s22p2 H H H H NH4+ + Oxygen in Hydronium 2s22p4 – 1e- 2s22p3 H3O+ H H H Oxygen has valence electrons 2s22p4 so would have 6 valence electrons, Since it is positively charged, Oxygen now has 2s22p3 or 5 valence electrons in H3O+

Tetrahedral Geometry for Anions As we take away an electron from the central atom if cation, we add an electron if anion. - Nitrogen in Nitranion 2s22p3 + 1e- 2s22p4 lone pairs H H NH2- - Oxygen in hydroxide 2s22p4 + 1e- 2s22p5 lone pairs H OH-

sp2 Hybridization for Planar Geometry sp2 hybridization occurs for central atoms with 3 valence electrons – Group 13 Al, Ga, In, Tl Also, we will talk more about later, can get sp2 hybridization if you have central atoms with 4 valence electrons but are forming a double bond. Ethylene, has 3 sp2 orbitals, and the second bond between the carbons is a bond between p orbitals.

sp Hybridization for Linear Geometry Example: Dimethylzinc Zinc has electronic configuration 1s22s22p63s23p64s23d10 So has valence electrons 4s2 3 4 4 4 4 3

sp Hybridization for Linear Geometry sp2 hybridization occurs for central atoms with 2 valence electrons – Group 12 Zn, Cd, Hg Also, we will talk more about later, can get sp2 hybridization if you have central atoms with 4 valence electrons but with a triple bond or 2 double bonds.

Participation of d Orbitals for Higher Geometries If n > 2 then you have d orbitals that can you can break the octet rule. You can have more than 8 electrons surrounding the inner atom by including d orbitals To include d orbitals and produce equal orbitals you then have to hybridize d orbitals with the s and p orbitals.

sp3d Hybridization for Trigonal Bipyrimidal Geometry Example: PCl5 Phosphorus (n = 3) has 3s23p3 as valence electrons, also has empty 3d orbitals available. Get sp3d hybridization for atoms with 5 valence electrons if also has steric number = 5 PCl5 Cl Cl Cl Cl Cl

sp3d2 Hybridization for Octahedral Geometry Example: SF6 Sulfur (n = 3) has 3s23p4 as valence electrons, also has empty 3d orbitals available. Get sp3d2 hybridization for atoms with 6 valence electrons if also has steric number = 6 Ground State S SF6

sp3d3 Hybridization for Pentagonal Bipyrimidal Geometry Example: IF7 Iodine (n = 5) has 5s25p5 as valence electrons, also has empty 4d orbitals available Get sp3d3 hybridization for atoms with 7 valence electrons if also has steric number = 7 IF7

Hybridization Example Provide the hybridization for the inner atoms for the following compounds: (CH3)2CO, BF4-, HCCCH3, SO3, IF5

I Clicker Question Which of the following statements are true: Trigonal bipyrimidal geometry requires sp3d hybridization around the central atom. AlCl3 exhibits sp3 hybridization around Al. SCl4 exhibits sp3d hybridization around S. Both (a) and (b) Both (a) and (c)

Bonding in Ethene Ethene has 12 valence electrons (2x4 + 4x1) The bonds between C-H are single bonds – these are called sigma (σ) bonds For double bonds, the first bond is a σ bond and the second bond is a pi (π) bond. The σ bonds are sp2 hybridized, and the π bond is the with the remaining p orbital in carbon.

Orbital Overlap in Ethene Shows you the overlap of various orbitals. Different views showing the electron density of the π bond shown in blue.

Looking at a π bond Views of a π bond from four different angles. π bonds have a side-by-side overlap. p orbitals also have a node at the nucleus. Electron density is concentrated above and below the bond axis.