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CHEM 1030 Week 11 Notes

by: Alyssa Anderson

CHEM 1030 Week 11 Notes CHEM 1030

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Alyssa Anderson

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These notes cover the material we went over in class 3/29/16 and 3/31/16.
Fundamentals Chemistry I
Dr. Streit
Class Notes
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This 3 page Class Notes was uploaded by Alyssa Anderson on Wednesday April 6, 2016. The Class Notes belongs to CHEM 1030 at a university taught by Dr. Streit in Spring 2016. Since its upload, it has received 13 views.

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Date Created: 04/06/16
Chemistry Notes Week 11 Hybridization of Atomic Orbitals A. Hybridization or mixing of atomic orbitals can account for observed bond angles in molecules that could not be described by the direct overlap of atomic orbitals B. Example: BeCl2. 1. Lewis structure and VSEPR theory predict Cl-Be-Cl (linear geometry) with a bond angle of 180* 2. Beryllium cannot form two bonds because it has a full 2s2 set 3. The two bonds formed would not be equivalent because it would become 2s1/2p1 4. Experimentally, the bonds are identical in length and strength. 5. Mixing one s orbital and one p orbital to yield two sp orbitals 6. The hybrid orbitals on Be each overlap with a singly occupied 3p orbital on a Cl atom 7. The mathematical combination of the quantum mechanical wave functions for an s orbital and a p orbital has one small lobe and one large lobe, and like any two electron domains on an atom, they re oriented in opposite directions with a 180* angle between them. 8. With two sp hybrid orbitals, each containing an unpaired electron, we can see how the Be atom is able to form two identical bonds with two Cl atoms. 9. Each of the singly occupied sp hybrid orbitals on the Be atom overlaps with the singly occupied 3p atomic orbitals on the Cl atom. C. Example: BF3. 1. Draw the Lewis diagram and count the number of electron domains. 2. Combine the necessary number of atomic orbitals to generate the required number of hybrid orbitals 3. Place electrons in the hybrid orbitals, putting one election in each orbital before pairing any electrons 4. The ground-state electron configuration of the B atom is [He] 2s2/2p1, containing just ONE unpaired electrons needed to explain the formation of three bonds 5. The ground-state and excited-state electron configurations can be represented as 2s2/2p1 —>(promoted to) 2s1/2p2 6. Because the three ones in BF3 are identical, we must hybridize the three singly occupied atomic orbitals (the one s and 2 p orbitals) to give the singly occupied hybrid orbitals: 2s1/2p2 —> (hybridization) sp2/sp2/sp2 7. Mixing of one s orbital and two p orbitals to yield three sp orbitals D. Example: CH4. 1. Draw the ground state orbital diagram for the central atom 2. Combine the necessary number of atomic orbitals, putting one election in each orbital before pairing any electrons. 3. Promote electrons from 2s2/2p2 —> 2s1/2p3 4. Hybridize the carbon atom from 2s1/2p3 —> sp3/sp3/sp3/sp3 5. Mixing of one s orbital and three p orbitals to yield four spy orbitals 6. The set of four spy orbitals on carbon, like any four electron domains on a central atom, assumes a tetrahedral arrangement 3. Rather than 2s1, 2p3, make it 2s4. (four electrons in the s block) E. NOTE 1. Recall that elements in the third period of the periodic table and beyond do not necessarily obey the octet rule because they have d orbitals that can hold additional electrons 2. In order to explain the bonding and geometry of molecules in which there are more than four electron domains, on the central atom, we must include d orbitals in our hybridization scheme F.Example: PCl5. 1. Draw the ground state electron orbital diagram for the central atom 2. Maximize the number of unpaired electrons by promotion 3. Hybrid orbitals on phosphorus overlap with the 3p orbitals on chlorine 4. Promote from 3s2/3p3 —> 3s1/3p3/3d1 5. Hybridize Hybridization In Molecules Containing Multiple Bonds A. The remaining unhybridized p orbital is perpendicular to the plane in which the atoms of the molecule lie B. Pi bonds restrict free rotation around the bond axis C. The acetylene molecule is linear with sp hybridized carbons D. Promotion of an electron maximizes the number of unpaired electrons E. Formation of the C-C pi bond in acetylene 1. 2 pi bonds 2. 2 between the two carbon atoms F. Worked Example 7.8 Molecular Orbital Theory A. The atomic orbitals involved in bonding actually combine to form new orbitals that are the “property” of the entire molecu;e, rather than of the individual atoms forming the bonds, called molecular orbitals B. In molecular orbital theory electrons shared by atoms in a molecule reside in the molecular orbitals


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