chem lecture 12

chem lecture 12 - 1st Year THE ^ CHEMISTRY STUDENT...

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Chemistry 6A 1 st Year THE ^ CHEMISTRY STUDENT EXPERIENCE Strategies for Success and Engagement Wednesday, November 12, 2008 7:00 PM – 9:00 PM 1205 Natural Science Building (Main Auditorium) Making the Transition from High School or College Study Methods: What’s Effective (and What’s Not!)? Test-taking Strategies The Big Picture 1
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Chemistry 6A 2 Which molecules is most polar? (a) SF 6 (b) CF 4 (c) NF 3 (d) PF 5 (e) BF 3
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Chemistry 6A 3 Appreciable overlap of two 1 s orbitals On approach, 1 s orbitals begin to overlap and e- experience attractive force of both protons From a distance, no interaction; each e- in H atom is associated with single proton σ -bond formation along internuclear axis
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Chemistry 6A How many sigma and how many pi-bonds are in melamine? (a) 15 σ , 0 π (b) 15 σ , 3 π (c) 12 σ , 3 π (d) 9 σ , 0 π (e) None of the above 4
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Chemistry 6A 5 When is it necessary to invoke the use of ‘hybridized’ atomic orbitals? How do Molecular Orbital theory yield deeper insights into the nature of the (covalent) chemical bond?
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Chemistry 6A 6
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Chemistry 6A 7 Typically, atomic orbitals do not provide the required geometry, so hybrid orbitals are constructed on an atom (via mathematical mixing) to reproduce the electronic arrangement that will yield the experimental shape of a molecule. Step: 1. Write Lewis structure for molecule 2. Deduce geometry from VSEPR 3. Deduce hybridization based on electron-group geometry The algebraic addition and subtraction of the wavefunctions describing two atomic orbitals produces a new hybrid orbital . Keep in mind: – The energy and number of orbitals is conserved. If you mix 2 atomic orbitals, you produce two hybrid orbitals! – The amplitude of the hybrid orbital is affected by the phase of the atomic orbitals from which it is derived.
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Chemistry 6A 8 # of hybrid orbitals on central atom must be consistent with the electron arrangement predicted by VSEPR.
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Chemistry 6A 9 Example (cont’d): 4 H (1 s 1 ) and 1 C (1 s 2 2 s 2 2 p 2 ) to form CH 4 We can generate 4 equivalent σ bonds in CH 4 by: Mixing the one 2 s and three 2 p atomic orbitals of carbon to form four sp 3 hybrid atomic orbitals of the same energy and spatial distribution .
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Chemistry 6A 10 4 hybrid atomic orbitals of the same energy and spatial distribution .
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Chemistry 6A 11 The in phase overlap of four H 1 s atomic orbitals with C’s four sp 3 hybrid atomic orbitals leads to bond formation in methane. The structure is tetrahedral, with H–C–H bond angles of ~ 109.5°
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Chemistry 6A 12 Example : 6 H (1 s 1 ) and 2 C (1 s 2 2 s 2 2 p 2 ) to form ethane, C 2 H 6 Each tetrahedral C center has four sp 3 hybrid orbitals, each containing a single electron The C-C σ -bond is formed by overlap of a sp 3 orbital on each C. Each of the six C-H σ -bonds (only one is shown here) is formed by overlap of the H 1 s orbital with a sp 3 orbital on one of the C.
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Chemistry 6A 13 Example : 3 H (1 s 1 ) and 1 N (1 s 2 2 s 2 2 p 3 ) to form ammonia, NH 3 Mix the 2 s and 2 p orbitals in N to form 4 sp 3 hybrid orbitals, which are occupied by five valence electrons: One
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This note was uploaded on 08/07/2009 for the course CHEM 6A taught by Professor Pomeroy during the Spring '08 term at UCSD.

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chem lecture 12 - 1st Year THE ^ CHEMISTRY STUDENT...

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