W2013CHM2311 Part 4b Notes

The bonding interac0on of carbon 2s with oxygen 2pz

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Unformatted text preview: for them to interact. However, H (1s) and F (2p) can interact according to energy requirements! Which F (2p) orbitals have the appropriate symmetry to interact with H (1s)? Step 3: Look at the symmetries of the AOs: Orbitals must have similar symmetry about the internuclear axis: No interac0on between px and s or py and s. Interac0on between pz and s is allowed. z z px s pz s C2 C∞ C∞ C∞ MO Diagram for HF σ* = c1ϕ1s – c2ϕ2pz energy 1s n.b. n.b. σ = c1ϕ2pz + c2ϕ1s n.b. H HF 2p 2s F MO Diagram for HF Due to energy separaBon, F (2s) cannot interact with H (1s). Due to inappropriate symmetry, F (2px), F (2py) cannot interact with H (1s). H (1s) combines with F (2pz) The bonding MO has greater F 2pz character because its energy is closer to the F 2pz The an0bonding MO has greater H 1s character because its energy is closer to H 1s. The F (2s), F (2px), and F (2py) orbitals become nonbonding molecular orbitals. They are drawn at the same energy level as the original AO’s and they keep the same shapes. MO Diagram for HF Characteris0cs of HF as determined by the MO Diagram: 1.  The fluorine atom on HF has 3 non- bonding pairs of electrons – these are lone pairs. 2. The bonding electrons are in a molecular orbital for which the fluorine 2pz orbital makes the predominant contribu0on. Thus, these electrons will be closer to F than to H, giving F par0al nega0ve character...
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This note was uploaded on 03/28/2014 for the course CHM 2311 taught by Professor Richardson during the Winter '09 term at University of Ottawa.

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