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CHEMICAL BONDING II MOLECULAR ORBITAL THEORY II

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C 2 + MOLECULE

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NO MOLECULE

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BOND ORDER Bond order = number of bonding e- minus number of antibonding e- 2 If the bond order is zero→ no bond! divide by two because of pairs of electrons. Bond order is an indication of strength. Larger bond order means greater bond strength. H 2 molecule: Bond order = (2-0)/2 = 1 and He 2 : Bond order = (2-2)/2 = 0 This implies it is not stable.

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BOND ORDER Calculate the Bond order for F 2, C 2 +, and NO

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BOND ORDER F:(10-8)/2 = 1 C : (7-4)/2 = 1.5 NO:(10-5)/2 = 2.5

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HOMONUCLEAR DIATOMIC MOLECULES those composed of two identical atoms. Li 2 Bond order = (2-0)/2 = 1. Li 2 is a stable molecule. Be 2 Bond order = (2-2)/2 = 0. This is not more stable than two Be atoms, so no molecule forms.

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FILLING THE DIAGRAM It gets slightly more complicated when we leave Be and move to 2p. The filling order for p’s is pi, pi, sigma all bonding, followed by pi, pi, sigma all antibonding. Hund's Rule and the Pauli Exclusion Principle still apply.

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General Energy Level Sequence for Filling Orbitals Using the MO Theory σ1s 2 σ1s 2 * σ2s 2 σ2s 2 * Π2pxy 4 σ2p 2 π2pxy 4 * σ2p 2 *

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MOLECULAR ORBITAL CONFIGURATIONS For period 2 diatomic molecules up to and including N 2 : σ1s 2 σ1s 2 * σ2s 2 σ2s 2 * π2pxy 4 σ2p 2 π2pxy 4* σ2p 2 * For period 2 diatomic molecules O 2, F 2, and Ne 2 (hypothetical): π2p and σ2p change order. σ1s 2 σ1s 2 * σ2s 2 σ2s 2 * σ2p 2 π2pxy 4 π2pxy 4* σ2p 2 *

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MOLECULAR ORBITAL CONFIGURATIONS Write the molecular orbital configurations for F 2, C 2 +, and NO.

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MOLECULAR ORBITAL CONFIGURATIONS F 2 : σ1s 2 σ1s 2 * σ2s 2 σ2s 2 * σ2p 2 π2pxy 4 π2pxy 4 * C 2 + : σ1s 2 σ1s 2 * σ2s 2 σ2s 2 * σ2p 2 π2pxy 3 NO: σ1s 2 σ1s 2 * σ2s 2 σ2s 2 * σ2p 2 π2pxy 4 π2pxy 1 *

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PRACTICE SIX Predict the bonding of B 2.

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PRACTICE SIX - ANSWER B – 1s 2 2s 2 2p 1 B B

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PARAMAGNETISM One of the most useful parts of this model is its ability to accurately predict paramagnetism and diamagnetism as well as bond order. B 2 is paramagnetic. That means that the pi orbitals are of LOWER energy than the sigmas and Hund’s rule demands that the 2 electrons fill the 2 bonding pi orbitals singly first before paring.

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PRACTICE SEVEN Write the appropriate energy diagram using the MO theory for the nitrogen molecule. Find the bond order for the molecule and indicate whether this substance is paramagnetic or diamagnetic.

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PRACTICE SEVEN N 2 electron configuration: 1s 2 2s 2 2p 3 Bond order = (10-4)/2 = 6 diamagnetic N N

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PRACTICE EIGHT For the species O 2, O 2 +, O 2 -, give the electron configuration and the bond order for each. Which has the strongest bond?

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PRACTICE EIGHT O 2 σ1s 2 σ1s 2 * σ2s 2 σ2s 2 * σ2p 2 π2pxy 4 π2pxy 2 * Bond order = (8-4)/2 = 2

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PRACTICE EIGHT

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O 2 + σ1s 2 σ1s 2 * σ2s 2 σ2s 2 * σ2p 2 π2pxy 4 π2pxy 1 * Bond order = (8-3)/2 = 2.5 O 2 - σ1s 2 σ1s 2 * σ2s 2 σ2s 2 * σ2p 2 π2pxy 4 π2pxy 3 * Bond order = (8-5)/2 = 1.5

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PRACTICE NINE Use the molecular orbital model to predict the bond order and magnetism of each of the following molecules. Ne 2 P 2

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PRACTICE NINE Ne 2 Bond order = (8 - 8)/2 = 0 Does not exist

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PRACTICE NINE P 2 Bond order = ( )/2 = 3 diamagnetic

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PRACTICE TEN Use the MO Model to predict the magnetism and bond order of the NO + and CN - ions.

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PRACTICE TEN Each has the same number of valence electrons = 10 and total electrons = 14 σ1s 2 σ1s 2 * σ2s 2 σ2s 2 * π2pxy 4 σ2p 2 Bond order = (10-4)/2 = 3 Both are diamagnetic.

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The Molecular Orbtial Energy-Level Diagrams, Bond Orders, Bond Energies, and Bond Lengths for the Diatomic Molecules B 2 Through F 2

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COMBINING LE AND MO MODELS LE model assumes that electrons are localized. This is not the case for some molecules. This is apparent with molecules for which we can draw several valid Lewis structures. So resonance was constructed. The best model is one with the simplicity of the LE model with the delocalization characteristics of the MO model. So we combine these two models to describe molecules that require resonance. In O 3 and NO 3 -1, the double bond changes position in the resonance structure.

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COMBINING LE AND MO MODELS Since a double bond requires one σ and one π bond, there is a σ bond between all bound atoms in each resonance structure. It is the π bond that has different locations. Conclusion: σ bonds in a molecule can be described as being localized with no apparent problems, but the π bonding must be treated as delocalized. So use LE to describe the σ bonds in structures with resonance, but use MO to describe π bonds in these same structures.

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