Molecular Orbital Theory: Build MOs for Diatomic Molecules

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SUMMARY

This discussion focuses on Molecular Orbital (MO) Theory, specifically the energy ordering of molecular orbitals in diatomic molecules such as B2, C2, N2, O2, and F2. It is established that the 2p∏ orbitals of B2, C2, and N2 are higher in energy than those of O2 and F2 due to increased s-p splitting and the effects of orbital repulsion. The bonding p-sigma orbitals are shifted above the bonding pi orbitals in the case of B, C, and N, while in O2 and F2, the distance between orbitals prevents this repulsion from altering their energy order. The absolute energies of these orbitals may not be accurately represented in MO diagrams.

PREREQUISITES
  • Understanding of Molecular Orbital Theory
  • Familiarity with diatomic molecules and their electronic configurations
  • Knowledge of orbital types: sigma and pi orbitals
  • Concept of s-p splitting in molecular bonding
NEXT STEPS
  • Study the differences in MO diagrams for diatomic molecules using the ChemWiki resource
  • Investigate the concept of s-p splitting and its implications in molecular bonding
  • Explore the role of orbital repulsion in determining molecular orbital energy levels
  • Learn about the specific electronic configurations of B2, C2, N2, O2, and F2
USEFUL FOR

Chemistry students, educators, and researchers interested in theoretical chemistry and molecular bonding, particularly those focusing on the intricacies of Molecular Orbital Theory.

MathewsMD
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http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Molecular_Orbital_Theory/How_to_Build_Molecular_Orbitals


Hi,

I'm currently trying to understand MO theory and how diagrams are made and interpreted. I stumbled across this website and it shows the general MO diagrams for different diatomic molecules.

It also shows the 2p∏ of B2, C2 and N2 as being higher in energy than compared to O2 and F2. I was trying to reason it out and considered that since there will be more ∏ antibonding orbitals for the O2 and F2 that this is the reason why. But aren't their structural constraints to this reasoning? Aren't ∏ orbitals of a specific energy due to their perpendicular orientation and their sideways overlapping? Why exactly is their energy lower in the MO diagrams? I'm assuming my explanation for this is incorrect so so any clarification on the above matters would be great!

Thank you! :)
 
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These pictures are only correct as far as the ordering of the sigma and pi orbitals are concerned however, the absolute energies of these orbitals is probably incorrect.
The point is that the s-p splitting increases in a period (that is from B to F, in our case).
The anti-bonding sigma formed from the s-orbitals and the bonding sigma from the p orbitals repell, that is why the bonding p-sigma is shifted above the bonding pi orbitals in case of B, C and N. In O2 and F2, the orbitals are too far appart, so that this repulsion won't change the order of the sigma and pi orbitals derived from p.
 
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DrDu said:
These pictures are only correct as far as the ordering of the sigma and pi orbitals are concerned however, the absolute energies of these orbitals is probably incorrect.
The point is that the s-p splitting increases in a period (that is from B to F, in our case).
The anti-bonding sigma formed from the s-orbitals and the bonding sigma from the p orbitals repell, that is why the bonding p-sigma is shifted above the bonding pi orbitals in case of B, C and N. In O2 and F2, the orbitals are too far appart, so that this repulsion won't change the order of the sigma and pi orbitals derived from p.

That was a great explanation. Thank you!
 

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