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MO bonding in F2 and O2 ­ Chemwiki
MO bonding in F2 and O2
Introduction
Molecular orbitals (MO) are constructed from atomic orbitals. They do not exist in real life, but are useful in illustrating bonding properties
of molecules. In O2 and F2, there is a crossover of the sigma and the pi ortbials: the relative energies of the sigma orbitals drop below that
of the pi orbitals'. Information from the MO diagram justify O2's stability and show that it's bonding order is 2. The LUMO (lowest
unoccupied molecular orbital) and HOMO (highest occupied molecular orbital) of difluoride's MO diagram help explain why the molecule
is very stable ­ the diagram also tells us that the bond order is 1.
1. Types of Interaction
The formation of F2 and O2 involves 2s­2s and 2p­2p orbital interactions. S­orbitals are independent of the angular part of the
wavefunction; so, they, as well as the 2pz orbital, show spherically symmetry about the nucleus. The in­phase and out­of­phase
combinations of the orbitals lead to the organization of the sigma g and sigma star u labels. However, the 2px and 2py atomic orbitals
combine to form pi and pi­star molecular orbitals because they are dependent on the angular part of the wavefunction.
2. MO Diagram:
omics.html
To construct a MO for a diatomic molecule, you must first write down the atomic orbitals of each atom and organize
them by energy levels. For instance, for O2 you write down all the atomic orbitals (which are 2p and 2s orbitals) and
list them from highest (on top) to lowest energy levels. Next, you will create the MOs from atomic orbitals, making sure
the number of MOs equal the number of atomic orbitals and for each pair of atomic orbitals there is one bonding MO
and on anti­bonding MO (bonding MOs are lower in energy than anti­bonding MOs). The last step is to fill in the MO's
with the valence electrons from both atoms. By looking at the orbital diagram, we can see that oxygen has BO of 2 because it has 10 bonding and 6 antibonding,
very stable. It's paramagnetic because it posses 2 unpaired electrons.
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For Difluorine, by counting the number bonding, 10, and number of antibonding, 8, give us the BO of 1. It is diamagnetic with no
unpaired electrons.
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2. 3. 4. 5. MO bonding in F2 and O2 ­ Chemwiki
References
1. Cotton, Albert F., Geoffrey Wilkinson, Carlos Murillo A., and Mantred Bochmann. Advanced Inorganic Chemistry. New
York, Chichester, Weinheim, Brisbane, Singapore, and Toronto: John Wiley and Sons, INC., 1999. Print.
Outside Links
http://www.metasynthesis.com/webbook/39_diatomics/diatomics.html
http://www.grandinetti.org/Teaching/Chem121/Lectures/MOTheory/
Problems
1. Draw molecular diagram of H2, B2, and N2.
2. Give bond order, and predict whether they are diamagnetic or paramagnetic for all the molecules above (question 1.)
3. What's the different between the MO diagram of all the above molecules (from question 1) compared to O2 and F2? (Note: interm of
orbitals of simmilar symmetry and energy.)
Answer:
1.
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3. The difference between molecules on the questions 1 and 2 are introduced somewhat at the beginning of the topic. It basically is the
mixing between orbitabls of simmilar symmetry and energy, and the result turns out to be different compare with F2 and O2. As you can
see from the MO diagram show on for all of these two cases, the energy of O2 and F2 have bi­bond higher energy than sigma­bond of the
2p orbital. This is also called sigma­bi crossover between N2 and O2.
Contributors
Dieu Huynh
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