The molecular orbital diagram for $\ce{O2}$ says that the sigma 2p bonding molecular orbital is lower in energy than the pi 2p bonding molecular orbital. Why is this not the case in the $\ce{B2}$ MO diagram?
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$\begingroup$ It is, though, isn't it? $\endgroup$– mikhailcaziOct 25, 2013 at 14:48
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1 Answer
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This topic is really fun. You see in case of boron molecule $\sigma_{2s}$ and $\sigma_{2p_z}$ (being on the intermolecular axis) will have the same symmetry and hence the orbitals will overlap and result in the formation of 2 bonded $sp_z$ orbitals. So the $\sigma_{2p_z}$ orbital will obtain somewhat $s$ character and it will become unstable or in other words will be of higher energy. The $\sigma_{2s}$ MO will obtain somewhat $p$ character. So in a energy level MO diagram $\sigma_{2p}$ bonding orbital is placed above the $\pi_{ 2p}$ orbital.
However, in case of $\ce{O2}$, due to the higher nuclear charge of oxygen, the energy difference of $2s$ and $2p$ is large so there is no such kind of intermixing between the bonded MOs. Whereas in boron, the energy difference of $2s$ and $2p$ is less, facilitating intermixing.
You can find this topic in any higher level chem books. Hope you got my point; thank you.
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1$\begingroup$ Could you provide an example or link to a book that contains this information? $\endgroup$ Jan 15, 2014 at 16:53
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$\begingroup$ Boron radius - 80 pm, Oxygen - 65 pm. Funny-funny-funny $\endgroup$ Oct 26, 2014 at 9:32
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