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Topic: MO diagram of BN  (Read 10972 times)

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Offline yiyo

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MO diagram of BN
« on: October 01, 2013, 10:16:00 AM »
I would like to ask what is the bond order of BN?
If I include the s-p mixing in this MO diagram, I think the electron configuration of BN should be (1sigma)2(2sigma)2(3pi)4, then all these MO orbitals are bonding orbitals, and the bond order of BN is 4.... Am I correct?

Offline Enthalpy

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Re: MO diagram of BN
« Reply #1 on: October 23, 2013, 06:22:38 AM »
Can diatomic BN exist?

I understand (do I?) this would correspond to a gas, but BN decomposes before boiling.
http://en.wikipedia.org/wiki/Boron_nitride

Offline Corribus

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Re: MO diagram of BN
« Reply #2 on: October 23, 2013, 11:27:28 AM »
In a hypothetical BN molecule, the 1s orbitals do not mix because of proximity. The 2s orbitals would mix to form filled sigma bonding and antibonding orbitals.  The two 2px orbitals (along the molecular axis) would mix to form a filled bonding sigma orbital (two electrons).  The px and py orbitals for both B and N would mix to form two mutually perpendicular pi bonding orbitals, each half filled.  The total bond order would be 2.  A similar molecular oribtal diagram could be constructed for an isoelectronic C2 molecule, although in the former case the bonding pi and sigma orbitals would tend to reside more on the nitrogen center, whereas in C2, electrons would be equally shared.

Interestingly, this prediction from simple MO theory doesn't agree very well with experiment.  Diatomic carbon, for example, has been experimentally observed to have a singlet ground state, whereas the above treatment predicts it to be a triplet (unpaired electrons in the pi sigma orbitals). The reason for this discrepency is that in homonuclear diatomics of the 2nd row N2, C2 and B2, the 2s sigma and 2p sigma orbitals are close enough in energy and of appropriate symmetry to mix efficiently, which raises the energy of the 2p (sigma) orbital and lowers the energy of the 2s (sigma) orbital.  The 2p sigma orbital is raised enough that it becomes higher energy than the 2p pi orbitals, and so the actual MO diagram would feature fully filled 2p pi orbitals and a completely empty 2p sigma, with a bond order of 2 and a total spin of 0 (singlet).  BN would seem to feature a similar reversal of the order of the sigma and pi orbital ordering, and in this case would also be expected to be a ground state singlet.  However experimentally the ground state of BN appears to be a triplet (see http://webbook.nist.gov/cgi/cbook.cgi?ID=C10043115&Mask=1000 and references therein), so the energies have to be close.  Sometimes it can be difficult in heteronuclear diatomics to determine what the order of the 2p sigma and 2p pi orbitals should be, as the degree of mixing depends closely on the energy difference between the atomic orbitals of the two nuclei; in the experiments linked to above, the BN so measured was trapped in an artificial matrix, so maybe this can influence the measurement.  If the orbitals are close enough in energy, the triplet state usually wins out, because there's a pairing energy required to make a singlet state.  Anyway, essentially this is a form of hybridization.  In homonuclear diatomics O2 and F2, the 2p and 2s atomic orbitals are far enough apart in energy that mixing is inefficient, and therefore the 2p sigma is lower in energy than the 2p pi orbitals.

Note also that the predicted MO diagram fro C2 is quite a bit different than that of stable acetylene.

Either way, the MO diagram would seem to suggest that BN and C2 are energetically stable (e.g., they have bond orders greater than zero).  However we know they don't really exist in isolated form under most standard conditions.  Their reactivity stems from the fact that reaction with other nearby molecules creates new chemical species with even higher stability.  Effectively, C2 and BN are just really strong bases / oxidants.  You might wonder why, then, homonuclear diatomics like N2 and O2 are very unreactive.  N2 is unreactive because it has a bond order of 3, and therefore even if you could surmount the necessary barrier of breaking a triple bond, the energetic payoff would be slight.  O2 is actually quite thermodynamically reactive (like C2) but reactions with oxygen are kinetically slow due to the fact that it is a ground state triplet - unlike C2 - and most other nearby molecules available for O2 to react with are ground state singlets.  I guess BN might be kinetically stable for the same reasons, were it abundant enough to be readily observed. However it has to be made synthetically, and because of the high stability of polyatomic crystalline BN due to resonance stabilization, it's not easy to observe diatomic BN experimentally. Diatomic carbon doesn't exist here on earth in free form, but in space it can be and has been observed quite frequency.

Anyway, I hope some of that rambling was of help.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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