Bond order of B2 molecule

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Belicia Tang 1B
Posts: 32
Joined: Wed Sep 21, 2016 3:00 pm

Bond order of B2 molecule

Postby Belicia Tang 1B » Thu Nov 03, 2016 10:22 pm

Hey guys,
So the bond order of B2 is equal to 1, which you can get by drawing the molecular orbital diagram and performing the equation Bond Order = .5 * (# of bonding electrons - # of antibonding electrons). However, when you draw the Lewis structure of B2, you get a triple bond. I always thought bond order corresponded to the number of bonds. That is, molecule held together by a single bond would have a bond order of 1, and a molecule held together by a double bond would have a bond order of 2, etc. Is B2 an exception to this rule?




Thanks!



EDIT: My mistake, B2 lewis structure is bounded by single bond. I don't know what I was thinking when I said triple bond, because the # of electrons don't even add up to 14. So the bond order of 1 is correct.
Last edited by Belicia Tang 1B on Fri Nov 11, 2016 12:36 pm, edited 1 time in total.

Jessie_Chen_2L
Posts: 20
Joined: Wed Sep 21, 2016 2:59 pm

Re: Bond order of B2 molecule

Postby Jessie_Chen_2L » Thu Nov 03, 2016 11:13 pm

The molecular orbital diagram for B2 isImage

You can see that there are two electrons in the σ 2s and two electrons in the σ*2s and two in the π 2p orbitals. We now know that there are two electrons in the antibonding orbitals and 4 in the bonding orbitals.

From this we can use the equation, Bond Order = 1/2(# of electrons in bonding orbitals - # of electrons in antibonding orbitals).

Bond Order = 1/2(4-2) = 1


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