Clarification of Pi Bonds and Rotation

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Nathan Tran 4K
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Clarification of Pi Bonds and Rotation

Postby Nathan Tran 4K » Wed Nov 21, 2018 4:04 pm

Can someone clarify the point Lavelle made about rotating with pi bonds? I recall he used two pencils to try to rotate the atoms, but I think he said something along the lines of "if there is a pi bond, it means there can be no rotation." Could someone clarify this concept?

Schuyler_Howell_4D
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Re: Clarification of Pi Bonds and Rotation

Postby Schuyler_Howell_4D » Wed Nov 21, 2018 4:16 pm

Pi bonds have electron density on either side of the internuclear axis. This means that because of electron repulsion and the regions of electron density formed by the already present sigma bond, the electrons are "locked" into a region where they can remain. There is no room for the electrons to move around the atom and therefore the bond does not rotate.

Vikramjeet Gill 1C
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Re: Clarification of Pi Bonds and Rotation

Postby Vikramjeet Gill 1C » Thu Nov 22, 2018 10:35 am

Pi bonds are formed by the overlapping of two 'spare' p orbitals. This forms a flat bond above and below the sigma bond, this pi bond is only stable in one orientation and if there is any bond rotation then this bond breaks. Pi bonds prevents rotation because of the electron overlap both above and below the plane of atoms.

Saleha_Mian_3E
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Joined: Fri Sep 28, 2018 12:25 am

Re: Clarification of Pi Bonds and Rotation

Postby Saleha_Mian_3E » Thu Nov 22, 2018 11:06 pm

What stood out to me is when he said they can not move like sigma bonds because of the density on both sides, which makes them lock into space since there is no way they can rotate. Hope that helps to visualize!


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