[patricio2626]

  Hi all!  I've tried searching various pages on the interent and this site, but to no avail.  I'd like to know why ethylene forms a 2sp2 and a 2p orbital instead of 2sp3, and also why acetylene forms 2sp and  2p and not simply 2sp3.  My text does not give the reasoning/rule behind this (gives the answers, but not the reason why this occurs), and I'm not sure if I'm going to have to know this/draw these valence bond structures for the General Chem I test (CHM 2045).  Even if I don't (not on the test review), I'd still like to know, I hate not "getting" this part of the material, it drives me nuts.  Thanks in advance, and this is a great site!

 -Patrick

[Yggdrasil]

Basically it has to do with the nature of double and triple bonds.  In general, hybrid orbitals can participate only in the formation of sigma bonds, which are a along the internuclear axis (the imaginary line connecting the nuclei of the two atoms).  Since there is only one internuclear axis, only one sigma bond is possible between two atoms.  In order to form double and triple bonds, the atoms must form pi bonds which lie outside of the internuclear axis. Since pi bonds are formed from two parallel p orbitals, the atoms must leave unhydridized p orbitals to allow the formation of pi bonds.  Therefore, ethylene, which contains a double bond, contains one sigma bond and one pi bond, so each carbon requires one unhybridized p orbital.  Acetylene's triple bond is composed of one sigma bond and two pi bonds, requiring each carbon to have two unhybridized p orbitals.

[patricio2626]

Wow, you really broke that down nicely!    Thank you, now I can figure it out if I draw the Lewis structure first!