[Lo.Lee.Ta.]

Hi, my school has never made us learn how sp3, sp2, and sp carbon orbitals look when bonded, but I feel like it's important to know...

But I'm sort of confused...

I started to try to draw sp2 carbons bonded together. My example sp2 molecule is ethene.

sp2 = (s orbital) + (p orbital) + (p orbital) = 3 sp2 hybridized orbitals arranged in a trigonal planar shape.



So, I know this picture is incorrect because there are only 6 electrons per carbon and only 2 hydrogens (in yellow). There should be 4 hydrogens in ethene.
But where would they bond to???
There should only be 3 orbitals per sp2 carbon, and I thought 2 of them have to be used to create the double bond.

Every double bond is made by two overlapping dumbell p orbitals, right?

..Ugh. Would you please let me know what's wrong with my picture and how it should really be drawn.
Confused 

Thanks! 

[Corribus]

Each carbon will have 1 p orbital (pz orbital) and 3 sp2 hybridized orbitals. Carbon has 4 valence electrons available for bonding.  One of the sp2 hybridized orbitals is bonded to the other carbon (C-C sigma bond), and the remaining two are bonded to hydrogens (C-H sigma bonds).  This gives two bonds between the carbons and 1 bond each between the carbons and hydrogens.

[danteOne]

One of your mistakes is thinking that a single pi orbital is actually two pi orbitals. Both the top node and the bottom node of you drew are part of the same pi orbital.
This picture I just drew shows a single pi orbital and none of the other orbitals in the atom.

[Lo.Lee.Ta.]

Corribus, I tried to draw what I think you were describing...? Is this right? :/

You said that there still is a p orbital for each carbon even though it is sp2...
Is there still that p orbital for an sp carbon?

I'm so used to the idea that sp3 has 4 orbitals, sp2 has 3 orbitals, sp has 2 orbitals, so how could this be right? 

How can the sp2 carbon here have 4 orbitals? 

Thanks! 

[Corribus]

Your diagram looks more or less right, but the C-C sigma bond should be oriented along the axis drawn between the two carbon nuclei, not above the plane as it appears in your drawing.  I'd gripe about some of the angles of the hydrogens but it would be unseemly of me to criticize your artistic skills.

Every carbon has four bonds. This is practically a rule that can't be broken as far as you should be concerned.

Maybe this will help.

A carbon atom has basically four orbitals that can participate in bonding: three p-orbitals (2px, 2py, 2pz) and one s-orbital (2s). 

An sp3 hybridized carbon takes all three p-orbitals and the s-orbital and combines them together to form four sp3 hybridized orbitals which all bond to adjacent atoms with sigma bonds. That's four bonds for the sp3 carbon, all of them sigmas.

An sp2 hybridized carbon takes two of the p-orbitals and the s-orbital and combines them together to form three sp2 hybridized orbitals which bond to adjacent atoms with sigma bonds; the remaining p orbital remains unhybridized and bonds to an adjacent carbon (which also must have a 'free' p orbital and thus must also be sp2 hybridized) using a pi-bond. That's four bonds for the sp2 carbon: 3 sigmas and 1 pi. And one of those sigmas is to the same carbon to which the pi bond is formed.

And an sp hybridized carbon takes one of the p-orbitals and the s-orbital and combines them together to form two sp hybridized orbitals, which bond to adjacent atoms with sigma bonds; the remaining two p-orbitals remain unhybridized and bond to an adjacent carbon (or separately to two adjacent carbons), which also must have at least one 'free' p-orbital (two if both are bonded to the same carbon) using pi-bonds. That's four bonds for the sp carbon: two sigmas and two pis. And one of those sigmas is to the same carbon to which the pi bonds are formed (in the case of a triple bond).

Make sense?

[trinitrotoluene]

http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/Hybrid_Orbitals