Hybrid versus non-hybrid orbitals

[yebo]

When atoms bond, they always form hybrid orbitals, right? So why am I running across questions about there being a s or p orbital in a double bond?

Example form EK 1001 Orgo #63:

The overlap of what two orbitals form the pi bond between carbon atoms in an alkene?
A. two p orbitals
B. two sp2 orbitals
C. two sp3 orbitals
D. two s orbitals

The correct answer is A.

I get that a double bond is made from two pi orbitals (but also an s orbital too, right?). However, I also read that according to VESPR, each carbon in an alkene reforms all of its orbitals to make three sp2 orbitals. So why wouldn't that double bond be resulting from the overlap of two sp2 orbitals? If the correct answer is two p orbitals, does that mean that there are hybrid and non-hybrid orbitals existing at the same time?

Hoping someone understands my confusion here...

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[swivelj]

Remember what defines a hybrid and I think this will be a little more clear. An sp3 is 1 part s character and 3 parts p... sp2 is 1 part s character and 2 parts p. sp bonds are 1 part s character and 1 part p. So if you have to use a pi orbital to make an Alkene that only leaves you with 1 part s and 2 parts p hence sp2. So all of your remaining sigma bonds will be of sp2 character in this description. The question also is pretty straightforward when it's asks about the pi bond. If it asked about the sigma bond of the Alkene it would be sp2... Hope this helps!

 

[ReggieD6868]

As indicated by the modifier double in double bond two types of bonds are being formed, a sigma and a pi bond.

For an easy example that helps to explain the concept in the EK question, let's look at ethene, H2CCH2. Each methylene group, CH2, is SP2 hybridized, that means it has 3 hybridized orbital and one lone p-orbital. When the two planar spy hybridized methylene groups join, they form a sigma bond. The sigma bond between each methylene group as well as the two sigma bonds to their respective hydrogens utilizes the three hybridized orbital, but each ch2 group still has a lone p-orbital. Since the two groups are planar, each of their lone p-orbitls are able to overlap, and this overlap forms the pi-bond portion of the double bond.

The only orbitals that form pi-bonds are UNhybridized p-orbitals, so that narrows down the answer choices without the aforementioned concept being considered. The correct answer is indeed choice A.

 

[yebo]

Ahhhh ok... I think I understand now.

So basically you only make enough hybrid orbitals to have one per bond/lone pair. The extra orbitals left over can then contribute to double/triple bonds. Which would mean that usually any sigma bond is going to be a hybrid and that a double or triple bond will just be one or two p orbitals. Correct?

 

[yebo]

Wait, wait. I was wrong. As I am going through more questions related to the one I posted I am seeing that something is off on my thinking here.

Take this question:
53. What types of bonds exist between the C atom and the N atom in HCN? (drawing of the atom shows a triple bond between the C and N)

A. 1 pi bond and two sigma bonds
B. 2 pi bonds and one sigma bond
C. 3 pi bonds and zero sigma bonds
D. 3 sigma bonds and zero pi bonds

So according to what I was understanding the C in HCN has sp hybridization since it has two bonds and no lone pairs. So why then isn't the C-N triple bond made up of 1 sp bond (for the sigma bond) and then two additional p bonds? Why do all of the answer options here NOT involve hybridized orbitals?

 

[swivelj]

Answer b says exactly what you just described, a sigma bond (an sp hybrid), and 2 pi bonds. Look sp hybrids form sigma bonds. Pi and sigma, name and define the type of bond you are looking at...... sp, sp2, and sp3 define the orbital of the electrons and describe to a degree their "shape". The question asks what types of bonds? Not what type of orbitals, there is no need to describe the hybridization to answer for what types of bonds. Understanding hybridization however can help you to determine what bonds are present and vice versa.

 

[yebo]

Got it! Yeah, you are right I was just confusing bonds with orbitals. Oops!