[vinci]

I haven't posted an introduction because I was a member here I think the admins just deleted my account because of inactivity, but just for the heck of it I'm a highschooler. PS I clearly understand I should've been giving a better introduction but I literally have 2 months and to cover 2 years worth of studying in it for my GCSE exams(im giving accelerated)
Now coming to the actual question:
Here is what I know:
S and P orbitals are hybridized to form new orbitals when with c2h4 and c2h6 because carbon only has two unpaired electrons. (1s2 2s2 2px1 2py1)
In case of c2h6 three p orbitals hybridize with one s orbital to make four sp3 orbits, which then join linearly(forming sigma bonds)
Now this is not the case of c2h4, in c2h4 it forms three sp2 orbitals and leaves one p orbital as-it-is, forming three sigma bonds and one pi bond; and I dont quite understand why not. Why does it form three sp2 orbitals and have one p orbital to make pi bond with another carbon in case of c2h4? Since the number of bonds are the same why not just make four sp3 orbitals(as it does in c2h6) and form sigma bonds (two with hydrogen and two with carbon)?

[orthoformate]

S and P orbitals are hybridized to form new orbitals when with c2h6

Correct. Ethane's bonds are all Sp3 hybridized

and c2h4

semi-correct. Ethylene's bonds to carbon and hydrogen are Sp₂ hybridized, but there is a p orbital for both carbons which is not hybridized (we'll get to this in a minute).

because carbon only has two unpaired electrons. (1s2 2s2 2px1 2py1)

This is not accurate. I'll do my best to explain at the bottom.

In case of c2h6 three p orbitals hybridize with one s orbital to make four sp3 orbits, which then join linearly(forming sigma
bonds)

Completely correct. Good work.

Now this is not the case of c2h4, in c2h4 it forms three sp2 orbitals and leaves one p orbital as-it-is, forming three sigma bonds and one pi bond; and I dont quite understand why not. Why does it form three sp2 orbitals and have one p orbital to make pi bond with another carbon in case of c2h4? Since the number of bonds are the same why not just make four sp3 orbitals(as it does in c2h6) and form sigma bonds (two with hydrogen and two with carbon)?

You are asking really great questions. As for the "Why does it form three sp2 orbitals and have one p orbital to make pi bond with another carbon in case of c2h4?" Imagine this scenario:

you have a ethane molecule, with "three p orbitals hybridize with one s orbital to make four sp3 orbits, which then join linearly(forming sigma bonds)" as you so accurately describe. The ethane molecule then loses a pair of electrons due to oxidation.
http://patentimages.storage.googleapis.com/US6310241B1/US06310241-20011030-C00001.png (look at the first equation). If this occurs then the 2 electrons and hydrogen removed from one carbon will have to be  replaced by the electrons on the other carbon. This causes the formation of a pi bond. This Pi bond is two electrons being shared across the top of a sigma bond.

https://en.wikipedia.org/wiki/Ethane
https://en.wikipedia.org/wiki/Ethylene

https://en.wikipedia.org/wiki/Pi_bond  Read this carefully!! it contains great information for your questions


this sharing occurs because carbon does not have any unpaired electrons and the carbon must fill it's octet while maintaining the correct amount of valence electrons. It is said that " the p orbital is left as it is" because now the molecule (ethylene) is planar, and there is an open p orbital that is not participating in hybridiztaion.

Hopefully this helps. You are asking really good questions that are advanced for your age. this stuff really needs to be drawn out for you. If we had a blackboard it would be a lot easier..

[vinci]

Thank you for the explanation. You've given a thorough explanation, it's just my ignorance with the subject that's being a boulder in the way. Hopefully, sooner or later, that will be eliminated.
If I try it in the simplest of terms it has got more to do with how ehtylene is formed (by the oxidation of ethane). When a molecule of C2h6 is oxidized as the first equation depicts, it takes away two electrons from each carbon(or an H atom with two electrons), now each carbon has to form another bond to make up for the 'lost-electrons' and one way it can do it is by forming another bond with the other carbon.
Now let's just wind the clock back a bit, we had four sp3 orbitals(in an c2h6), three of which already have bonded and there's one sp3 orbital left(from where the carbon-hydrogen bond was broken because of oxidation of C2h6). Common sense would say that now since each carbon has an 'idle' sp3 orbital these should just join linearly and form another sigma bond(as I had speculated) but that is not the case and which is exactly what bothers me.
I think(and I can be absolutely incorrect with this hypothesis), that another sp3-sp3 bond isn't possible because there's another sp3-sp3 bond and it would repulse another one being there pushing away the newly formed bond and making it a pi bond. But this hypothesis doesn't explain how an sp3 orbital(left by the taking away of hydrogen from each carbon) turns into a p orbital for the pi bond and what happens to the 's part of sp3' orbital. I understand that pi bonds are merely orbitals overlapping sideways, so couldn't it just be the sp3 orbitals making the pi bonds? why did it have to be the p orbitals doing that? I know I am complicating things here, and which reminds me of this (https://www.youtube.com/watch?v=wMFPe-DwULM)
That there is no answer to a 'Why'. Is it the case with my question too or is there actually an explanation?

[AWK]

Carbon atom may show 3 types of hybridization sp, sp² and sp³. Sigma-bonds (σ) are formed only from sp^x orbitals. Pi-bonds (π) are formed only from non-hybridized orbitals of p-electrons. Sometimes hybridized orbitals or p-electrons do not form bonds.
Then we have radicals (stable eg.: (C₆H₅)₃C· or unstable eg.:   :CH₂)

[vinci]

Carbon atom may show 3 types of hybridization sp, sp² and sp³. Sigma-bonds (σ) are formed only from sp^x orbitals. Pi-bonds (π) are formed only from non-hybridized orbitals of p-electrons. Sometimes hybridized orbitals or p-electrons do not form bonds.
Then we have radicals (stable eg.: (C₆H₅)₃C· or unstable eg.:   :CH₂)

Isn't it that sigma bonds can be formed with simply p orbitals too? (If the join linearly). I remember reading that somewhere. Please correct me if I'm wrong. I understand, that answer to this it is just the way it is. Right?

[AWK]

Theoretically - yes , but carbon is not a case (p orbital symmetric with respect to the axis joining the two nuclear centers - such carbon orbital forms hybrids).

[orthoformate]

If I try it in the simplest of terms it has got more to do with how ehtylene is formed (by the oxidation of ethane). When a molecule of C2h6 is oxidized as the first equation depicts, it takes away two electrons from each carbon(or an H atom with two electrons), now each carbon has to form another bond to make up for the 'lost-electrons' and one way it can do it is by forming another bond with the other carbon.

You are starting to get it. What happens is that 2 electrons are removed from ethane, not from each carbon. Also 2 hydrogen's are removed from ethane as you can see from the equation for oxidation( I said 1 hydrogen in the previous post, my mistake). This means that the remaining pair of electrons must be shared between the carbons as it was once shared between that carbon and one of the hydrogens.

Draw ethane. erase one of the bonds to hydrogen and remove two hydrogens. you will see that if you take this new lone pair and move it over in between the carbons to form a double bond, ethylene will be formed.

Now let's just wind the clock back a bit, we had four sp3 orbitals(in an c2h6), three of which already have bonded and there's one sp3 orbital left(from where the carbon-hydrogen bond was broken because of oxidation of C2h6). Common sense would say that now since each carbon has an 'idle' sp3 orbital these should just join linearly and form another sigma bond(as I had speculated) but that is not the case and which is exactly what bothers me.

One reason it doesn't form a sigma bond is that there is already a sigma bond there. meaning there are already electrons directly in between the two carbons atoms (Physicists would kill me for this explanation, but it's a good learning tool). If you were to try to push more electrons into that space you would meet with firm resistance because doing so would destabilize the molecule.
Because of this the electrons lay over top of the sigma bond forming the pi bond. does this make sense?

in my mind you are way ahead of your age. The reason I say Physicists or even some Chemists might be upset is because electrons do not "sit in between the atoms" they move around.

I think(and I can be absolutely incorrect with this hypothesis), that another sp3-sp3 bond isn't possible because there's another sp3-sp3 bond and it would repulse another one being there pushing away the newly formed bond and making it a pi bond. But this hypothesis doesn't explain how an sp3 orbital(left by the taking away of hydrogen from each carbon) turns into a p orbital for the pi bond and what happens to the 's part of sp3' orbital.

it is said that a carbon atom is sp₃ hybridized when it is tetrahedral in shape. Meaning that all three p orbitals making up the x,y,and z axis's are involved in bonding. When those 2 electrons are removed, and you do the drawing I mentioned earlier and remove the protons as well, then make the double bond, the molecule is said to be sp₂ hybridized. This is because now one axis is not involved in bonding, let's say that is the z axis. your molecules is not planar, as in, in the plane of the page.

I think what you are getting tripped up on is the terminology. that Pz orbital (axis) was hybridized with the other axis in bonding, now there is nothing on the Pz axis, so you cannot refer to it as sp₃ anymore?

I understand that pi bonds are merely orbitals overlapping sideways, so couldn't it just be the sp3 orbitals making the pi bonds? why did it have to be the p orbitals doing that? I know I am complicating things here, and which reminds me of this
(https://www.youtube.com/watch?v=wMFPe-DwULM) That there is no answer to a 'Why'. Is it the case with my question too or is there actually an explanation?

Because now the z axis is not involved so you don't refer to it as sp₃.

I hope this helps, you are asking lots of good questions that are hard to explain. I'm glad AWK has answered as well.