Yes, the "Cancer Curer"'s explanation is a mess and full of nonsense.
Obviously, he has limited understanding of what he is writing. A shame.
[...]
In the same sense, every two electrons is ONE sigma bond. Every four electrons is TWO sigma bonds and ONE pi bond.
[...]
1 sigma bond between carbon and oxygen (sp hybrid orbitals)
There is one carbon-oxygen bond. There are two sigma bonds here and one pi bond.
[...]
1 pi bond between carbon and oxygen (p orbital)
There is one carbon-oxygen bond. There are two sigma bonds here and one pi bond.
[...]
You must be kidding me
!!!! Are you talking of 3 bonds formed by 4 electrons? This case is extremely rare
http://en.wikipedia.org/wiki/Three-center_four-electron_bond, and for COCl
2, it is certainly not the case.
I have the feeling that you don't realise that a double bond (like the C=O of the example) is made of one sigma bond (along the bond axis) and one pi bond (on the side of the axis) only.
You do not hybridise C-Cl bonds and C-O bonds separately. Carbon's orbitals are hybridised only once, at the time of bond formation.
We have only 3 sigma bonds totally. Two C-Cl and one C-O, and one pi bond between C and O. Carbon uses 3 sp2 orbitals for bonding with all atoms, as hybrid orbitals are used only for sigma bonding. And the one p-orbital that is free on the Carbon is used for pi-bonding.
I agree with Schrödinger.
Let's explain hybridisation again:* An isolated carbon atom possesses four atomic orbitals: 2s, 2p
x, 2p
y and 2p
z. The 2s is spherical and the 2p have two lobes pointing towards one direction: x, y and z. So, the angles between the 2p orbitals are 90º.
* However, when that carbon atom is place inside a molecule, the angles observed are never 90º. They are closer to: 109.5º, 120º and 180º. In order to explain this difference of angles between the expected 90º and the observed angles, the theory of hybridisation was proposed. It consists in mixing 2s and 2p orbitals together at the beginning to form new sp, sp
2 or sp
3 orbitals (depending on the case). Then, the angles expected and observed agree (180º for sp, 120º for sp
2 and 109.5º for sp
3).
The name sp means that the orbital 2s and only one orbital 2p mixed to form two hybridised sp orbitals. As a consequence, two 2p orbitals are left unhybridised and available to make two pi bonds.
The name sp
2 means that the one orbital 2s and
two orbitals 2p mixed to form three hybridised sp
2 orbitals. As a consequence, there is one 2p orbital left unhybridised and available to make one pi bond.
Let's go to the example COCl
2:
* The carbon atom is hybridised sp
2. It has three orbitals sp
2 and one orbital 2p
z. It form three sigma bonds with the two Cl and O atoms, using the sp
2 orbitals. The 2p
z orbital make one pi bond with the oxygen atom.
* The oxygen atom is also hybridised sp
2. It has three orbitals sp
2 and one orbital 2p
z. The sp
2 orbitals form one sigma bond with the carbon and contain two lone pairs. The 2p
z orbital make one pi bond with the 2p
z from the carbon.
* The chlorine atoms are hybridised sp
3: it has four orbitals sp
3. They are used to make one sigma bond with the carbon and contain three lone pairs.
I know how the book got the sigma bonds and pi bonds for the answer, but what I don’t understand is how to find out whether or not the sigma or pi bond is a sp hybrid orbital or a p orbital, which is indicated in the parentheses.
Can you please explain to me how to recognize and be able to label what is a sp and p?
A pi bond always comes from two p orbitals of two adjacent atoms. It can't arise from one s orbital, it is impossible by symmetry.
Look at that short video of C
2H
4:
http://www.youtube.com/watch?v=C2W-yDPcpl4* In blue colour, you see the pi bond between the two carbon atoms. Notice how it is on the side of the C=C bond, and not directly in between the two C atoms.
* In green colour, you have the carbon sp
2 orbitals that form sigma bonds between the C and H atoms. See how the sigma bonds are directly in between the two atoms they are binding.
* In purple, you have the hydrogen 1s orbitals.