Can anyone explain orbital pi and sigma bonding stuff?

I have to determine how many Pi and Sigma bonds there are for the following compounds:
CH2Cl2 C2H3Cl (Carbons double bonded)

If some one could maybe run through the steps you use to determine this, I would really appreciate it, I thought I had a grasp on it but apparently ehh, not so much.



The only time you have pi bonds are with double and triple-bonded atoms. Single bonds are sigma. The theory begind this is that sigma bonds are formed from hybridization of each atom's outer orbitals and extend in the space between the two bonded atoms. To have a sigma bond, orbitals MUST overlap each other. By contrast, pi bonds are not hybrids and extended above and below the two atoms being bonded in a figure-8 shape. The orbitals involved in pi bonds do NOT overlap with each other. When you have a double bond, you have a sigma bond in between the atoms AND a pi bond over and under them. When you TWO separate sets of pi bonds above and below the atoms.

For an example, carbon is atomic number 6. It has a 1s orbital containing 2 electrons (NOT valence), and a valence shell consisting of 2 electrons in the 2s orbital and 2 electrons, each occupying a separate p orbital of the 2p level. The energy between the 2s and the 2p isn't that great, so when hybridization happens, one of the 2s electrons gets bumped up to a p shell, so that there is 1 2s electron and 3 2p electrons.

Many times in organic chemistry, you will see carbon atoms containing four bonds around them. They have mashed up their 1 2s orbital and 3 3p orbitals to create FOUR hybrid sp3 orbitals. This is the case in methane, CH4, so that carbon extends out its four sp3 orbitals and forms sigma bonds with four hydrogens. For a more complicated example, if you take ethene, H2C=CH2, each carbon is sp2-hybridized, meaning that it has taken 2 of its p orbitals and its only valence s orbital, mashed them together, and created 3 new hybrid orbitals that are composed of 1/3 s character and 2/3 p character. Two of these sp2 orbitals will form bonds with hydrogen, and the remaining will bond with the other carbon. What causes the double bond? Well, each carbon didn't mash up its remaining p orbital, and these two p orbitals (each containing 1 electron) align in a plane and forma pi bond with each other.

In answer to the compounds you give:
H2C=CCl2: Each carbon is sp2 hybridized, with two of its sp2 orbitals forming sigma bonds with hydrogens and chlorines respectively, and the remaining sp2s overlap to form a sigma bond with each other. The one p orbital on each C that wasn't mashed-up aligns in the plane with the other carbon's C to form a pi bond. Hydrogen CANNOT hybridize its orbitals - it only has 1s - so its 1s orbital bonds with a carbon sp2 hydrid.

HClC=CCl2: same principle as before - each carbon is sp2 hybridized, forming a sigma bond between it and hydrogen, chlorine, and other carbon. The remaining p orbitals on each carbon form a pi bond above and below the molecule.
1. Draw the molecular structure.
2.Count the single bonds. They areall sigma bonds.
3.Count the double bonds. Here among the two, one is sigma, the other a pi.
4. Count the no. of triple bonds. In each triple bond, one is a sigma and the other two are pi.
5. Count all sigma and all pi.
Remember sigma are the stronger bonds
Hope this helps...

The answers post by the user, for information only, does not guarantee the right.

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