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When we calculate the electrons and orbital numbers in $\ce{CH4}$ $sp^3$ hybridization we have taken 3 electrons from $2p$ of $\ce{C}$ and 1 electron from $2s$ of $\ce{C}$ but in $\ce{C2H4}$ we have $2sp^2$ (2 electrons from $2p$ and 1 from $2s$).

  1. Why do $\ce{C}$'s atomic orbitals make $2sp^2$ in $\ce{C2H4}$ and $sp^3$ in $\ce{CH4}$?
  2. How we associate the numbers of electrons from $2p$ and $2s$?
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How much do you know about VSEPR theory, especially about sigma and pi bonds? What do you mean by $2sp^2$? –  ManishEarth Nov 12 '12 at 3:56
    
I know how to explain molecules in VSEPR and know sigma is first bond for C-C pi is used to form second bond. Sigma is from sp3 hybrid molecular orbital and pi formed in sp2. (these what i know about pi and sigma) 2sp2 is used in my book "Solomons and Fryhle Organic Chemistry" when explainin the excited state and hybridization energy diagram. –  ordinary chemistry student Nov 12 '12 at 5:07

2 Answers 2

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This can be explained in terms of orbital overlap and therefore bond strength.

Consider a single carbon in an ethene molecule with two sp3 hybridised carbon atoms. The orbitals on the carbon repel each other and arrange themselves into a tetrahedron. Two of the four sp3 hybrids are used to form sigma bonds with hydrogen atoms. The carbon is then left to share its remaining two sp3 hybridized orbitals with its neighbouring carbon. Draw this system and you can see that two sigma bonds can be formed, though they are "bent" as the sp3 orbitals on the carbon are at a 109.5 degree angle to one another.

Now consider a single sp2 hybridised carbon in an ethene molecules with two sp2 hybridised carbon atoms. Again two orbitals are used to bond with hydrogen. This time we have an sp2 orbital pointing directly at the neighbouring carbon with which to form a strong sigma bond. The remaining p orbital which is orthogonal to the C-C sp bond can also forms a C-C pi bonding interaction.

Draw these orbital systems and compare.

  • The sp2 hybridised system constructs a strong linear sigma bond.
  • The sp2 hybridised system better shields the nuclei of the bonding atoms.
  • The sp2 hybridised system provides more orbital overlap, leading to stronger bonding and lower free energy.
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So, in ethylene we got two "2sp2" due to two carbon atoms constructing pi bonds. –  ordinary chemistry student Nov 13 '12 at 14:05
    
The fast and simple answer to part 1. of your question is that the orbitals which result in the strongest bonding will arise. –  Horba Nov 13 '12 at 14:29

I agree with Horba's answer. However the usual case is that the bond structure of a given molecule is already known and then the hybridization is chosen to fit the observed structure.

Consider $\ce{CH4}$ and $\ce{C2H4}$. It is clear that in methane the central carbon has four bonds and hence will need hybridization that gives four bonds. Thus we get $sp^3$. Ethene, on the other hand, has two carbons each of which needs three bonds. So we get $sp^2$.

In other words, hybridization is not a predictive scheme, it is an explanatory scheme.

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Ethene, on the other hand, has two carbons each of which needs three bonds. -- this is prone to misinterpretation--technically, a double bond is two bonds.. Could you clarify that? –  ManishEarth Nov 13 '12 at 9:22

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