So, four orbitals (one 2s + three 2p) are mixed and the result is four sp 3orbitals. The number of the hybrid orbitals is always the same as the number of orbitals that are mixed. So, in the next step, the s and p orbitals of the excited state carbon are hybridized to form four identical in size, shape and energy orbitals. Pay attention that the electron goes uphill as the p subshell is higher in energy than the s subshell and this is not energetically favorable, but we will see how it is compensated in the next step when orbitals are mixed (hybridized). This leads to the excited state of the carbon: In the first step, one electron jumps from the 2s to the 2p orbital. Now, let’s see how that happens by looking at methane as an example. Hybridization is a theory that is used to explain certain molecular geometries that would have not been possible otherwise. And this is where we get into the need of a theory that can help us explain the known geometry and valency of the carbon atom in many organic molecules.
You can see from the electron configuration that it is impossible to make four, identical in bond length, energy, and everything else (degenerate) bonds because one of the orbitals is a spherical s, and the other three are p orbitals. The valence electrons are the ones in the 2s and 2p orbitals and these are the ones that participate in bonding and chemical reactions. So, in order to predict the valency and geometry of the carbon atom, we are going to look at its electron configuration and the orbitals. Remember also that covalent bonds form as a result of orbital overlapping and sharing two electrons between the atoms. A reminder that in tetrahedral geometry, all the angels are 109.5 o and the bonds have identical length. It is confirmed experimentally that the carbon atom in methane (CH 4) and other alkanes has a tetrahedral geometry. Therefore, the number of orbitals used in the hybridization is the number of #\mathbf(sigma)# bonds made around the central atom.Let’s start first by answering this question: Why do we need the hybridization theory? That is, you get five #sp^3d# orbitals, for example. You can see that the number of orbitals used in the hybridization spits out the same number of hybridized orbitals that can #sigma# bond. #sp^3d^2# hybridization in #"SF"_6# corresponds with six #sigma# bonds around one sulfur. #sp^3d# hybridization in #"PF"_5# corresponds with five #sigma# bonds around one phosphorus. #sp^3# hybridization in methane corresponds with four #sigma# bonds around one carbon. #sp^2# hybridization in ethene corresponds with three #sigma# bonds around one carbon. #sp# hybridization in acetylene corresponds with two #sigma# bonds around one carbon. One pure double bond has one #sigma# and one #pi# bond, and one pure triple bond has one #sigma# and two #pi# bonds.įrom knowing the hybridization of the central atom, we can determine the number of #sigma# bonds around the central atom, but no more than that without more information. If you want to get the number of #pi# bonds, know what compound you're looking at, and just count them. Without knowing more context, it's not possible to know the number of #pi# bonds.
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