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    Senior Member euclid2's Avatar
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    chemistry

    The theory of hybridization of atomic orbitals was developed to explain molecular geometry. Sketch and name the shape of each of the following hybrid orbitals of a carbon atom in a compound:

    a) $\displaystyle sp$
    b) $\displaystyle sp^2$
    c) $\displaystyle sp^3$
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    Recall the valence electron configuration of carbon: $\displaystyle 2s^2 \ 2p^2$

    a) The hybridization of one $\displaystyle s$ and one $\displaystyle p$ forms two $\displaystyle sp$ orbitals which leaves two unhybridized $\displaystyle 2p$ orbitals, all half-filled. Thus when forming bonds, the two $\displaystyle 2p$ orbitals are involved in forming $\displaystyle \pi$-bonds and the $\displaystyle sp$ orbital would form a $\displaystyle \sigma$-bond. This is why we normally associated sp hybrid orbitals with triple bonds.

    Consider: $\displaystyle \text{H}-\text{C}\equiv \text{C} - \text{H}$

    b) Try using similar reasoning as above. One $\displaystyle s$ and two $\displaystyle p$ orbitals form 3 $\displaystyle sp^2$ hybrid orbitals which leave 1 unhybridized p orbital. The unhybridized p orbital would form a $\displaystyle \pi$-bond with another half-filled p orbital and a $\displaystyle \sigma$-bond when looking at the $\displaystyle sp^2$ orbital. Thus, we normally associate double bonds with $\displaystyle sp^2$ hybrid orbitals.

    Consider: $\displaystyle \text{H}_2\text{C}= \text{O}$

    c) Consider: $\displaystyle \text{C}\text{H}_4$
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