What is the hybridization for C2H6? my teacher wrote sp-s sigma bond [C-H] and sp-sp sigma bond[C-C]

Mar 1, 2016

${\text{C"_2"H}}_{6}$ has an $s {p}^{3}$ hybridization on each carbon because of the four electron groups surrounding each carbon. It made four identical bonds in a perfect tetrahedral geometry, which means it needed four identical orbitals to make those bonds.

Each carbon has to hybridize one $2 s$ and three $2 p$ orbitals in order to generate four identical $s {p}^{3}$ orbitals that are compatible in symmetry with hydrogen's $1 s$ orbitals. Therefore, each $\text{C"-"H}$ bond in ${\text{C"_2"H}}_{6}$ is between an $s {p}^{3}$ of carbon (YELLOW) and a $1 s$ of hydrogen (BLUE sphere), i.e. an $s {p}^{3} - s$ connection (the YELLOW/BLUE overlap in (b)), and each $\text{C"-"C}$ bond is an $s {p}^{3} - s {p}^{3}$ connection (the YELLOW overlap in (b)).

IF YOUR TEACHER WAS CORRECT ON THE HYBRIDIZATIONS

On the other hand, ${\text{C"_2"H}}_{2}$ would match your teacher's observations.

With two electron groups on a POLYatomic molecule, each carbon requires only two $s p$ lobes and hence only one $s p$ hybridized orbital to bond with the other carbon AND a single hydrogen. In this case we can see that:

• The $s p$ of carbon 1 (YELLOW dumbbell) overlaps with one hydrogen's $1 s$ (BLUE sphere) to make a $\setminus m a t h b f \left(\sigma\right)$ bond.
• The $s p$ of carbon 1 (YELLOW dumbbell) overlaps with the $s p$ of carbon 2 (YELLOW dumbbell) and vice versa to make a $\setminus m a t h b f \left(\sigma\right)$ bond.
• The $s p$ of carbon 2 (YELLOW dumbbell) overlaps with the second hydrogen's $1 s$ (BLUE sphere) to make a $\setminus m a t h b f \left(\sigma\right)$ bond.

Each triple bond incorporates an additional ${p}_{x} \text{/} {p}_{x}$ and ${p}_{y} \text{/} {p}_{y}$ overlap between carbons 1 and 2, accounting for two $\pi$ bonds (i.e. $p - p$ connections).

Hence, when including those two $p - p$ $\pi$ bonds with the $s p - s p$ $\sigma$ bond between carbons 1 and 2, we have accounted for the the triple bond between carbons 1 and 2.

(One triple bond = 1 $\sigma$ + 2 $\pi$ bonds)