# In what way does increasing hybridization correlate with decreasing s character?

Jun 22, 2017

I think you mean this with respect to alkanes, alkenes, and alkynes. Consider ethane, ethene (ethylene), and ethyne (acetylene).

We know that bond lengths for comparable SIGMA bonds (e.g. the first bond in a $\text{C"-"C}$, $\text{C"="C}$, and $\text{C"-="C}$ here) decrease as the OVERALL bond order increases.

Well, from left to right, we have $s {p}^{3}$, $s {p}^{2}$, and $s p$ hybridization.

Hybridization involves the mixing of more than one orbital shape to construct identical, totally symmetric orbitals. This literally means we have:

• $s {p}^{3} = s + p + p + p$ $\leftrightarrow$ bb(25% s) + 75% p
• $s {p}^{2} = s + p + p \text{ "" }$ $\leftrightarrow$ bb(33% s) + 67% p
• $s p = s + p \text{ "" "" "" }$ $\leftrightarrow$ bb(50% s) + 50% p

We thus have an increase in $s$ character from left to right in the hydrocarbons shown above, correlating with the decrease in the number of electron groups.

This decrease in bond length is due to the closeness of the $s$ orbital to the nucleus. Consider the radial density distributions of the $2 s$ vs. the $2 p$ orbitals:

That little bump at the far left means that the $2 s$ orbital is more penetrating than the $2 p$, i.e. more core-like. That means the optimal bond distance (that balances nuclear repulsions and electron-nucleus attractions) is shorter.

Hence, a hybrid orbital with more $s$ character leads to a shorter sigma bond.