Chapter 3, Problem 3.26P

Interpretation Introduction

Interpretation:

Among molecules A and B, the molecule which has longer $\text{C=C}$ distance is to be determined. Also the molecule which has stronger $\text{C=C}$ bonds is to be determined.

Concept introduction:

Hybridization affects both bond length and bond strength. The two $\text{sp}$ hybrid atomic orbitals are formed by mixing one $\text{s}$ orbital and one of the three $\text{p}$ orbitals from the valence shell. Each $\text{sp}$ hybrid orbital formed has $\text{50\%}$ $\text{s}$ character and $\text{50\%}$ $\text{p}$ character. Three ${\text{sp}}^{\text{2}}$ hybrid atomic orbitals are formed by mixing one $\text{s}$ orbital and two of the three $\text{p}$ orbitals from the valence shell. Each ${\text{sp}}^{\text{2}}$ hybrid orbital formed has $\text{33}\text{.33\%}$$\text{s}$ character and $\text{66}\text{.67\%}$ $\text{p}$ character. Four ${\text{sp}}^{\text{3}}$ hybrid atomic orbitals are formed by mixing the one $\text{s}$ orbital and three $\text{p}$ orbitals from the valence shell. Each ${\text{sp}}^{\text{3}}$ hybrid orbital formed has $\text{25\%}$ $\text{s}$ character and $\text{75\%}$ $\text{p}$ character. As the $\text{s}$ character increases, the hybrid atomic orbital becomes more compact. The bonds formed by shorter atomic orbitals are shorter in length. The shorter bonds connecting the same atoms are stronger. Therefore, as the hybridization of an atom changes from ${\text{sp}}^{\text{3}}$ to ${\text{sp}}^{\text{2}}$ to $\text{sp}$, its bonds become shorter and stronger.