Chapter 3, Problem 3.52P

Interpretation Introduction

(a)

Interpretation:

Which of the two molecules shown has the stronger $\text{C-Cl}$ bond is to be determined.

Concept introduction:

The hybridization of the two atoms forming the bond affects the bond strength. As the $\text{s}$ character increases, the hybrid atomic orbital becomes more compact. The bonds formed by more compact atomic orbitals are shorter in length. The shorter bonds connecting the same atoms are stronger. The $\text{s}$ character of the carbon atoms in hybrid orbital increases in the order ${\text{sp}}^{\text{3}}{\text{< sp}}^{\text{2}}\text{< sp}$.

Interpretation Introduction

(b)

Interpretation:

Which of the two molecules shown has a shorter $\text{C-Cl}$ bond is to be determined.

Concept introduction:

Hybridization affects bond length. 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 $\text{s}$ character of the carbon atoms, in hybrid orbital, increases in the order ${\text{sp}}^{\text{3}}{\text{< sp}}^{\text{2}}\text{< sp}$.

Interpretation Introduction

(c)

Interpretation:

The chlorine atom, in which of the two molecules shown, has greater concentration of negative charge is to be determined.

Concept introduction:

Hybridization affects bond length. 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 $\text{s}$ character of the carbon atoms in hybrid orbital increases in the order ${\text{sp}}^{\text{3}}{\text{< sp}}^{\text{2}}\text{< sp}$. As the s character of the hybrid orbital increases, the electrons in it are held closer to the nucleus. The atom then behaves as if its electronegativity has increased, and the negative charge on it becomes more concentrated.