Chapter 6, Problem 6.13QP

Interpretation Introduction

Interpretation: From the given data of $\text{AlF}$, dipole moment should be determined.

Concept introduction:

Conversion formula

• $\text{1}\text{D}\text{=3}\text{.336}{\text{×10}}^{\text{-30}}\text{Cm}$
• $\text{1}\stackrel{}{\stackrel{\text{°}}{\text{A}}}{\text{=1×10}}^{\text{-10}}\text{m}$
• $\text{1}\text{pm =}\text{1}{\text{.0 ×10}}^{\text{-12}}\text{m}$

• $\begin{array}{l}\text{Dipolemoment,}\text{μ}\text{=}\text{Q}\text{×}\text{r}\\ \text{r=}\text{bond}\text{distance}\\ \text{ Q=}\text{Partial charge}\end{array}$

• • ${\text{1e}}^{\text{-}}=\text{1}\text{.66022}{\text{×10}}^{\text{-19}}\text{C}$

To determine: The dipole moment of $\text{AlF}$.