Chapter 6, Problem 6.14QP

Interpretation Introduction

Interpretation: It should be determined that the percent ionic character for the bond in $\text{BrF}$.

Concept introduction:

Conversion formula

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

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

• • ${\text{1e}}^{\text{-}}=\text{1}\text{.66022}{\text{×10}}^{\text{-19}}\text{C}$
  • $\text{Percent}\text{ionic}\text{charector=}\frac{\text{Measured}\text{dipole}\text{moment}}{\text{Expected}\text{dipole}\text{moment}}\text{×100}\text{\%}$

To determine: Percent ionic character of the bond in $\text{BrF}$.