$$\% \text{ ionic character} = \frac{\mu_{obs}}{\mu_{calc}} \times 100$$
where \(\mu_{obs}\) is the observed dipole moment and \(\mu_{calc}\) is the calculated dipole moment for a purely covalent bond.
The calculated dipole moment for a purely covalent bond can be calculated using the following equation:
$$\mu_{calc} = q \times d$$
where \(q\) is the charge of each atom and \(d\) is the bond distance.
In this case, we have a bond distance of 161 pm and a dipole moment of 0.380 D. The charge of each atom is +1 for hydrogen and -1 for fluorine.
Substituting these values into the equations, we get:
$$\mu_{calc} = (1.602 \times 10^{-19} \text{ C}) \times (161 \times 10^{-12} \text{ m}) = 2.58 \times 10^{-29} \text{ C m}$$
$$\% \text{ ionic character} = \frac{0.380 \text{ D}}{2.58 \times 10^{-29} \text{ C m}} \times 100 = 14.7\%$$
Therefore, the percent ionic character of the bond is 14.7%.
