Question

The dipole moment of HBr is 2.602 ×〖10〗^(-30)C m and its bond length is 141 pm. Calculate its percentage ionic character.

Answer 1

Question#46595 – Chemistry – Inorganic Chemistry

Question:

The dipole moment of HBr is $2.602 \times [10] \cdot (0.30) \, \text{C m}$ and its bond length is 141 pm. Calculate its percentage ionic character.

Answer:

For a given bond, the percent ionic character is given simply by

$$I = \frac{100 \mu_{obs}}{\mu_{ionic}}$$

Where $\mu_{obs}$ is actual or calculated dipole moment and $\mu_{ionic}$ is a dipole moment if the bond would be $100\%$ ionic. Simple formula for calculating the percent of ionic character (if $\mu$ expressed in D and bond length r expressed in pm):

$$I = \frac{10^4 \mu_{obs}}{4.8032r}$$

Where 4.8032 esu·Å is a fundamental charge.

To calculate dipole moment in D, we can divide $2.620 \cdot 10^{-30} \, \text{C·m}$ to $3.34 \cdot 10^{-30} \, \text{C·m/D}$:

$$\mu_{obs} = \frac{2.602 \cdot 10^{-30} \, \text{C·m}}{3.34 \cdot 10^{-30} \, \frac{\text{C·m}}{\text{D}}} = 0.78 \, \text{D}$$

Now using formula above:

$$I = \frac{10^4 \mu_{obs}}{4.8032r} = \frac{10^4 \times 0.78}{4.8032 \times 141} = 11.6\%$$