Question

Bond distance in $HF$ is $9.17 \times 10^{-11}\, m$. Dipole moment of $HF$ is $6.10^4 \times 10^{-30}\, Cm$ . The percentage ionic character in $HF$ will be : (electron charge $= 1.60 \times 10^{-19}\, C$)

• A. $61.0\%$
• B. $38.0\%$
• C. $35.5\%$
• D. $41.5\%$

Answer 1

The Correct Option is D

To find the percentage ionic character of the HF molecule, we need to use the following formula:

\text{Ionic Character (\%)} = \left( \frac{\text{Measured Dipole Moment}}{\text{Dipole Moment if 100\% Ionic}} \right) \times 100\%

First, we calculate the dipole moment if HF were 100% ionic. For a completely ionic bond, the charge separation is equal to one full electron charge (e = 1.60 \times 10^{-19}\, C) and the bond distance is given as 9.17 \times 10^{-11}\, m. Hence,

\text{Dipole Moment if 100\% Ionic} = e \times \text{Bond Distance} = (1.60 \times 10^{-19}) \times (9.17 \times 10^{-11})\, Cm

Calculating this gives:

\text{Dipole Moment if 100\% Ionic} = 1.467 \times 10^{-29}\, Cm

The measured dipole moment of HF is given as 6.10^4 \times 10^{-30}\, Cm. Converting this to standard form:

6.10^4 \times 10^{-30}\, Cm = 6.10 \times 10^{-26}\, Cm

Now substitute these values into the formula for ionic character:

\text{Ionic Character (\%)} = \left( \frac{6.10 \times 10^{-26}}{1.467 \times 10^{-29}} \right) \times 100\% = 0.415 \times 100\% = 41.5\%

Therefore, the percentage ionic character of HF is 41.5\%.

Thus, the correct answer is:

41.5\%

This matches the correct option given in the question.