Bond distance in HF is $$9.17 \times 10^{-11}$$ m. Dipole moment of HF is $$6.104 \times 10^{-30}$$ Cm. The percentage ionic character in HF will be : (electron charge = $$1.60 \times 10^{-19}$$ C)

The percentage ionic character in a bond is given by the formula:

$$ \text{Percentage ionic character} = \left( \frac{\mu_{\text{actual}}}{\mu_{\text{ionic}}} \right) \times 100\% $$

where $$\mu_{\text{actual}}$$ is the measured dipole moment and $$\mu_{\text{ionic}}$$ is the dipole moment if the bond were completely ionic. For a completely ionic bond, $$\mu_{\text{ionic}}$$ is calculated as the product of the charge of an electron and the bond distance.

Given:

• Bond distance, $$d = 9.17 \times 10^{-11}$$ m
• Dipole moment of HF, $$\mu_{\text{actual}} = 6.104 \times 10^{-30}$$ Cm
• Electron charge, $$e = 1.60 \times 10^{-19}$$ C

First, calculate $$\mu_{\text{ionic}}$$:

$$ \mu_{\text{ionic}} = e \times d = (1.60 \times 10^{-19}) \times (9.17 \times 10^{-11}) $$

Multiply the coefficients: $$1.60 \times 9.17 = 14.672$$.

Multiply the exponents: $$10^{-19} \times 10^{-11} = 10^{-30}$$.

So,

$$ \mu_{\text{ionic}} = 14.672 \times 10^{-30} \text{ Cm} $$

The given values for electron charge and bond distance have three significant figures. Therefore, $$\mu_{\text{ionic}}$$ should be rounded to three significant figures. Since $$14.672$$ rounds to $$14.7$$, we have:

$$ \mu_{\text{ionic}} = 14.7 \times 10^{-30} \text{ Cm} $$

Now, substitute the values into the percentage ionic character formula:

$$ \text{Percentage ionic character} = \left( \frac{6.104 \times 10^{-30}}{14.7 \times 10^{-30}} \right) \times 100\% $$

The $$10^{-30}$$ terms cancel out:

$$ = \left( \frac{6.104}{14.7} \right) \times 100\% $$

Now, divide $$6.104$$ by $$14.7$$:

$$ \frac{6.104}{14.7} = ? $$

Perform the division step by step. Multiply both numerator and denominator by 10 to simplify:

$$ \frac{6.104 \times 10}{14.7 \times 10} = \frac{60.84}{147} $$

Now, divide $$60.84$$ by $$147$$:

$$ 147 \times 0.4 = 58.8 $$

$$ 60.84 - 58.8 = 2.04 $$

Bring down a zero: $$20.4$$ (since $$60.84$$ has two decimal places).

$$ 147 \times 0.01 = 1.47 $$

$$ 2.04 - 1.47 = 0.57 $$ (but note we are working with 20.4, so adjust)

Actually, $$147 \times 0.414 = 147 \times 0.4 = 58.8$$, $$147 \times 0.014 = 2.058$$, total $$58.8 + 2.058 = 60.858$$

$$ 60.84 - 60.858 = -0.018 $$ (slight overestimate, so adjust)

Better to compute directly:

$$ \frac{60.84}{147} \approx 0.41402 $$

After detailed calculation (using long division or calculator), $$6.104 \div 14.7 \approx 0.415238$$.

Thus,

$$ \frac{6.104}{14.7} \approx 0.415238 $$

Now multiply by $$100\%$$:

$$ 0.415238 \times 100 = 41.5238\% $$

Rounding to three significant figures (as per input data), $$41.5238\% \approx 41.5\%$$.

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

Hence, the correct answer is Option D.