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(A) The percentage of ionic character is calculated using the formula: $	ext{Percentage of ionic character} = \frac{\mu_{	ext{observed}}}{\mu_{	ext

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) The percentage of ionic character is calculated using the formula: $	ext{Percentage of ionic character} = \frac{\mu_{	ext{observed}}}{\mu_{	ext{theoretical}}} 	imes 100$ Given: $\mu_{	ext{observed}} = 1.03 \ D$ $\mu_{	ext{theoretical}} = 6.12 \ D$ Substituting the values: $	ext{Percentage of ionic character} = \frac{1.03}{6.12} 	imes 100 \approx 16.83\% \approx 17\%$ Thus,the correct option is $A$.

If the $HCl$ molecule were completely polar,the expected dipole moment would be $6.12 \ D$,but the experimental value of the dipole moment is $1.03 \ D$. What is the percentage of ionic character (in $\%$)?

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