If HCl were purely ionic,its expected dipole | vedclass

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Question

(A) The percentage ionic character is calculated using the formula: $	ext{Percentage ionic character} = \left( \frac{	ext{Experimental dipole moment

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) The percentage ionic character is calculated using the formula: $	ext{Percentage ionic character} = \left( \frac{	ext{Experimental dipole moment}}{	ext{Theoretical dipole moment}} ight) 	imes 100$ Given: $	ext{Experimental dipole moment} = 1.03 \, D$ $	ext{Theoretical dipole moment} = 6.12 \, D$ Substituting the values: $	ext{Percentage ionic character} = \left( \frac{1.03}{6.12} ight) 	imes 100 \approx 16.83\% \approx 17\%$ Therefore,the correct option is $A$.

If $HCl$ were purely ionic,its expected dipole moment would be $6.12 \, D$. However,the experimental dipole moment is $1.03 \, D$. Calculate the percentage ionic character. (in $\%$)

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