Bond distance in HF is 9.17 times 10^{-11} m | vedclass

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(D) Given:Charge $e = 1.60 	imes 10^{-19} \ C$Bond distance $d = 9.17 	imes 10^{-11} \ m$Observed dipole moment $\mu_{obs} = 6.104 	imes 10^{-30} \

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$41.5$

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(D) Given:Charge $e = 1.60 	imes 10^{-19} \ C$Bond distance $d = 9.17 	imes 10^{-11} \ m$Observed dipole moment $\mu_{obs} = 6.104 	imes 10^{-30} \ Cm$Calculate theoretical dipole moment for $100\%$ ionic character $(\mu_{theo})$:$\mu_{theo} = e 	imes d$$\mu_{theo} = 1.60 	imes 10^{-19} 	imes 9.17 	imes 10^{-11} = 14.672 	imes 10^{-30} \ Cm$Calculate percentage ionic character:$	ext{Percentage ionic character} = \frac{\mu_{obs}}{\mu_{theo}} 	imes 100$$= \frac{6.104 	imes 10^{-30}}{14.672 	imes 10^{-30}} 	imes 100$$= 0.416 	imes 100 \approx 41.5\%$

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 $= 1.60 \times 10^{-19} \ C$)

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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 $= 1.60 \times 10^{-19} \ C$)