Bond distance in HF is 9.17×10−11m Dipole moment of HF is 6.104×10−30C.m The percentage ionic character in HF will be (electron charge =1.60×10−19C)

μcal=e×d=1.60×10−19×9.17×10−11m=14.672×10−30C.mμobs=6.104×10−30C.m% of ionic character μobsμcal×100=6.104×10−3014.672×10−30×100=41.6%

Bond distance in HF is 9.17×10−11m Dipole moment of HF is 6.104×10−30C.m The percentage ionic character in HF will be (electron charge =1.60×10−19C)

μcal=e×d=1.60×10−19×9.17×10−11m=14.672×10−30C.mμobs=6.104×10−30C.m% of ionic character μobsμcal×100=6.104×10−3014.672×10−30×100=41.6%

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

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41.6%

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Detailed Solution

$μ_{c a l} = e \times d = 1.60 \times 10^{- 19} \times 9.17 \times 10^{- 11} m = 14.672 \times 10^{- 30} C . m$

$μ_{o b s} = 6.104 \times 10^{- 30} C . m$

% of ionic character $\frac{μ_{o b s}}{μ_{c a l}} \times 100 = \frac{6.104 \times 10^{- 30}}{14.672 \times 10^{- 30}} \times 100 = 41.6 %$

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Bond distance in HF is 9.17×10−11m Dipole moment of HF is 6.104×10−30C.m The percentage ionic character in HF will be (electron charge =1.60×10−19C)

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