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

Question

Bond distance in HF is 9.17 × 10-11 m. Dipole moment of HF is 6.104 × 10-30 Cm . The percentage ionic character in HF will be: (electron charge = 1.60 × 10-19 C) (A) 61.0 % (B) 38.0 % (C) 35.5 % (D) 41.5 %

Tardigrade Admin · Accepted Answer

Givene e = 1.60 × 10-19 C d=9.17×10-11 m From μ=e× d μ=1.60×10-19×9.17×10-11 =14.672×10-30 % ionic character =(Observed dipole moment/Dipole moment for 100 % ionic bond) =(6.104×10-30/14.672×10-30)×100 =41.5 %

Q. Bond distance in $H F$ is $9.17 \times 1 0^{- 11} m$ . Dipole moment of $H F$ is $6.1 0^{4} \times 1 0^{- 30} C m$ . The percentage ionic character in $H F$ will be : (electron charge $= 1.60 \times 1 0^{- 19} C$ )

$61.0%$

$38.0%$

$35.5%$

$41.5%$

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

61.0%

38.0%

35.5%

41.5%

Givene e=1.60×10−19C d=9.17×10−11m From μ=e×d μ=1.60×10−19×9.17×10−11 =14.672×10−30 % ionic character =Dipolemomentfor100%ionicbondObserveddipolemoment​ =14.672×10−306.104×10−30​×100 =41.5%

Q. Bond distance in $H F$ is $9.17 \times 1 0^{- 11} m$ . Dipole moment of $H F$ is $6.1 0^{4} \times 1 0^{- 30} C m$ . The percentage ionic character in $H F$ will be : (electron charge $= 1.60 \times 1 0^{- 19} C$ )

$61.0%$

$38.0%$

$35.5%$

$41.5%$