SATHEE: Chemical Bonding And Molecular Structure Question 433

Chemical Bonding And Molecular Structure Question 433

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

Options:

A) 61.0%

B) 38.0%

C) 35.5%

D) 41.5%

Show Answer

Answer:

Correct Answer: D

Solution:

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

$d = 9.17 \times 10^{- 11} m$ From $μ = e \times d$

$μ = 1.60 \times 10^{- 19} \times 9.17 \times 10^{- 11}$

$= 14.672 \times 10^{- 30}$ % ionic character $= \frac{O b s e r v e d d i p o l e m o m e n t}{C a l c u l a t e d D i p o l e m o m e n t} \times 100$

$= \frac{6.104 \times 10^{- 30}}{14.672 \times 10^{- 30}}$

$= 41.5$ %

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Chemical Bonding And Molecular Structure Question 433

Question: 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 )

Options:

Answer:

Solution:

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