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}}Cm. $ 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 $ \mu =e\times d $
$ \mu =1.60\times {10^{-19}}\times 9.17\times {10^{-11}} $
$ =14.672\times {10^{-30}} $ % ionic character $ =\frac{Observeddipolemoment}{CalculatedDipolemoment}\times 100 $
$ =\frac{6.104\times {10^{-30}}}{14.672\times {10^{-30}}} $
$ =41.5% $ %
