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Chemistry Question on Chemical bonding and molecular structure

Bond distance in HFHF is 9.17×10−11 m9.17 \times 10^{-11}\, m. Dipole moment of HFHF is 6.104×10−30 Cm6.10^4 \times 10^{-30}\, Cm . The percentage ionic character in HFHF will be : (electron charge =1.60×10−19 C= 1.60 \times 10^{-19}\, C)

A

61.0%61.0\%

B

38.0%38.0\%

C

35.5%35.5\%

D

41.5%41.5\%

Answer

41.5%41.5\%

Explanation

Solution

Givene e=1.60×10−19 Ce = 1.60 \times 10^{-19}\,C
d=9.17×10−11 md=9.17\times10^{-11}\,m
From μ=e×d\mu=e\times d
μ=1.60×10−19×9.17×10−11\mu=1.60\times10^{-19}\times9.17\times10^{-11}
=14.672×10−30=14.672\times10^{-30}
%\% ionic character
=Observed dipole momentDipole moment for 100% ionic bond=\frac{Observed\, dipole\, moment}{Dipole\, moment\, for\, 100\%\,ionic\, bond}
=6.104×10−3014.672×10−30×100=\frac{6.104\times10^{-30}}{14.672\times10^{-30}}\times100
=41.5%=41.5\%