Chemistry · 2013
JEE Main · 23 April 2013, Shift 1 · Q56
Bond distance in HF is 9.17 × 10^-11 m. Dipole moment of HF is 6.104 × 10^-30 Cm. The percent ionic character in HF will be: (electron charge =1.60 ×…
Bond distance in HF is $\displaystyle 9.17 \times 10^{-11} \mathrm{~m}$.
Dipole moment of HF is $\displaystyle 6.104 \times 10^{-30} \mathrm{Cm}$.
The percent ionic character in HF will be :
(electron charge $\displaystyle =1.60 \times 10^{-19} \mathrm{C}$ )
Official answer
(4)
41.$\displaystyle 5$ %
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