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JEE Mains · Chemistry · STD 11 - 4. Chemical bonding and molecular structure
Bond distance in \(HF\) is \(9.17\times10^{-11}\, m\). Dipole moment of \(HF\) is \(6.104\times10^{- 30}\, Cm\). The percentage ionic character in \(HF\) will be : .............. \(\%\) ( electron charge \(= 1.60\times10^{-19}\, C\))
- A \(61\)
- B \(38\)
- C \(35.5\)
- D \(41.5\)
Answer & Solution
Correct Answer
(D) \(41.5\)
Step-by-step Solution
Detailed explanation
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…
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