question_answer3) Bond distance in HF is \[9.17\times {{10}^{-11}}\operatorname{m}.\] Dipole moment of HF is \[6.104\times {{10}^{-30}}\operatorname{Cm}.\] The percent ionic character in HF will be (electron charge = \[6.104\times {{10}^{-19}}\operatorname{C}\])
JEE Main Online Paper ( Held On 23 April 2013 )