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JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
                    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 )

    A)                   61.0 %

    B)                                                          38.0%

    C)                                          35.5 %                

    D)                                          41.5 %

    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 \[=\frac{\text{Observed dipole moment}}{\text{Dipole moment}\,\text{for}\,\text{100 }\!\!%\!\!\text{ }\,\text{ionic}\,\text{bond}}\] \[=41.5%\]


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