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JEE Main & Advanced Chemistry Chemical Bonding and Molecular Structure / रासायनिक आबंधन एवं आणविक संरचना Question Bank Self Evaluation Test - Chemical Bonding and Molecular Structure

  • question_answer
    Bond distance in HF is \[~9.17\times {{10}^{-11}}m.\] Dipole moment of HF is \[6.104\times {{10}^{-30}}Cm.\] The percentage ionic character in HF will be: (electron charge\[=1.60\times {{10}^{-19}}C\])

    A) 61.0%              

    B) 38.0%

    C) 35.5%                          

    D) 41.5%

    Correct Answer: D

    Solution :

    [d] 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 \[=\frac{Observed\text{ }dipole\text{ }moment}{Calculated\text{ }Dipole\text{ }moment}\times 100\] \[=\frac{6.104\times {{10}^{-30}}}{14.672\times {{10}^{-30}}}\] \[=41.5%\] %


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