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BCECE Engineering BCECE Engineering Solved Paper-2015

  • question_answer
    If the dipole moment of \[\text{HBr}\]is  \[2.60\times {{10}^{-30}}\text{Cm}\]and the interatomic spacing is \[1.41\overset{\text{o}}{\mathop{\text{A}}}\,,\]then the per cent ionic character of \[\text{HBr}\]is

    A) \[16.23%\]                         

    B) \[13.21%\]

    C) \[11.50%\]         

    D) \[15.81%\]

    Correct Answer: C

    Solution :

    Theoretical value of dipole moment of a 100% ionic character = e x d \[=(1.60\times {{10}^{-19}}C)(1.41\times {{10}^{-10}}m)\] \[=2.26\times {{10}^{-29}}Cm\] Observed value of dipole moment \[=2.60\times {{10}^{-30}}Cm\] \[\therefore \]  Per cent ionic character \[\text{=}\frac{\text{Obersved value}}{\text{Theortical value}}\text{ }\!\!\times\!\!\text{ 100}\] \[=\frac{2.60\times {{10}^{-30}}}{2.26\times {{10}^{-29}}}\times 100=11.5%\]


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