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AMU Medical AMU Solved Paper-2008

  • question_answer
    If the electronegativity difference between two atoms A and B is 2.0, then the percentage of covalent character in the molecule is

    A)  54%                                      

    B)  46%

    C)                  23%                                      

    D)                  72%

    Correct Answer: B

    Solution :

                     According to Hannay and Smith equation \[\therefore \] %  ionic character\[=16\,({{x}_{A}}-{{x}_{B}})+3.5\,{{({{x}_{A}}-{{x}_{B}})}^{2}}\] where, \[{{x}_{A}}\] and \[{{x}_{B}}\] are the electronegativites of the atoms A and B respectively. \[\therefore \] % ionic character \[=16\,(2)+3.5{{(2)}^{2}}\]                                                 = 32 + 14 = 46%


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