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VMMC VMMC Medical 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 \[=16\left( 2 \right)+3.5{{\left( 2 \right)}^{2}}\] \[=32+14=46%\]


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