JEE Main & Advanced Chemistry Classification of Elements and Periodicity in Properties / तत्त्वों का वर्गीकरण एवं गुणों में आवर्ति Sample Paper Topic Test - Periodic Table

  • 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 :

    [b] The percentage ionic character is related to electro negativity difference between the two atoms. It is given by Hannay and Smith relationship % age ionic character\[=16\,({{X}_{A}}-{{X}_{B}})+3.5\,{{({{X}_{A}}-{{X}_{B}})}^{2}}\] Here \[{{X}_{A}}\] and \[{{X}_{B}}\] are electro negativities of the two atoms A and B respectively. \[\therefore \] % ionic character \[=16(2)+3.5\,{{(2)}^{2}}=46%\]


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