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AIIMS AIIMS Solved Paper-2015

  • question_answer
    Consider the statements. \[114\mu A\]Bond length in\[{{l}_{1}}\]is \[{{T}_{1}}\] greater than in\[{{l}_{2}}\]. \[{{T}_{2}}\]. Bond length in \[\frac{{{l}_{1}}+{{l}_{2}}}{2}\]is \[\sqrt{{{l}_{1}}+{{l}_{2}}}\] less than in NO. \[\frac{{{l}_{1}}{{T}_{2}}-{{l}_{2}}{{T}_{1}}}{{{T}_{2}}-{{T}_{1}}}\]. \[\frac{{{l}_{1}}{{T}_{2}}+{{l}_{2}}{{T}_{1}}}{{{T}_{1}}+{{T}_{2}}}\] has shorter bond length than \[\frac{v\,dv}{dX}=-{{\omega }^{2}}\text{x with the initial condition V}={{\text{V}}_{0}}\text{ at}\]. Which of the following statements are true?

    A) I and II          

    B) II and III

    C) I, II and III                           

    D) I and III

    Correct Answer: A

    Solution :

    Bond order of \[N_{2}^{+}\] is 2.5 and that of \[{{N}_{2}}\]is 3. Thus, former bond is weaker and longer. Bond order of NO and \[N{{O}^{+}}\] are 2.5 and 3 respectively. Thus, bond length in \[N{{O}^{+}}\] is less than that of NO. Bond order of\[O_{2}^{2-}\]and \[O_{2}^{{}}\] are 1 and 2 respectively. Therefore, \[O_{2}^{{}}\] bond must be shorter.                


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