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JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    Increasing order of bond strength of \[{{O}_{2}},O_{2}^{-}\,O_{2}^{2-}\] and \[O_{2}^{+}\] is   AIEEE  Solved  Paper-2002

    A) \[O_{2}^{+}<{{O}_{2}}<O_{2}^{-}<O_{2}^{2-}\]   

    B) \[{{O}_{2}}<O_{2}^{+}<O_{2}^{-}<O_{2}^{2-}\]

    C) \[O_{2}^{-}<O_{2}^{2-}<O_{2}^{+}<{{O}_{2}}\]

    D) \[O_{2}^{2-}<O_{2}^{-}<{{O}_{2}}<O_{2}^{+}\]

    Correct Answer: D

    Solution :

    Higher the bond order greater the bond strength. Higher the bond order greater the bond strength
    Species Bond order
    \[O_{2}^{2-}\] 1
    \[O_{2}^{-}\] 1.5
    \[{{O}_{2}}\] 2.0
    \[O_{2}^{+}\] 2.5
    \[{{O}_{2}}\] molecule has 16 electrons. The electronic configuration of \[{{O}_{2}}\] molecule is              \[[\sigma {{(1s)}^{2}}],\,\,[\overset{*}{\mathop{\sigma }}\,{{(1s)}^{2}}],\,\,[\sigma {{(2s)}^{2}}],\,\,[\overset{*}{\mathop{\sigma }}\,{{(2s)}^{2}}],\,\,[\sigma {{(2{{p}_{z}})}^{2}}]\], \[{{[\pi {{(2{{p}_{x}})}^{2}}=\pi {{(2{{p}_{y}})}^{2}}]}^{2}},\,\,[\pi \,{{(2{{p}_{x}})}^{1}}={{\pi }^{*}}{{(2{{p}_{y}})}^{1}}]\] Here, \[{{N}_{b}}=10\] and \[{{N}_{a}}=6\] and bond order\[=\frac{1}{2}({{N}_{b}}-{{N}_{a}})=\frac{1}{2}(10-6)=2\] \[O_{2}^{10-}\] ion (superoxide ion) is formed by the gain of one electron by \[{{O}_{2}}\] molecules electronic configuration will be \[[\sigma {{(1s)}^{2}}]\],       \[[{{\sigma }^{*}}{{(1s)}^{2}}]\],     \[[\sigma {{(2s)}^{2}}]\],   \[[{{\sigma }^{*}}{{(2s)}^{2}}]\], \[{{[\pi {{(2{{p}_{x}})}^{2}}=\pi (2{{p}_{y}})]}^{2}}\],  \[[{{\pi }^{*}}{{(2{{p}_{x}})}^{2}}={{\pi }^{*}}{{(2{{p}_{y}})}^{1}}]\] Here, \[{{N}_{b}}=10\],                    \[{{N}_{a}}=7\] Bond order \[=\frac{1}{2}({{N}_{b}}-{{N}_{a}})=\frac{1}{2}(10-7)=\frac{3}{2}=1.5\] \[O_{2}^{-}\] (peroxide ion) is formed when 0^ gains two electrons. Electronic configuration will be              \[[\sigma {{(1s)}^{2}}],\,\,[{{\sigma }^{*}}{{(1s)}^{2}}],\,\,[\sigma {{(2s)}^{2}}],\,\,{{[{{\sigma }^{*}}(2s)]}^{2}},\,\,\,{{[\sigma (2{{p}_{z}})]}^{2}},\]              \[{{[\pi {{(2{{p}_{x}})}^{2}}=\pi (2{{p}_{y}})]}^{2}},{{[{{\pi }^{*}}{{(2{{p}_{x}})}^{2}}={{\pi }^{*}}(2{{p}_{y}})]}^{2}}\] Here \[{{N}_{b}}=10\], and \[{{N}_{a}}=8\] Bond order \[=\frac{1}{2}({{N}_{b}}-{{N}_{a}})=\frac{1}{2}(10-8)=1\] \[{{O}_{2}}^{+}\] ion is formed by the loss of one electron from \[{{O}_{2}}\] molecule              \[[\sigma {{(1s)}^{2}}][\sigma {{(1s)}^{2}}]{{[\sigma (2s)]}^{2}}[{{\sigma }^{*}}{{(2s)}^{2}}]{{[\sigma (2{{p}_{z}})]}^{2}}{{[\pi (2{{p}_{x}})]}^{2}}\] \[{{[\pi (2{{p}_{y}})]}^{2}}[{{\pi }^{*}}{{(2{{p}_{x}})}^{1}}]\] Here, \[{{N}_{b}}=10\] and\[{{N}_{a}}=5\] so Bond order \[=\frac{1}{2}({{N}_{b}}-{{N}_{a}})\]                    \[=\frac{1}{2}(10-5)=\frac{5}{2}=2.5\] So, the species can be arrange on the basis of their increasing bond order as follows
    Species Bond order
    \[O_{2}^{2-}\] 1
    \[O_{2}^{-}\] 1.5
    \[{{O}_{2}}\] 2.0
    \[O_{2}^{+}\] 2.5
    As we know that higher the bond order, higher will be the bond strength thus order in accordance with bond strength is                    \[O_{2}^{2-}<{{O}_{2}}^{-}<{{O}_{2}}<O_{2}^{+}\].


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