11th Class Chemistry Chemical Bonding and Molecular Structure

  • question_answer 76)   Using molecular orbital theory, compare the bond energy and magnetic character of \[O_{2}^{+}\] and \[O_{2}^{-}\] species.

    Answer:

      \[O_{2}^{+}:15\] \[KK\sigma {{(2s)}^{2}}{{\sigma }^{*}}{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}{{\pi }^{*}}{{(2{{p}_{x}})}^{1}}\] Bond order \[=\frac{8-3}{2}=2.5\]It is paramagnetic \[O_{2}^{-}:17\]\[KK\sigma {{(2s)}^{2}}{{\sigma }^{*}}{{(2s)}^{2}}\sigma (2{{p}_{z}})\pi {{(2{{p}_{x}})}^{2}}\]                 \[\pi {{(2{{p}_{y}})}^{2}}{{\pi }^{*}}{{(2{{p}_{x}})}^{2}}{{\pi }^{*}}{{(2{{p}_{y}})}^{1}}\] Bond order = \[\frac{8-5}{2}=1.5\]It is paramagnetic Bond order \[\propto \] Bond energy Thus, Bond energy of \[O_{2}^{+}\]> Bond energy of\[O_{2}^{-}\].


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