Discussion Forum

Which of the following species is not paramagnetic?

Answer:

Explanation:

To identify the magnetic nature we need to check the molecular orbital configuration. If all orbitals are fully occupied, species is diamagnetic while when one or  more molecular orbitals is/are singly occupied , species is paramagnetic

(a) $NO (7+8=15)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\pi2p_x^2$

           $=\pi2p_y^2,\pi2p_z^2,\pi^{*}2p_x^1$ $=\pi^{*}2p_y^0$

   One unpaired electron is present.

              Hence , it is paramagnetic

(b)  $CO(6+8=14)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\pi2p_x^2$

      $=\pi2p_y^2,\sigma2p_z^2$     

    No unpaired electron is present

   Hence , it is diamagenetic

(c)    $O_{2}(8+8=16)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\sigma2p_z^2,\pi2p_x^2$

        $=\pi2p_x^2,\pi^{*}2p_x^1$ $=\pi^{*}2p_y^1$

  Two unpaired electrons are present. Hence it is paramagnetic

(d)      $B_{2}(5+5)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\pi2p_x^1$

          $=\pi2p_x^2,\pi^{*}2p_x^1$

           $=\pi2p_y^1$

     Two unpaired electrond are present. Hence it is paramagnetic