A) diamagnetic
B) paramagnetic
C) ferromagnetic
D) ferromagnetic
Correct Answer: B
Solution :
According to MOT (molecular orbital theory) oxygen (at. no. 8) has the following orbital configuration \[{{O}_{2}}(16{{e}^{-}})=\sigma 1{{s}^{2}}{{\sigma }^{\centerdot }}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{\centerdot }}2{{s}^{2}}\sigma 2p_{z}^{2}\] \[(\pi 2{{p}_{x}}^{2}\pi 2{{p}_{y}}^{2}){{\pi }^{\centerdot }}2{{p}_{x}}^{1}{{\pi }^{*}}2{{p}_{y}}^{1}\] As oxygen has got two unpaired electrons, it is paramagnetic. The sequence of energies of homonuclear diatomic molecule (except\[{{\text{O}}_{\text{2}}}\] and\[{{\text{F}}_{2}}\]) is as follows \[\sigma \,1s\,<{{\sigma }^{*}}\,1s<\sigma 2s<{{\sigma }^{*}}\,2s\] \[<\left[ \begin{matrix} \pi & 2{{p}_{x}} \\ \pi & 2{{p}_{y}} \\ \end{matrix} \right]<\sigma 2{{p}_{z}}<\left[ \begin{matrix} {{\pi }^{*}} & 2{{p}_{x}} \\ {{\pi }^{*}} & 2{{p}_{y}} \\ \end{matrix} \right]<{{\sigma }^{*}}2{{p}_{z}}\] For oxygen and fluorine \[\sigma \,1s\,<{{\sigma }^{*}}1s<\sigma 2s<{{\sigma }^{*}}2s<\sigma 2{{p}_{z}}\] \[<\left[ \begin{matrix} \pi & 2{{p}_{x}} \\ \pi & 2{{p}_{y}} \\ \end{matrix} \right]<\left[ \begin{matrix} \pi & 2{{p}_{x}} \\ \pi & 2{{p}_{y}} \\ \end{matrix} \right]<{{\sigma }^{*}}2{{p}_{z}}\]
You need to login to perform this action.
You will be redirected in
3 sec
