A) \[{{B}_{2}}\]
B) \[{{C}_{2}}\]
C) \[{{N}_{2}}\]
D) \[{{F}_{2}}\]
Correct Answer: A
Solution :
Paramagnetic property arise through unpaired electron. \[{{B}_{2}}\] molecule have the unpaired electron so it show paramagnetism. \[{{B}_{2}}\to \]\[\underset{\text{(2 unpaired electron)}}{\mathop{\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}},\pi 2{{p}_{x}}^{1}=\pi 2{{p}_{y}}^{1}}}\,\] \[{{C}_{2}}\to \] \[\underset{\text{(No unpaired electron)}}{\mathop{\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}},\pi 2{{p}_{x}}^{2}.\pi 2{{p}_{y}}^{2}}}\,\] \[{{N}_{2}}\to \] \[\underset{\text{(No unpaired electron)}}{\mathop{\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}},\sigma 2{{p}_{x}}^{2},\pi 2{{p}_{y}}^{2}\pi 2{{p}_{z}}^{2}}}\,\] \[{{F}_{2}}\to \]\[\underset{\text{(No unpaired electron)}}{\mathop{\sigma {{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2{{p}_{x}}^{2},\pi 2{{p}_{y}}^{2},\pi 2{{p}_{z}}^{2},}}\,\] \[{{\pi }^{*}}2{{p}_{y}}^{2},{{\pi }^{*}}2{{p}_{z}}^{2}\] So only \[{{B}_{2}}\] exist unpaired electron and show the paramagnetism.
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