A) \[{{F}_{2}}\]
B) \[{{B}_{2}}\]
C) \[L{{i}_{2}}\]
D) \[{{N}_{2}}\]
Correct Answer: B
Solution :
Molecules in which unpaired electrons are present, are paramagnetic while that have absence of unpaired electrons, are diamagnetic. \[{{F}_{2}}(18)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}}\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2},\] \[\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{2}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{2}\] (No unpaired electron, so diamagnetic) \[{{B}_{2}}(10)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}}\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\pi 2p_{x}^{1}\approx \pi 2p_{y}^{1}\] (Two unpaired electrons, paramagnetic) \[L{{i}_{2}}(6)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}}\] (No unpaired electrons, diamagnetic) \[{{N}_{2}}(14)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\pi 2p_{x}^{2}\] \[\approx \pi 2p_{y}^{2},\sigma 2p_{z}^{2}\] (No unpaired electrons, diamagnetic)
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