A) \[{{B}_{2}}\]
B) \[{{N}_{2}}\]
C) \[He_{2}^{+}\]
D) \[N{{e}_{2}}\]
Correct Answer: D
Solution :
\[N{{e}_{2}}\] molecule does not exist, because it has zero bond order. A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule. \[{{B}_{2}}\] (10 electrons) \[=\sigma 1{{s}^{2}}\overset{*}{\mathop{\sigma }}\,1{{s}^{2}}\sigma 2{{s}^{2}}\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\pi 2p_{x}^{1}\] \[=\pi 2p{{y}^{1}}\] Bond order \[=\frac{1}{2}({{N}_{b}}-{{N}_{a}})\] \[=\frac{6-4}{2}=1\] \[He_{2}^{+}\] (3 electrons) \[=\sigma 1{{s}^{2}}\overset{*}{\mathop{\sigma }}\,1{{s}^{1}}\] Bond order \[=\frac{1}{2}({{N}_{b}}-{{N}_{a}})=\frac{1}{2}\times (2-1)\] \[=\frac{1}{2}\] = 0.5 \[{{N}_{2}}\] (14 electrons) \[=\sigma 1{{s}^{2}}\overset{*}{\mathop{\sigma }}\,1{{s}^{2}}\sigma 2{{s}^{2}}\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\sigma 2p_{z}^{2}\pi 2p_{x}^{2}=\pi 2p_{y}^{2}\] Bond order \[=\frac{1}{2}(10-4)=3\] \[N{{e}_{2}}\] (20 electrons) \[=\sigma 1{{s}^{2}}\overset{*}{\mathop{\sigma }}\,1{{s}^{2}}\], \[\sigma 2{{s}^{2}}\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2}\,\,\pi 2p_{x}^{2}=\pi 2p_{y}^{2}\] \[{{\pi }^{*}}2p_{x}^{2}={{\pi }^{*}}2p_{y}^{2}{{\sigma }^{*}}2p_{z}^{2}\] Bond order \[=\frac{1}{2}(10-10)=0\] Hence, \[N{{e}_{2}}\] molecule does not exist.
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