A) \[{{N}_{2}}\]
B) \[{{O}_{2}}\]
C) \[H{{e}_{2}}\]
D) \[{{H}_{2}}\]
Correct Answer: A
Solution :
Molecular orbital configuration of \[{{N}_{2}}=14{{e}^{-}}=\sigma 1{{s}^{2}},\,\,\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\] \[\sigma 2{{p}_{x}}^{2},\pi 2{{p}_{y}}^{2},\pi 2{{p}_{z}}^{2}\] Bond order \[=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-4}{2}=3\] Similarly for \[{{O}_{2}}\] molecule, bond order \[=2\] For \[H{{e}_{2}}\] molecule, bond order \[=0\] For \[{{H}_{2}}\] molecule bond order \[=1\] So, \[{{N}_{2}}\] has highest bond order.
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