A) \[{{N}_{3}}\]
B) \[{{O}_{2}}\]
C) \[H{{e}_{2}}\]
D) \[{{H}_{2}}\]
Correct Answer: A
Solution :
Molecular orbital configuration of \[{{N}_{2}}=14{{e}^{-}}=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\] \[\sigma 2p_{x}^{2},\pi 2p_{y}^{2},\pi 2p_{z}^{2}\] 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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