A) 3
B) 0
C) 1
D) 2
Correct Answer: A
Solution :
Bond order \[=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] where,\[{{N}_{b}}=\]Number of bonding electrons \[{{N}_{a}}=\]Number of antibonding electrons \[{{N}_{2}}(14)=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},\pi 2p_{y}^{2}\] \[\approx \pi 2p_{z}^{2},\sigma 2p_{x}^{2}\] Bond order of\[{{N}_{2}}=\frac{10-4}{2}=3\]You need to login to perform this action.
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