A) 1
B) 2
C) 1.5
D) 2.5
Correct Answer: B
Solution :
\[{{O}_{2}}(16)=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p_{x}^{2},\pi 2p_{y}^{2}\] \[\approx \pi 2p_{z}^{2}{{\pi }^{*}}2p_{y}^{1}\approx {{\pi }^{*}}2p_{z}^{1}\] Number of bonding electrons\[-\] Bond order \[=\frac{Number\text{ }of\text{ }antibondine\text{ }electron}{2}\] \[=\frac{10-6}{2}=\frac{4}{2}=2\]
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