A) 2
B) 1.5
C) 2.5
D) 0
Correct Answer: B
Solution :
Number of electrons in\[O_{2}^{-}=8+8+1=17\] MO configuration of \[O_{2}^{-}=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\] \[\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\] \[\approx \pi 2p_{y}^{2},{{\pi }^{*}}2p_{x}^{2}\approx {{\pi }^{*}}2p_{y}^{1}\] Here, \[{{\text{N}}_{\text{b}}}\](bonding electrons) = 10 \[{{N}_{a}}\] (antibonding electrons) = 7 \[\because \] Bond order\[=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] \[\therefore \] Bond order \[=\frac{10-7}{2}=15\]
You need to login to perform this action.
You will be redirected in
3 sec
