A) \[O_{2}^{+}<O_{2}^{2-}<O_{2}^{-}<{{O}_{2}}\]
B) \[O_{2}^{2-}<O_{2}^{-}<{{O}_{2}}<O_{2}^{+}\]
C) \[{{O}_{2}}<O_{2}^{+}<O_{2}^{2-}<O_{2}^{-}\]
D) \[O_{2}^{2-}<O_{2}^{-}<O_{2}^{+}<{{O}_{2}}\]
Correct Answer: B
Solution :
The MO configuration of 02 can be written as \[{{O}_{2}}(8+8=16)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p_{z}^{2},\] \[\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\,\,\overset{*}{\mathop{\pi }}\,2p_{x}^{1}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] \[\therefore \] Bond order \[\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-6}{2}=2\] Similarly, \[O_{2}^{+}(8+8-1=15)\], \[BO=\frac{10-5}{2}=2.5\] \[O_{2}^{-}(8+8+1=17)\], \[BO=\frac{10-7}{2}=1.5\] \[O_{2}^{2-}(8+8+2=18)\], \[BO=\frac{10-8}{2}=1\] Thus, the order of bond order (BO) and bond dissociation energy is \[O_{2}^{2-}<O_{2}^{-}<{{O}_{2}}<O_{2}^{+}\]
You need to login to perform this action.
You will be redirected in
3 sec
