A) \[O_{2}^{+}>{{O}_{2}}>O_{2}^{2-}>O_{2}^{-}\]
B) \[O_{2}^{2-}>O_{2}^{-}>{{O}_{2}}>O_{2}^{+}\]
C) \[O_{2}^{+}>{{O}_{2}}>O_{2}^{-}>O_{2}^{2-}\]
D) \[{{O}_{2}}>O_{2}^{+}>O_{2}^{-}>O_{2}^{2-}\]
Correct Answer: C
Solution :
Electronic configuration of\[O_{2}^{+}\]: \[\sigma 1{{s}^{2}},\,\,\sigma *1{{s}^{2}},\,\,\sigma 2{{s}^{2}},\,\,\sigma *2{{s}^{2}},\,\,\sigma 2p_{x}^{2},\,\,\pi 2_{y}^{2}\] \[=\pi 2p_{z}^{2},\,\,\pi *2p_{y}^{1}=\pi *2p_{z}^{0}\] \[\therefore \]bond order of\[O_{2}^{+}=\frac{1}{2}\](bonding electron - anti bonding electron) \[=\frac{1}{2}(10-5)\] \[=\frac{1}{2}\times 5=2.5\] Electronic configuration of\[O_{2}^{2-}\]: \[\sigma 1{{s}^{2}},\,\,\sigma *1{{s}^{2}},\,\,\sigma 2{{s}^{2}},\,\,\sigma *2{{s}^{2}},\,\,\sigma 2p_{x}^{2},\,\,\pi 2_{y}^{2}\] \[=\pi 2p_{z}^{2},\,\,\pi *2p_{y}^{1}=\pi *2p_{z}^{1}\] \[\therefore \]bond order of\[{{O}_{2}}=\frac{1}{2}(10-6)\]\[=\frac{1}{2}\times 4=2\] Electronic configuration of\[O_{2}^{2-}\]: \[\sigma 1{{s}^{2}},\,\,\sigma *1{{s}^{2}},\,\,\sigma 2{{s}^{2}},\,\,\sigma *2{{s}^{2}},\,\,\sigma 2p_{x}^{2},\,\,\pi 2_{y}^{2}\] \[=\pi 2p_{z}^{2},\,\,\pi *2p_{y}^{1}=\pi *2p_{z}^{2}\] \[\therefore \]bond order of\[O_{2}^{2-}=\frac{1}{2}(10-8)\] \[=\frac{1}{2}\times 2=1\] Electronic configuration of\[O_{2}^{-}\]: \[\sigma 1{{s}^{2}},\,\,\sigma *1{{s}^{2}},\,\,\sigma 2{{s}^{2}},\,\,\sigma *2{{s}^{2}},\,\,\sigma 2p_{x}^{2},\,\,\pi 2_{y}^{2}\] \[=\pi 2p_{z}^{2},\,\,\pi *2p_{y}^{1}=\pi *2p_{z}^{1}\] \[\therefore \]bond order of\[O_{2}^{-}=\frac{1}{2}(10-7)\]\[=\frac{1}{2}\times 3=1.5\] Hence, the bond order of\[O_{2}^{+},\,\,{{O}_{2}},\,\,O_{2}^{2-}\]and\[O_{2}^{-}\]varies in the order: \[O_{2}^{+}>{{O}_{2}}>O_{2}^{-}>O_{2}^{2-}\]
You need to login to perform this action.
You will be redirected in
3 sec
