A) \[O_{2}^{-}>{{O}_{2}}>O_{2}^{+}\]
B) \[O_{2}^{-}<{{O}_{2}}<O_{2}^{+}\]
C) \[O_{2}^{-}>{{O}_{2}}<O_{2}^{+}\]
D) \[O_{2}^{-}<{{O}_{2}}>O_{2}^{+}\]
Correct Answer: D
Solution :
| Bond order of \[O_{2}^{-}\] |
| \[O_{2}^{-}=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}}\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\sigma 2p_{z}^{2}(\pi 2p_{x}^{2}=\pi 2p_{y}^{2})\]\[({{\pi }^{*}}2p_{x}^{2}={{\pi }^{*}}2p_{y}^{1})\] |
| Bond order |
| \[\frac{\text{number of electrons in BMO - number of elections ABMO}}{2}\]\[=\frac{10-7}{2}=\frac{3}{2}=1.5\] |
| \[O_{2}^{+}=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\sigma 2p_{z}^{2}\] |
| \[(\pi 2p_{x}^{2}=\pi 2p_{y}^{2})({{\pi }^{*}}2p_{x}^{1}={{\pi }^{*}}2p_{y}^{0})\] |
| \[BO=\frac{10-5}{2}=\frac{5}{2}=2.5\] |
| \[{{O}_{2}}=\sigma 1{{s}^{2}}\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\sigma 2p_{z}^{2}\,\,\,\,(\pi 2p_{x}^{2}=\pi 2p_{y}^{2})\] \[(\overset{*}{\mathop{\pi }}\,2p_{x}^{1}=\overset{*}{\mathop{\pi }}\,2p_{y}^{1})\] |
| \[BO=\frac{10-6}{2}=\frac{4}{2}=2\] |
| So, the correct sequence is |
| \[O_{2}^{-}<{{O}_{2}}<O_{2}^{+}\] |
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