A) \[O_{2}^{+}\]
B) \[O_{2}^{-}\]
C) \[O_{2}^{2-}\]
D) \[O_{2}^{{}}\]
Correct Answer: B
Solution :
| [a] MO configuration of \[O_{4}^{+}(8+8-1=15)\] |
| \[=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2s,\sigma 2p_{z}^{2},\] |
| \[\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{0}\] |
| \[BO=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] |
| (where, \[{{N}_{b}}=\] number of electrons in bonding molecular orbital \[{{N}_{a}}=\] number of electrons in anti-bonding molecular orbital |
| \[\therefore \] \[BO=\frac{10-5}{2}=2.5\]Similarly, |
| [b]\[O_{2}^{-}(8+8+1=17)\] |
| so\[BO=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-7}{2}=1.5\] |
| [c]\[O_{2}^{2=}(8+8+2=18)\] |
| \[BO=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-8}{2}=1\] |
| [d]\[{{O}_{2}}(8+8=16)\] |
| \[BO=\frac{10-6}{2}=2\] |
| Thus, \[O_{2}^{-}\] shows the bond order 1.5. |
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