A) \[N_{2}^{2-}<N_{2}^{-}<{{N}_{2}}\]
B) \[{{N}_{2}}<N_{2}^{2-}<N_{2}^{-}\]
C) \[N_{2}^{-}<N_{2}^{2-}<{{N}_{2}}\]
D) \[N_{2}^{-}<{{N}_{2}}<N_{2}^{2-}\]
Correct Answer: A
Solution :
| Molecular orbital configuration of |
| \[N_{2}^{2-}=\sigma 1{{s}^{2}}\sigma *1{{s}^{2}}\sigma 2{{s}^{2}}\sigma *2{{s}^{2}}\,\,\,\,\sigma 2p_{z}^{2}\] |
| \[\pi 2p_{x}^{2}=\pi 2p_{y}^{2}\,\pi *2p_{x}^{1}=\pi *2p_{y}^{1}\] |
| Bond order \[=\frac{10-6}{2}=2\] |
| \[N_{2}^{-}=\sigma 1{{s}^{2}}\sigma *1{{s}^{2}}\sigma 2{{s}^{2}}\sigma *2{{s}^{2}}\,\,\,\sigma 2p_{z}^{2}\] |
| \[\pi 2p_{x}^{2}=\pi 2p_{y}^{2}\pi *2p_{x}^{1}=\pi *2p_{y}^{0}\] |
| Bond order \[=\frac{10-5}{2}=2.5\] |
| \[{{N}_{2}}=\sigma 1{{s}^{2}}\sigma *1{{s}^{2}}\sigma 2{{s}^{2}}\sigma *2{{s}^{2}}\pi 2p_{x}^{2}=\pi 2p_{y}^{2}\,\sigma 2p_{z}^{2}\] |
| Bond order \[=\frac{10-4}{2}=3\] |
| \[\therefore \] The correct order is \[N_{2}^{2-}<N_{2}^{-}<{{N}_{2}}\] |
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