A) \[NO\]
B) \[CO\]
C) \[{{O}_{2}}\]
D) \[{{B}_{2}}\]
Correct Answer: B
Solution :
| Those species, which have one or more than one unpaired electron, are paramagnetic in nature. The electronic configuration of given species are as follows: |
| [a] \[NO=15\] |
| \[=\sigma 1s_{,}^{2}{{\sigma }^{*}}1s_{,}^{2}\sigma 2s_{,}^{2}{{\sigma }^{*}}2{{s}^{2}}\sigma 2p_{z}^{2}\pi 2p_{x}^{2}=\pi 2p_{y}^{2}{{\pi }^{*}}2p_{x}^{1}\] |
| [b] \[CO=14\] |
| \[1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\sigma 2p_{z}^{2}\pi 2p_{x}^{2}=\pi 2p_{y}^{2}\] |
| [c] \[{{O}_{2}}=16=\sigma 1{{s}^{2}}\sigma *1{{s}^{2}}\sigma 2{{s}^{2}}\sigma *2{{s}^{2}}\sigma 2p_{z}^{2}\] |
| \[\pi 2p_{x}^{2}=\pi 2p_{y}^{2}\pi *2p_{x}^{1}=\pi *2p_{y}^{1}\] |
| [d]\[{{B}_{2}}=10=\sigma 1{{s}^{2}}\sigma *1{{s}^{2}}\sigma 2{{s}^{2}}\sigma *2{{s}^{2}}\pi 2p_{x}^{1}=\pi 2p_{y}^{1}\] |
| As all the electrons in \[CO\]are paired, so it is not a paramagnetic species. |
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