A) NO
B) \[N{{O}^{+}}\]
C) \[N{{O}^{-}}\]
D) \[{{N}_{2}}\]
Correct Answer: A
Solution :
(a,c) NO = (7 + 8 = 15) \[=KK=\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\pi 2p_{x}^{2},\pi 2p_{x}^{2},\,_{2}^{*}2p_{z}^{1}\] \[N{{O}^{+}}=\] (7 + 8 - 1 = 14) \[=KK\,\,\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\pi 2p_{x}^{2},\pi 2p_{y}^{2},2p_{z}^{2}\] \[N{{O}^{-}}=\] (7 + 8 + 1 = 16) \[=KK\,\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p_{z}^{2},\pi 2p_{x}^{2},\pi 2p_{y}^{2},\] \[{{\pi }^{*}}2{{p}_{x}}^{1},{{\pi }^{*}}2p_{y}^{1}\] \[{{N}_{2}}=\] (7 + 7 = 14) \[=KK,\,\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\pi 2p_{x}^{2},\pi 2p_{y}^{2},\sigma 2p_{z}^{2}\] Due to presence of unpaired electrons in NO and \[N{{O}^{-}}\] both of these species are paramagnetic in nature.
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