A) among the three, the one isoelectronic with\[{{N}_{2}}\]alone is diamagnetic
B) among the three, the one isoelectronic with\[{{O}_{2}}\]alone is paramagnetic
C) all the three species are paramagnetic
D) the three overlaps are stronger than\[{{N}_{2}}\]or \[{{O}_{2}}\]
Correct Answer: A
Solution :
The molecular orbital configuration of \[NO,N{{O}^{+}}\]and\[N{{O}^{-}}\]can be written as \[NO(7+8=15)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\text{ }\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\text{ }\sigma 2p_{z}^{2},\] \[\underset{(paramagnetic)}{\mathop{\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1}}}\,\] \[N{{O}^{+}}(7+8+1=19)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\] \[\pi 2P_{x}^{2}\approx \pi 2p_{y}^{2},\sigma 2p_{z}^{2}\] (\[\therefore \]It is diamagnetic as no unpaired electron is present). \[N{{O}^{-}}(7+8+1=16)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\] \[\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] (\[\because \]It is paramagnetic as contains unpaired electrons). \[N{{O}^{+}}\]is isoelectronic with\[{{N}_{2}},\]ie, 14 electrons and\[N{{O}^{-}}\]is isoelectronic with\[{{O}_{2}},\]ie, 16 electrons but\[N{{O}^{-}}\]and NO both are paramagnetic.
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