Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
The species having bond order three are (1) \[C{{N}^{-}}\] (2) \[O_{2}^{-}\] (3) \[N{{O}^{+}}\] (4) \[C{{N}^{+}}\]A) 1, 2 and 3 are correct.
B) 1 and 2 are correct.
C) 2 and 4 are correct.
D) 1 and 3 are correct.
Correct Answer: D
Solution :
\[C{{N}^{-}}(6+7+1=14)=\] \[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2}\] \[B.O=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-4}{2}=3\] \[O_{2}^{-}(8+8+1=17)=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\] \[\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},{{\pi }^{*}}2P_{y}^{1}\] \[B.O=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-7}{2}=1.5\] \[N{{O}^{+}}(7+8-1=14)=\] \[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p_{z}^{2},\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2}\] \[B.O.=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-4}{2}=3\] \[C{{N}^{+}}(6+7-1=12)\] \[=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},\sigma 2{{s}^{2}},\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2}\] \[B.O.=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{8-4}{2}=2\] \[\therefore \]\[C{{N}^{-}}\]and\[N{{O}^{+}}\]have bond order three.
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