A) \[{{N}_{2}}\]
B) \[{{O}_{2}}\]
C) \[H{{e}_{2}}\]
D) \[{{H}_{2}}\]
Correct Answer: A
Solution :
\[{{N}_{2}}(7+7=14{{e}^{-}})=K{{K}^{*}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\pi 2p_{x}^{2}\] \[\approx \pi 2p_{y}^{2},\sigma 2p_{z}^{2}\] \[BO=\frac{1}{2}(8-2)=3\] \[{{O}_{2}}(8+8=16{{e}^{-}})=K{{K}^{*}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p_{z}^{2},\] \[\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},{{\pi }^{*}}2p_{x}^{1}\approx {{\pi }^{*}}2p_{y}^{1}\] \[BO=\frac{1}{2}(8-4)=2\] \[H{{e}_{2}}(2+2=4{{e}^{-}})=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}}\] \[BO=\frac{1}{2}(2-2)=0\] \[{{H}_{2}}(1+1=2{{e}^{-}})=\sigma 1{{s}^{2}}\] \[BO=\frac{1}{2}(2-0)=1\] Hence, \[{{N}_{2}}\]has the highest bond order.
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