A) Paramagnetic and bond order \[<O_{2}^{-}\]
B) Paramagnetic and bond order \[>O_{2}^{-}\]
C) Diamagnetic and bond order \[<O_{2}^{-}\]
D) Diamagnetic and bond order \[>O_{2}^{-}\]
Correct Answer: B
Solution :
\[{{O}_{2}}(16)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2},\] \[\pi 2p_{x}^{2}=\pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1}=\overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] Bond order \[=\frac{10-6}{2}=2\] Due to presence of unpaired electrons, \[{{O}_{2}}\] is paramagnetic. \[O_{2}^{-}(17)=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2},\] \[\pi 2p_{x}^{2}=\pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{2}=\overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] Bond order \[=\frac{10-7}{2}=1.5\] \[\therefore \] \[BO\]of oxygen \[>O_{2}^{-}\]
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