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: C
Solution :
\[{{O}_{2}}:\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2p_{z}^{2},\] \[\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1}\approx \overset{*}{\mathop{\pi }}\,2p_{y}^{1}\] Bond order\[=\frac{{{N}_{b}}-{{N}_{a}}}{2}=\frac{10-6}{2}=2\] Paramagnetic (because two unpaired electrons are present.) \[O_{2}^{+}:\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}},\sigma 2p_{z}^{2},\] \[\pi 2p_{x}^{2},\pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1}\] Bond order \[=\frac{10-5}{2}=2.5\] Paramagnetic (because one unpaired electron is present.)
You need to login to perform this action.
You will be redirected in
3 sec
