JEE Main 2013 — Chemistry Inorganic Chemistry

M.O. Electronic configuration.

${O}_{2} =\sigma 1{s}^{2} <\overset{*}{\sigma } 1{s}^{2} <\sigma 2{s}^{2} <\overset{*}{\sigma } 2{s}^{2} <\sigma 2{p}_{z}^{2} <\pi 2{p}_{y}^{2} =\pi 2{p}_{x}^{2} <\overset{*}{\pi } 2{p}_{y}^{1} =\overset{*}{\pi } 2{p}_{x}^{1}$O2​=σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz2​<π2py2​=π2px2​<π∗2py1​=π∗2px1​ Paramagnetic

${{O}_{2}}^{-2}$O2​−2=σ1 s $^{2}$2 < σ * 1 s$^{ }$$^{2}$2 <σ2 s $^{2}$2$^{ }$ < σ * 2 s $^{2}$2 <σ2 p $_{z}$z​ $^{2}$2 <π2 p $_{y}$y​ $^{2}$2 =π2 p $^{2}$2$_{x}$x​ < π * 2 p $_{y}$y​ $^{2}$2 = π * 2 p $_{x}$x​ $^{2}$2$^{ }$ Diamagnetic

${C}_{2}=\sigma 1{s}^{2} <\overset{*}{\sigma } 1{s}^{2} <\sigma 2{s}^{2} <\overset{*}{\sigma } 2{s}^{2} <\pi 2{p}_{x}^{2} =\pi 2{p}_{y}^{2}$C2​=σ1s2<σ∗1s2<σ2s2<σ∗2s2<π2px2​=π2py2​ Diamagnetic

${N}_{2} =\sigma 1{s}^{2} <\overset{*}{\sigma } 1{s}^{2} <\sigma 2{s}^{2} <\overset{*}{\sigma } 2{s}^{2} <\pi 2{p}_{x}^{2} =\pi 2{p}_{y}^{2} <\sigma 2{p}_{z}^{2}$N2​=σ1s2<σ∗1s2<σ2s2<σ∗2s2<π2px2​=π2py2​<σ2pz2​ Diamagnetic

${O}_{2}$O2​ has unpaired e$^{-}$− so it is paramagnetic.