JEE Main 2019 — Chemistry Inorganic Chemistry

$${O}_{2}^{-} =\sigma 1{s}^{2} <\overset{*}{\sigma } 1{s}^{2} <\sigma 2{s}^{2} <\overset{*}{\sigma } 2{s}^{2} <\sigma 2{p}_{x}^{2} =\pi 2{p}_{y}^{2} =\pi 2{p}_{z}^{2} <\overset{*}{\pi } 2{p}_{y}^{2} =\overset{*}{\pi } 2{p}_{z}^{1}$$O2−​=σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2px2​=π2py2​=π2pz2​<π∗2py2​=π∗2pz1​

Bond Order $=\frac{ 6-3}{2} =1.5$=26−3​=1.5

$${O}_{2} =\sigma 1{s}^{2} <\overset{*}{\sigma } 1{s}^{2} <\sigma 2{s}^{2} <\overset{*}{\sigma } 2{s}^{2} <\sigma 2{p}_{x}^{2} <\pi 2{p}_{y}^{2} =\pi 2{p}_{z}^{2} <\overset{*}{\pi } 2{p}_{y}^{1} =\overset{*}{\pi } 2{p}_{z}^{1}$$O2​=σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2px2​<π2py2​=π2pz2​<π∗2py1​=π∗2pz1​

Bond order $=\frac{ 6-2}{2} =2$=26−2​=2

$${N}_{2} =\sigma 1{s}^{2} <\overset{*}{\sigma } 1{s}^{2} <\sigma 2{s}^{2} <\overset{*}{\sigma } 2{s}^{2} <\pi 2{p}_{z}^{2} =\pi 2{p}_{z}^{2} <\sigma 2{p}_{x}^{2}$$N2​=σ1s2<σ∗1s2<σ2s2<σ∗2s2<π2pz2​=π2pz2​<σ2px2​

Bond order $=\frac{ 6-0}{2} =3$=26−0​=3

According to $MOT$MOT ; electronic configuration of ${N}_{2}^{+}$N2+​ is:

$${N}_{2}^{+} =\sigma 1{s}^{2} <\overset{*}{\sigma } 1{s}^{2} <\sigma 2{s}^{2} <\overset{*}{\sigma } 2{s}^{2} <\pi 2{p}_{x}^{2} =\pi 2{p}_{y}^{2} <\sigma 2{p}_{z}^{1}$$N2+​=σ1s2<σ∗1s2<σ2s2<σ∗2s2<π2px2​=π2py2​<σ2pz1​

Bond order $=\frac{1}{2}(9-4)=2.5$=21​(9−4)=2.5

i.e., $2\pi$2π bonds and $0.5 \sigma$0.5σ bonds.