NEET UG 2015 — Chemistry Inorganic Chemistry

$$B.O=\frac{{N}_{b} -{N}_{a}}{2}$$B.O=2Nb​−Na​​

$${O}_{2}^{-} \Rightarrow \sigma 1{s}^{2} <\overset{*}{\sigma } 1{s}^{2} <\sigma 2{s}^{2} <\overset{*}{\sigma } 2{s}^{2} <\sigma 2{p}_{z}^{2} <\pi 2{p}_{x}^{2} =\pi 2{p}_{y}^{2} <\overset{*}{\pi } 2{p}_{x}^{2} =\overset{*}{\pi } 2{p}_{y}^{1}$$O2−​⇒σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz2​<π2px2​=π2py2​<π∗2px2​=π∗2py1​

B.O for ${O}_{2}^{-}=1.5\Rightarrow \frac{10 - 7}{2}=1.5$O2−​=1.5⇒210−7​=1.5

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

B.O for ${O}_{2}=2.0\Rightarrow \frac{10 - 6}{2}=2$O2​=2.0⇒210−6​=2

$${O}_{2}^{+} \Rightarrow \sigma 1{s}^{2} <\overset{*}{\sigma } 1{s}^{2} <\sigma 2{s}^{2} <\overset{*}{\sigma } 2{s}^{2} <\sigma 2{p}_{z}^{2} <\pi 2{p}_{x}^{2} <\pi 2{p}_{y}^{2} <\overset{*}{\pi } 2{p}_{x}^{1} <\overset{*}{\pi } 2{p}_{y}^{0}$$O2+​⇒σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz2​<π2px2​<π2py2​<π∗2px1​<π∗2py0​

B.O for ${O}_{2}^{+}=2.5\Rightarrow \frac{10 - 5}{2}=2.5$O2+​=2.5⇒210−5​=2.5

As per M.O.T electronic configurations

$${O}_{2}^{-}<{O}_{2}<{O}_{2}^{+}$$O2−​<O2​<O2+​