JEE Main 2017 — Chemistry Inorganic Chemistry

Number of ${e}^{-}$e−

$$\mathrm{CO}=14,{\mathrm{NO}}^{+}=14$$CO=14,NO+=14

${O}_{2}^{2-}=18, {B}_{2}=10$O22−​=18,B2​=10.

According to $\mathrm{MOT}$MOT,

$$\mathrm{CO}:- \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2}<\sigma 2{{p}_{z}}^{2}$$CO:−σ1s2<σ∗1s2<σ2s2<σ∗2s2<π2px​2=π2py​2<σ2pz​2

no unpaired e$^{-}$− so diamagnetic

$${B}_{2} :- \sigma 1{s}^{2} <\overset{*}{\sigma } 1{s}^{2} <\sigma 2{s}^{2} <\overset{*}{\sigma } 2{s}^{2} <\pi 2{p}_{x} {}^{1} =\pi 2{p}_{y} {}^{1}$$B2​:−σ1s2<σ∗1s2<σ2s2<σ∗2s2<π2px​1=π2py​1

$2$2 unpaired ${e}^{-}$e−,so paramagnetic

$${O}_{2}^{-2}:- \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\sigma 2{{p}_{z}}^{2}<\pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2}<\overset{*}{\pi }2{{p}_{x}}^{2}=\overset{*}{\pi }2{{p}_{y}}^{2}$$O2−2​:−σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz​2<π2px​2=π2py​2<π∗2px​2=π∗2py​2

no unpaired ${e}^{-}$e−, so diamagnetic

$${\mathrm{NO}}^{+}:- \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2}<\sigma 2{{p}_{z}}^{2}$$NO+:−σ1s2<σ∗1s2<σ2s2<σ∗2s2<π2px​2=π2py​2<σ2pz​2

no unpaired ${e}^{-}$e−, so diamagnetic.