JEE Main 2022 — Chemistry Inorganic Chemistry

The species that contain unpaired electrons are paramagnetic.

Electronic configuration of ${\mathrm{Li}}_{2}$Li2​ molecule is ${(\sigma 1s)}^{2}{({\sigma }^{*}1s)}^{2}{(\sigma 2s)}^{2}$(σ1s)2(σ∗1s)2(σ2s)2.

The electronic configuration of ${B}_{2}$B2​ molecule is

$${(\sigma 1s)}^{2}{({\sigma }^{⋆}1s)}^{2}{(\sigma 2s)}^{2}{({\sigma }^{⋆}2s)}^{2}(\pi 2{{p}_{x}}^{1}=\pi 2{{p}_{y}}^{1})$$(σ1s)2(σ⋆1s)2(σ2s)2(σ⋆2s)2(π2px​1=π2py​1)

The presence of unpaired electrons in the two $\pi$π bonding molecular orbitals explains its paramagnetic nature.

The electronic configuration of ${C}_{2}$C2​ $\mathrm{KK}\sigma 2{s}^{2}{\sigma }^{*}2{s}^{2}\pi 2{p}_{x}^{2}\pi 2{p}_{y}^{2}$KKσ2s2σ∗2s2π2px2​π2py2​

$${O}_{2}:\mathrm{KK}{({\sigma }_{2s})}^{2}{({\sigma }_{2s})}^{2}{({\sigma }_{2{p}_{z}})}^{2}{({\pi }_{2{p}_{x}})}^{2}={({\pi }_{2{p}_{y}})}^{2}{(\pi {*}_{2{p}_{x}})}^{1}={(\pi {*}_{2{p}_{y}})}^{1}$$O2​:KK(σ2s​)2(σ2s​)2(σ2pz​​)2(π2px​​)2=(π2py​​)2(π∗2px​​)1=(π∗2py​​)1

Hence, ${O}_{2}$O2​ molecule is paramagnetic.

The electronic configuration of ${C}_{2}^{-}$C2−​ is ${(\sigma 1s)}^{2}{({\sigma }^{⋆}1s)}^{2}{(\sigma 2s)}^{2}{({\sigma }^{⋆}2s)}^{2}(\pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2})\sigma 2{{p}_{z}}^{1}$(σ1s)2(σ⋆1s)2(σ2s)2(σ⋆2s)2(π2px​2=π2py​2)σ2pz​1

Hence, it is paramagnetic substance.

$${O}_{2}^{2-}:\mathrm{KK}{({\sigma }_{2s})}^{2}{({\sigma }_{2s})}^{2}{({\sigma }_{2{p}_{z}})}^{2}{({\pi }_{2{p}_{x}})}^{2}={({\pi }_{2{p}_{y}})}^{2}{(\pi {*}_{2{p}_{x}})}^{2}={(\pi {*}_{2{p}_{y}})}^{2}$$O22−​:KK(σ2s​)2(σ2s​)2(σ2pz​​)2(π2px​​)2=(π2py​​)2(π∗2px​​)2=(π∗2py​​)2

$${O}_{2}^{+}:\mathrm{KK}{({\sigma }_{2s})}^{2}{({\sigma }_{2s})}^{2}{({\sigma }_{2{p}_{z}})}^{2}{({\pi }_{2{p}_{x}})}^{2}={({\pi }_{2{p}_{y}})}^{2}{(\pi {*}_{2{p}_{x}})}^{1}=(\pi {*}_{2{p}_{y}})$$O2+​:KK(σ2s​)2(σ2s​)2(σ2pz​​)2(π2px​​)2=(π2py​​)2(π∗2px​​)1=(π∗2py​​)

Hence, the above species is paramagnetic.