JEE Main 1-Sep-2021 Shift 2 Solved Paper

For the oxides Li2O,CaO,Na2O2,KO2,MgO and K2O ,
Li2O⇒O2−
CaO⇒O2−
Na2O2⇒O22–
KO2⇒O2−
MgO⇒O2−
K2O⇒O2−
∴ Electronic configuration of O22− molecule is
σ1s2,σ*1s2,σ2s2,σ*2s2, σ2pz2,π2px2 =π2py2,π*2px2 =π*2py2
Hence, O22− is diamagnetic in nature as no unpaired electron is present.
Electronic configuration of O2− is : 1s2,2s2,2p6
Hence, O2− is also diamagnetic.
Electronic configuration of O2− is :
σ1s2,σ*1s2,σ2s2, σ*2s2,σ2px2,π2py2 =π2pz′2π*2py2 =π*2pz1
Since, unpaired electron is present in π*2pz orbital.
Therefore, O2− is paramagnetic in nature.
∴ Only KO2 is paramagnetic oxide.