Consider the following statement.I. When O2 is converted into O22+ bond order decreases.II. O2 molecule is paramagnetic because it contains two unpaired electrons in π*2px and π*2py molecular orbitals.III. The bond length in NO is greater than NO+.Choose the correct option.

Statement II and III are correct.The molecular orbital configuration of O2  molecule is as follows:   σ   1s2,   σ   *1s2,   σ   2s2,    σ   *2s2,   σ   2p2z,π2px2    =π2py2      ,π*2px1=π*2py1There are 10 bonding and 6 nonbonding electrons in the orbitals according to the molecular orbital configuration.Therefore,  Bond order =12[Bonding-antibonding]= 12[10−6]=12(4)=2Thus, the bond order O2 is 2. The molecular orbital configuration of O2+2 is σ1s2,σ*1s2,σ2s2,σ*2s2,σ2pz2,π2px2=π2py2There are 10 bonding  and 4 nonbonding electrons in the orbitals according to the molecular orbital configuration.Therefore, Bond order =12[Bonding-antibonding]=12[10−4]=3.Hence statement I is incorrect.