Consider the following statement.
I. When ${\mathrm{O}}_2$ is converted into ${\mathrm{O}}_2^{2+}$ bond order decreases.
II. ${\mathrm{O}}_2$ molecule is paramagnetic because it contains two unpaired electrons in ${\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}$ and ${\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}$ molecular orbitals.
III. The bond length in NO is greater than ${\mathrm{NO}}^{+}$.
Choose the correct option.

detailed solution

Correct option is C

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.