The bond order of a molecule is calculated as the difference between the number of bonding electrons and the number of antibonding electrons, divided by 2. For NO, the bond order is given as 2.5.NO has 11 valence electrons 5 from nitrogen and 6 from oxygen . When we form NO+ ion, we remove one electron, resulting in 10 valence electrons. For NO- ion, we add one electron, resulting in 12 valence electrons.To determine the bond order of NO+ and NO-, we can use the molecular orbital MO diagram for NO. The MO diagram for NO is similar to that of N2 and O2, with the following order of molecular orbitals: 1s, *1s, 2s, *2s, 2px=2py, 2pz, *2px=*2py, and *2pz.For NO+ 10 valence electrons :- 2 electrons in 1s- 2 electrons in *1s- 2 electrons in 2s- 2 electrons in *2s- 2 electrons in 2px and 2py 1 electron in each Bond order of NO+ = 8 bonding electrons - 2 antibonding electrons / 2 = 6 / 2 = 3For NO- 12 valence electrons :- 2 electrons in 1s- 2 electrons in *1s- 2 electrons in 2s- 2 electrons in *2s- 4 electrons in 2px and 2py 2 electrons in each - 1 electron in 2pzBond order of NO- = 10 bonding electrons - 3 antibonding electrons / 2 = 7 / 2 = 3.5So, the bond order of NO+ ion is 3, and the bond order of NO- ion is 3.5.