Given that, the bond order of dioxygen is m. Electronic configuration of O2(Z=16)=[σ(1s)]2[σ*(1s)]2[σ(2s)]2[σ(*s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π*(2px)][π*(2py)] Bond order of oxygen =
1
2
(10−6)=2 Thus, value of m=2 Now, electronic configuration of N2+is [σ(1s)]2[σ*(1s)]2[σ(2s)]2[σ(‌*2s)]2[π(2px)]2[π(2py)]2[σ(2pz] Bond order of N2+=
9−4
2
=
5
2
=2.5 and electronic configuration of C22− is [σ(1s)]2[σ*(1s)]2[σ(2s)]2[σ(‌*2s)]2‌[π(2px)]2[π(2py)]2[σ(2pz)]2 Bond order of C22−=
10−4
2
=
6
2
=3 The bond order of N2+ and C22− in term of m is