Which of the following orders is correct for the bond dissociation energy of ${\mathrm{O}}_2,{\mathrm{O}}_2^{+},{\mathrm{O}}_2^{-}\text{and }{\mathrm{O}}_2^{2-}$?

Bond Order =$\frac{\mathrm{Bonding} \mathrm{electrons}-\mathrm{Antibonding} \mathrm{electrons}}{2}$

${\mathrm{O}}_2^{2-}$→ Total electrons = 16 + 2 = 18
Configuration:
σ(1s)²σ^∗(1s)²σ(2s)²σ^∗(2s)²σ(2p_z)²π(2p_x)²π(2p_y)²π^∗(2p_x)²π^∗(2p_y)²
Bond Order=$\frac{10-8}{2}$=1

${\mathrm{O}}_2^{-}$→ Total electrons = 16 +1 = 17
Configuration:
σ(1s)²σ^∗(1s)²σ(2s)²σ^∗(2s)²σ(2p_z)²π(2p_x)²π(2p_y)²π^∗(2p_x)²π^∗(2p_y)¹
Bond Order=$\frac{10-7}{2}$=1.5

O₂→ Total electrons = 8 + 8 = 16
Configuration:
σ(1s)²σ^∗(1s)²σ(2s)²σ^∗(2s)²σ(2p_z)²π(2p_x)²π(2p_y)²π^∗(2p_x)¹π^∗(2p^y)¹
Bond Order=$\frac{10-6}{2}$=2

${\mathrm{O}}_2^{+}$→ Total electrons = 16 -1 = 15
Configuration:
σ(1s)²σ^∗(1s)²σ(2s)²σ^∗(2s)²σ(2p_z)²π(2p_x)²π(2p_y)²π^∗(2p_x)¹
Bond Order=$\frac{10-5}{2}$=2.5

Hence, the correct bond order in the following species is:
 

${\mathrm{O}}_2^{+}>{\mathrm{O}}_2>{\mathrm{O}}_2^{-}>{\mathrm{O}}_2^{2-}$

Bond order ∝ Bond dissociation enthalpy

Order: ${\mathrm{O}}_2^{+}>{\mathrm{O}}_2>{\mathrm{O}}_2^{-}>{\mathrm{O}}_2^{2-}$