Question

What is the bond order in ${{\mathrm{O}}_2}^{+}$?

Answer 1

Bond Order

• Bond order is defined as one-half the difference between the number of electrons present in the bonding and the antibonding orbitals.
• The mathematical formula is as follows,
• Bond order (B.O.)$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ (number of electrons in the bonding orbitals) - (number of electrons in the antibonding orbitals)

Example

• The electronic configuration is as follows
• $\sigma 1s^{\mathit{2}}\mathit{,}\mathit{}\sigma \mathit{*}1s^{\mathit{2}}\mathit{,}\mathit{}\sigma \mathit{2}{s}^{\mathit{2}}\mathit{,}\mathit{}\sigma \mathit{*}\mathit{2}{s}^{\mathit{2}}\mathit{,}\mathit{}\sigma 2{{\mathit{p}}_{\mathit{z}}}^{\mathit{2}}\mathit{,}\mathit{}\pi 2{{\mathit{p}}_{\mathit{y}}}^{\mathit{2}}\mathit{,}\mathit{}\pi 2{{\mathit{p}}_{\mathit{x}}}^{\mathit{2}}\mathit{,}\mathit{}\pi \mathit{*}2{{\mathit{p}}_{\mathit{y}}}^{\mathit{1}}$
• Bond order of ${{\mathrm{O}}_2}^{+}$$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ (number of electrons in the bonding orbitals) - (number of electrons in the antibonding orbitals)
• Bond order ${{\mathrm{O}}_2}^{+}$$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.(10-5)$
• Bond order ${{\mathrm{O}}_2}^{+}$$=2.5$