Question

What is the bond order of $\mathrm{NO}$ and ${\mathrm{NO}}^{+}$?

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)

Bond order of $\mathrm{NO}$

• The electronic configuration is as follows
• $\mathrm{\sigma }1{\mathrm{s}}^2,\mathrm{\sigma }*1{\mathrm{s}}^2,\mathrm{\sigma }2{\mathrm{s}}^2,\mathrm{\sigma }*2{\mathrm{s}}^2,\mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{x}}}^2,\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2,\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{z}}}^2,\mathrm{\pi }*2{{\mathrm{p}}_{\mathrm{x}}}^1$
• Bond order $\mathrm{NO}$$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.(10-5)$
• Bond order $\mathrm{NO}$$=2.5$

Bond order of ${\mathrm{NO}}^{+}$

• The electronic configuration is as follows
• $\mathrm{\sigma }1{\mathrm{s}}^2,\mathrm{\sigma }*1{\mathrm{s}}^2,\mathrm{\sigma }2{\mathrm{s}}^2,\mathrm{\sigma }*2{\mathrm{s}}^2,\mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{x}}}^2,\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2,\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{z}}}^2$
• Bond order of ${\mathrm{NO}}^{+}$$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ (number of electrons in the bonding orbitals) - (number of electrons in the antibonding orbitals)
• Bond order ${\mathrm{NO}}^{+}$$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.(10-4)$
• Bond order ${\mathrm{NO}}^{+}$$=3$