Question

Arrange the following from lowest to highest ionization energy: ${\mathbf{\text{O}}}_{\mathbf{1}}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{,}{\mathbf{\text{O}}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{,}{{\mathbf{\text{O}}}_{\mathbf{2}}}^{\mathbf{+}}$. Explain your answer.

Short answer

Lowest to highest form of Ionization energy:$O<O_2^{-}<O_2<O_2^{+}$

Step-by-step solution

Electronic configuration

Electronic configuration of O atom:$1s^{2}2s^{2}2p^4$

Stability

The stability of the molecule depends on Bond order. More the stability of molecule more is the ionization energy.

Bond order of $O_2$:$\frac{1}{2}(8-4)=2$

Bond order of $O_2^{-}$:$\frac{1}{2}(8-5)=1.5$

Bond order of $O_2^{+}$:$\frac{1}{2}(8-3)=2.5$

Bond order of O: 1

Therefore, the increasing form of ionization energy will be$O<O_2^{-}<O_2<O_2^{+}$