Question

Use the MO model to determine which of the following has the smallest ionization energy: ${\mathit{N}}_{\mathbf{2}}\mathbf{,}{\mathit{O}}_{\mathbf{2}}\mathbf{,}{\mathit{N}}_{\mathbf{2}}^{\mathbf{2}\mathbf{-}}\mathbf{,}{\mathit{N}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{,}{\mathit{O}}_{\mathbf{2}}^{\mathbf{+}}$. Explain your answer.

Short answer

Some amount of energy that is required to remove the outermost electron from the shell, is known as ionization energy of a molecule. By calculating the bond order, we can determine the order of ionization energy as they are directly related to each other.

Step-by-step solution

Molecular orbital configuration

The configuration of $N_2$,$O_2$, $N_2^{2-}$, $N_2^{-}$, $O_2^{+}$ according to the molecular orbital theory will be:

$N_2$=$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2(\sigma 2p_z{)}^2$

$O_2$=$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2(\sigma 2p_z{)}^2({\pi }^{*}2p_x{)}^1({\pi }^{*}2p_y{)}^1$

$N_2^{2-}$=$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2(\sigma 2p_z{)}^2({\pi }^{*}2p_x{)}^1({\pi }^{*}2p_y{)}^1$

$N_2^{-}$=$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2(\sigma 2p_z{)}^2({\pi }^{*}2p_x{)}^1$

$O_2^{+}$=$(\sigma 1s{)}^2({\sigma }^{*}1s{)}^2(\sigma 2s{)}^2({\sigma }^{*}2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2(\sigma 2p_z{)}^2({\pi }^{*}2p_x{)}^1$

Bond order

$$\begin{gathered} Bondorderof{N}_2=\frac{No.ofbondingelectrons-no.ofnonbondingelectrons}{2} \\ =\frac{10-4}{2} \\ =3 \end{gathered}$$

$$\begin{gathered} Bondorderof{O}_2=\frac{No.ofbondingelectrons-no.ofnonbondingelectrons}{2} \\ =\frac{10-6}{2} \\ =2 \end{gathered}$$

$$\begin{gathered} Bondorderof{N}_2^{2-}=\frac{No.ofbondingelectrons-no.ofnonbondingelectrons}{2} \\ =\frac{10-6}{2} \\ =2 \end{gathered}$$

$$\begin{gathered} Bondorderof{N}_2^{-}=\frac{No.ofbondingelectrons-no.ofnonbondingelectrons}{2} \\ =\frac{10-5}{2} \\ =2.5 \end{gathered}$$

role="math" localid="1663923494412" $$\begin{gathered} Bondorderof{O}_2^{+}=\frac{No.ofbondingelectrons-no.ofnonbondingelectrons}{2} \\ =\frac{10-5}{2} \\ =2.5 \end{gathered}$$

Smallest ionization energy

As ionization energy is directly related to the bond order, so the smallest ionization energy will be possessed by $O_2$, $N_2^{2-}$as they have smallest bond order of 2.