Based on MO theory, draw a molecular orbital diagram for N2+, N2, and N2-. Using your diagram answer the following questions:
Which molecule has the shorter bond length?
What is the bond order of each molecule?
Which is the most stable?
Which are paramagnetic and which are diamagnetic?
Write the electron configuration for each of these molecules.
The correct answer and explanation is :
Molecular Orbital (MO) Diagram for $\text{N}_2^+$, $\text{N}_2$, and $\text{N}_2^-$
To analyze the bond properties, stability, and magnetic behavior of the molecules $\text{N}_2^+$, $\text{N}_2$, and $\text{N}_2^-$, we need to consider their molecular orbital diagrams.
Molecular Orbitals of $\text{N}_2$:
For nitrogen ($\text{N}_2$), each nitrogen atom has 7 electrons, leading to 14 electrons in total for the $\text{N}_2$ molecule. In molecular orbital theory, the molecular orbitals are formed by the combination of atomic orbitals, and the energy levels follow this order for diatomic molecules from $Z = 8$ to $Z = 14$:
- $\sigma_{1s}$, $\sigma_{1s}^*$
- $\sigma_{2s}$, $\sigma_{2s}^*$
- $\sigma_{2p_z}$, $\pi_{2p_x} = \pi_{2p_y}$, $\pi_{2p_x}^* = \pi_{2p_y}^$, $\sigma_{2p_z}^$
The electron configuration for $\text{N}2$ with 14 electrons is: $\sigma{1s}^2 \, \sigma_{1s}^{2} \, \sigma_{2s}^2 \, \sigma_{2s}^{2} \, \pi_{2p_x}^2 \, \pi_{2p_y}^2 \, \sigma_{2p_z}^2$
Molecular Orbitals of $\text{N}_2^+$ (one electron removed):
For $\text{N}2^+$, we remove one electron from the highest occupied molecular orbital (HOMO), which is $\pi{2p_x}$ or $\pi_{2p_y}$:
$\sigma_{1s}^2 \, \sigma_{1s}^{2} \, \sigma_{2s}^2 \, \sigma_{2s}^{2} \, \pi_{2p_x}^1 \, \pi_{2p_y}^2 \, \sigma_{2p_z}^2$
Molecular Orbitals of $\text{N}_2^-$ (one electron added):
For $\text{N}2^-$, we add one electron to the lowest unoccupied molecular orbital (LUMO), which is the $\pi{2p_x}^$ or $\pi_{2p_y}^$ orbital:
$\sigma_{1s}^2 \, \sigma_{1s}^{2} \, \sigma_{2s}^2 \, \sigma_{2s}^{2} \, \pi_{2p_x}^2 \, \pi_{2p_y}^2 \, \sigma_{2p_z}^2 \, \pi_{2p_x}^{*1}$
Key Questions Answered:
- Which molecule has the shorter bond length?
The bond length decreases with increasing bond order. The bond order is calculated as: $$
\text{Bond order} = \frac{1}{2} \left( \text{(Number of bonding electrons)} – \text{(Number of antibonding electrons)} \right)
$$
- For $\text{N}_2$, the bond order is: $$
\text{Bond order} = \frac{1}{2} \left( 10 – 4 \right) = 3
$$ - For $\text{N}_2^+$, the bond order is: $$
\text{Bond order} = \frac{1}{2} \left( 10 – 4 \right) = 3
$$ - For $\text{N}_2^-$, the bond order is: $$
\text{Bond order} = \frac{1}{2} \left( 10 – 5 \right) = 2.5
$$ Hence, $\text{N}_2$ and $\text{N}_2^+$ have the shortest bond length, with $\text{N}_2^-$ having a longer bond due to a lower bond order.
- What is the bond order of each molecule?
- $\text{N}_2$ and $\text{N}_2^+$ have a bond order of 3.
- $\text{N}_2^-$ has a bond order of 2.5.
- Which is the most stable?
The molecule with the highest bond order is the most stable, because higher bond order means more bonding electrons and fewer antibonding electrons. Thus, $\text{N}_2$ and $\text{N}_2^+$ are the most stable, with $\text{N}_2^+$ being slightly less stable due to the removal of an electron, which reduces electron-electron repulsion. - Which are paramagnetic and which are diamagnetic?
- Paramagnetic molecules have unpaired electrons, and diamagnetic molecules have all paired electrons.
- $\text{N}_2$ is diamagnetic because all its electrons are paired.
- $\text{N}_2^+$ is paramagnetic because it has one unpaired electron in a $\pi^*$ orbital.
- $\text{N}_2^-$ is paramagnetic because it has one unpaired electron in a $\pi^*$ orbital.
- Electron configurations:
- $\text{N}2$: $\sigma{1s}^2 \, \sigma_{1s}^{2} \, \sigma_{2s}^2 \, \sigma_{2s}^{2} \, \pi_{2p_x}^2 \, \pi_{2p_y}^2 \, \sigma_{2p_z}^2$
- $\text{N}2^+$: $\sigma{1s}^2 \, \sigma_{1s}^{2} \, \sigma_{2s}^2 \, \sigma_{2s}^{2} \, \pi_{2p_x}^1 \, \pi_{2p_y}^2 \, \sigma_{2p_z}^2$
- $\text{N}2^-$: $\sigma{1s}^2 \, \sigma_{1s}^{2} \, \sigma_{2s}^2 \, \sigma_{2s}^{2} \, \pi_{2p_x}^2 \, \pi_{2p_y}^2 \, \sigma_{2p_z}^2 \, \pi_{2p_x}^{*1}$
Conclusion:
- The bond lengths are shortest in $\text{N}_2$ and $\text{N}_2^+$, with $\text{N}_2^-$ having a longer bond.
- The bond orders follow the trend $\text{N}_2 = \text{N}_2^+ > \text{N}_2^-$.
- $\text{N}_2$ and $\text{N}_2^+$ are the most stable, with $\text{N}_2^+$ slightly less stable than $\text{N}_2$.
- $\text{N}_2$ is diamagnetic, while both $\text{N}_2^+$ and $\text{N}_2^-$ are paramagnetic.