Question

${\mathrm{N}}_2(\mathrm{g})$ has an exceptionally high bond energy. Would you expect either ${\mathrm{N}}_2^{-}$ or ${\mathrm{N}}_2^{2-}$ to be a stable diatomic species in the gaseous state? Explain.

Short answer

While N2 is very stable due to its triple bond, adding extra electrons to form N2- or N2^2- is predicted to decrease its stability because the extra electrons go into antibonding orbitals, which weaken bonds. Therefore, neither N2- nor N2^2- is expected to be a stable diatomic species in the gaseous state.

Step-by-step solution

Understanding Bond Energy of N2

First, it's important to understand that the high bond energy of N2 (nitrogen gas) comes from the triple bond between the two nitrogen atoms. This triple bond is particularly stable, which makes N2 a very stable molecule.

Predict the Effect of Extra Electrons

Next, consider what happens when an electron is added to N2 to form N2-. The added electron will go into an antibonding orbital, which tends to weaken bonds rather than strengthen them. Therefore, N2- is predicted to be less stable than N2.

Consider the stability of N2^2-

Now consider N2^2-. Two extra electrons mean even more antibonding character, which is predicted to further weaken the bond and decrease stability. Therefore, N2^2- is expected to be even less stable than N2-.

Key concepts

Bond Energy

Bond energy is a measure of the strength of a chemical bond. It represents the amount of energy required to break a bond between atoms in a molecule. The higher the bond energy, the more stable the molecule is typically considered. For example, in the case of nitrogen gas oindent ${\text{N}}_2$, the bond energy is exceptionally high due to its triple bond, which consists of one sigma bond and two pi bonds.

• The triple bond contributes significantly to the overall stability of ${\text{N}}_2$, requiring more energy to break apart than a single or double bond would.
• This strong bond is why ${\text{N}}_2$ is relatively inert and doesn't react easily with other substances.

Whenever the bond energy is altered, it affects the stability of the molecule. It is crucial to understand this when analyzing how additional electrons might impact bond stability, as with ${\text{N}}_2^{-}$ and ${\text{N}}_2^{2-}$.

Antibonding Orbital

Antibonding orbitals are molecular orbitals where electron presence results in a decreased bond order and increased molecular instability. When electrons fill these orbitals, they counteract the stabilizing effects of bonding orbitals.

• Consider ${\text{N}}_2$, where adding extra electrons to create ${\text{N}}_2^{-}$ or ${\text{N}}_2^{2-}$ results in electrons occupying antibonding orbitals.
• These added electrons effectively "cancel out" part of the strong triple bond, reducing the bond order.
• In ${\text{N}}_2$, the bond order is 3, but in ${\text{N}}_2^{-}$, with one electron in an antibonding orbital, the bond order decreases to approximately 2.5.
• For ${\text{N}}_2^{2-}$, the bond order drops further to around 2, making it even less stable.

Understanding the role of antibonding orbitals is essential for predicting the stability and reactivity of molecules undergoing electron additions, as the presence of electrons in these orbitals weakens the bond.

Molecular Stability

Molecular stability is often influenced by the bond energy and the occupancy of bonding vs. antibonding orbitals. A stable molecule balances these factors in a way that maximizes bonding interactions while minimizing antibonding ones.

• For ${\text{N}}_2$, high bond energy provides a great deal of stability.
• In ${\text{N}}_2^{-}$, the addition of an electron into an antibonding orbital decreases stability due to reduced bond order.
• Further addition of an electron to form ${\text{N}}_2^{2-}$ reduces bond order even more, potentially leading to a molecule too unstable to exist under normal conditions.

The takeaway here is that molecular stability is heavily reliant on electron distribution within molecular orbitals. A molecule’s ability to maintain high bond energy without excessive antibonding effects, as seen in ${\text{N}}_2$ compared to ${\text{N}}_2^{-}$ and ${\text{N}}_2^{2-}$, is crucial for its stability in the gaseous state.