Answer :
To determine which molecules or ions exhibit delocalized bonding, let's analyze each of the given species:
### Analysis of [tex]\( \text{NO}_2^- \)[/tex]:
Nitrite ion ([tex]\( \text{NO}_2^- \)[/tex]) has a resonance structure. In [tex]\( \text{NO}_2^- \)[/tex], the nitrogen and oxygen atoms share electrons in such a way that the bonding electrons are delocalized over the entire ion. This means that instead of being fixed between individual atoms, the bonding electrons are spread out across the [tex]\( \text{NO}_2^- \)[/tex] molecule. This delocalization is a characteristic of resonant structures, confirming that [tex]\( \text{NO}_2^- \)[/tex] exhibits delocalized bonding.
### Analysis of [tex]\( \text{NH}_4^+ \)[/tex]:
Ammonium ion ([tex]\( \text{NH}_4^+ \)[/tex]) does not exhibit delocalized bonding. In [tex]\( \text{NH}_4^+ \)[/tex], the nitrogen atom forms four localized sigma bonds with hydrogen atoms. There are no resonance structures or delocalized electrons in the ammonium ion, so it does not exhibit delocalized bonding.
### Analysis of [tex]\( \text{N}_3^- \)[/tex]:
Azide ion ([tex]\( \text{N}_3^- \)[/tex]) has a resonance structure where the electrons are delocalized over the three nitrogen atoms. This ion can be represented by multiple resonance structures where the positions of the double and single bonds are switched. The delocalization of electrons across different resonance structures means that [tex]\( \text{N}_3^- \)[/tex] exhibits delocalized bonding.
### Conclusion:
From our analysis, we see that:
- [tex]\( \text{NO}_2^- \)[/tex] exhibits delocalized bonding.
- [tex]\( \text{NH}_4^+ \)[/tex] does not exhibit delocalized bonding.
- [tex]\( \text{N}_3^- \)[/tex] exhibits delocalized bonding.
Therefore, the molecules or ions that exhibit delocalized bonding are [tex]\( \text{NO}_2^- \)[/tex] and [tex]\( \text{N}_3^- \)[/tex].
The correct answer is:
[tex]$ \text{NO}_2^- \text{ and } \text{N}_3^- $[/tex]
### Analysis of [tex]\( \text{NO}_2^- \)[/tex]:
Nitrite ion ([tex]\( \text{NO}_2^- \)[/tex]) has a resonance structure. In [tex]\( \text{NO}_2^- \)[/tex], the nitrogen and oxygen atoms share electrons in such a way that the bonding electrons are delocalized over the entire ion. This means that instead of being fixed between individual atoms, the bonding electrons are spread out across the [tex]\( \text{NO}_2^- \)[/tex] molecule. This delocalization is a characteristic of resonant structures, confirming that [tex]\( \text{NO}_2^- \)[/tex] exhibits delocalized bonding.
### Analysis of [tex]\( \text{NH}_4^+ \)[/tex]:
Ammonium ion ([tex]\( \text{NH}_4^+ \)[/tex]) does not exhibit delocalized bonding. In [tex]\( \text{NH}_4^+ \)[/tex], the nitrogen atom forms four localized sigma bonds with hydrogen atoms. There are no resonance structures or delocalized electrons in the ammonium ion, so it does not exhibit delocalized bonding.
### Analysis of [tex]\( \text{N}_3^- \)[/tex]:
Azide ion ([tex]\( \text{N}_3^- \)[/tex]) has a resonance structure where the electrons are delocalized over the three nitrogen atoms. This ion can be represented by multiple resonance structures where the positions of the double and single bonds are switched. The delocalization of electrons across different resonance structures means that [tex]\( \text{N}_3^- \)[/tex] exhibits delocalized bonding.
### Conclusion:
From our analysis, we see that:
- [tex]\( \text{NO}_2^- \)[/tex] exhibits delocalized bonding.
- [tex]\( \text{NH}_4^+ \)[/tex] does not exhibit delocalized bonding.
- [tex]\( \text{N}_3^- \)[/tex] exhibits delocalized bonding.
Therefore, the molecules or ions that exhibit delocalized bonding are [tex]\( \text{NO}_2^- \)[/tex] and [tex]\( \text{N}_3^- \)[/tex].
The correct answer is:
[tex]$ \text{NO}_2^- \text{ and } \text{N}_3^- $[/tex]
