Question

How many resonance structures does ${{\mathrm{N}}_3}^{-}$ have?

Answer 1

Resonance structures:

• The delocalization of electrons in a polyatomic ion or molecule is described by resonance structures, which are collections of Lewis structures.
• Due to the presence of partial charges and fractional bonds, a single Lewis structure frequently fails to explain the bonding in a molecule/polyatomic ion. Chemical bonding is described using resonance structures in such circumstances.
• ${{\mathrm{N}}_3}^{-}$is known as an azide ion with a negative charge on it.
• Electronic configuration of nitrogen$=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^3$, therefore, number of valence electron$=5$
• The total number of the valence electrons in azide ion$=5\times 3+1\Rightarrow 16$. (one extra electron is due to the presence of a negative charge).
• Now we have to align the single, double, and triple bonds between the three atoms, and then give the atoms the remaining electrons in such a way that all the three atoms possess octet configuration.
• There are four resonance structures possible for azide ion but the fourth one does not fulfill the octet condition. so it can not be considered the resonance structure.

• Therefore, the three possible resonance structures of azide ion are,

• Now calculate the formal charge on the nitrogen atom, $\mathrm{Formal}\mathrm{charge}=\mathrm{Valence}\mathrm{electron}\mathrm{of}\mathrm{atom}-\frac{\mathrm{Bonding}\mathrm{electron}}{2}-\mathrm{Non}\mathrm{bonding}\mathrm{electron}$
• The formal charge on nitrogen,$\mathrm{Formal}\mathrm{charge}=5-\frac{4}{2}-4\Rightarrow -1$

Therefore, only three resonance structures are possible for ${{\mathrm{N}}_3}^{-}$.