Question

What is the hybridization of atomic orbitals of nitrogen in ${\mathrm{NO}}_2^{+}$, ${\mathrm{NO}}_3^{-}$and ${\mathrm{NH}}_4^{+}$?

Answer 1

Step1: Finding Hybridization

• Hybridization is the concept of mixing several atomic orbitals to form a new hybrid orbital with a new hybridization contributed by all atomic orbitals.
• In order to find the hybridization, we need to calculate the steric number of the given compound.
• The steric number is the total number of sigma bonds and lone pairs in that atom we are considering.
• $Stericnumber=numberofsigmabonds+numberoflonepairs$
• The relation between steric number and hybridization is given below.

Step 2: Hybridization of ${\mathbf{NO}}_{\mathbf{2}}^{\mathbf{+}}$,${\mathbf{NO}}_{\mathbf{3}}^{\mathbf{-}}$ and ${\mathbf{NH}}_{\mathbf{4}}^{\mathbf{+}}$

${\mathbf{NO}}_{\mathbf{2}}^{\mathbf{+}}$

• Here there is one positive charge on nitrogen(one electron is lost). Thus the lone pair is absent on nitrogen.
• There will be $(5-1)=4$ valence electron on nitrogen. So there are two sigma bond and two pi bonds in structure as follows,

• Thus the steric number is $(2+0)=2$
• Therefore the hybridization is $\mathbf{sp}$.

${\mathbf{NO}}_{\mathbf{3}}^{\mathbf{-}}$

• Number of valence electron in nitrogen is $(5-1)=6$.
• Thus there will be three sigma bonds and one pi bonds present. And no lone pairs on nitrogen.

• So the steric number is $(3+0)=3$
• Therefore hybridization of nitrogen in ${\mathrm{NO}}_3^{-}$ is ${\mathbf{sp}}^{\mathbf{2}}$

${\mathbf{NH}}_{\mathbf{4}}^{\mathbf{+}}$

• Here there is a total of four sigma bonds and no lone pairs present as follows,

• The steric number is $(4+0)=4$
• Thus the hybridization is ${\mathbf{sp}}^{\mathbf{3}}$.

Therefore the hybridization of nitrogen in ${\mathbf{NO}}_{\mathbf{2}}^{\mathbf{+}}$,${\mathbf{NO}}_{\mathbf{3}}^{\mathbf{-}}$and ${\mathbf{NH}}_{\mathbf{4}}^{\mathbf{+}}$ are $\mathbf{sp}$,${\mathbf{sp}}^{\mathbf{2}}$, and ${\mathbf{sp}}^{\mathbf{3}}$respectively.