Q7P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the hybridization of nitrogen in each of the following: (a) NO; (b) ${\mathbf{NO}}_{\mathbf{2}}$ ; (c) ${{\mathbf{NO}}^{\mathbf{.}}}_{\mathbf{2}}$ ?

Step-by-Step Solution

Answer

By using this formula, we can find the hybridization of Nitrogen

No. of Hybrid Orbitals=No. of lone pair(N)+ No. of sigma bond+ No.of unpaired electron

Step1: hybridization of N in NO

The hybridization of nitrogen in NO is ${\mathbf{sp}}^{\mathbf{2}}$

No. of lone pair of nitrogen=1

No. of sigma bond=1

No. of unpaired electron=1

 

By substituting

No. of Hybrid Orbitals=No.of lone pair(N)+ No. of sigma bond+ No. of unpaired electron.

                                     =1+1+1=3( ${\mathbf{sp}}^{\mathbf{2}}$)

This means that the hybridization of N in NO is ${\mathbf{sp}}^{\mathbf{2}}$ .

Step2: Hybridization of N in NO 2

The hybridization of nitrogen in ${\mathbf{NO}}_{\mathbf{2}}$  is ${\mathbf{sp}}^{\mathbf{2}}$ 

 

 

No. of lone pair of nitrogen=0

No. of sigma bond=2

No. of unpaired electron=1

 

By substituting

No. of Hybrid Orbitals=No. of lone pair(N)+ No.of sigma bond+ No. of unpaired electron.

                                     =0+2+1=3( ${\mathbf{sp}}^{\mathbf{2}}$)

This means that the hybridization of N in ${\mathbf{NO}}_{\mathbf{2}}$  is ${\mathbf{sp}}^{\mathbf{2}}$ .

Step3: hybridization of N in NO 2

The hybridization of nitrogen in ${\mathbf{NO}}_{\mathbf{2}}$  is  ${\mathbf{sp}}^{\mathbf{2}}$

No. of lone pair of nitrogen=1

No. of sigma bond=2

No. of unpaired electron=0

 

By substituting

No. of Hybrid Orbitals=No. of lone pair(N)+ No. of sigma bond+ No. of unpaired electron.

                                     =1+2+0=3( ${\mathbf{sp}}^{\mathbf{2}}$)

This means that the hybridization of N in  ${\mathbf{NO}}_{\mathbf{2}}$ is ${\mathbf{sp}}^{\mathbf{2}}$ .