Q8P - Chemistry: Molecular Nature Of Matter And Change

Question

What is the hybridization of carbon in each of the following: (a) ${\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}$ ; (b) ${\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}$ ;(c) ${\mathbf{NCO}}^{\mathbf{-}}$ ?

Step-by-Step Solution

Answer

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

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

Step1: hybridization of C in CO 3 2 -

The hybridization of C in  ${\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}$ is  ${\mathbf{sp}}^{\mathbf{2}}$

 

 No. of lone pair of carbon=0

No. of sigma bond=3

No. of unpaired electron=0

By substituting

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

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

This means that the hybridization of C in ${\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}$ is ${\mathbf{sp}}^{\mathbf{2}}$ .

Step2: Hybridization of C in C 2 O 4 2 -

The hybridization of C in ${\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_4^{\mathbf{2}\mathbf{-}}$  is  ${\mathbf{sp}}^{\mathbf{2}}$ 

 

No. of lone pair of carbon=0

No. of sigma bond=3

No. of unpaired electron=0

 

By substituting

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

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

This means that the hybridization of C in ${\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}$  is ${\mathbf{sp}}^{\mathbf{2}}$ .

Step3: hybridization of C in NCO -

The hybridization of C in  ${\mathbf{NCO}}^{\mathbf{-}}$ is sp

 

 No. of lone pair of carbon=0

No. of sigma bond=2

No. of unpaired electron=0

 

By substituting

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

                                     =0+2+0=2(sp)

This means that the hybridization of C in NCO^- is sp