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) role="math" localid="1656658782748" ${\mathbf{NCO}}^{\mathbf{-}}$?

Short 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.

Step-by-step solution

hybridization of C in CO32-

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 role="math" localid="1656659597767" ${\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}$is role="math" localid="1656659444438" ${\mathbf{sp}}^{\mathbf{2}}$ .

Hybridization of C in  C2O42-

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 thehybridization of C in ${\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}$ is ${\mathbf{sp}}^{\mathbf{2}}$.

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 thatthe hybridization of C in NCO^- is sp