Determine if each compound or ion below has a dipole moment.

(a)Carbonate ion$\left({{\mathbf{CO}}_3}^{2-}\right)$

(b) $\mathbf{-}\underset{\mathrm{..}}{\overset{\mathrm{..}}{O}}\mathbf{-}$

(c)$\stackrel{+}{C}{\left({\mathbf{CH}}_3\right)}_3$

Earlier, carbocation was known as carbonium ions.[U1] It can be defined as the ion in which a carbon atom has a positive charge. There are three types of carbocations:

1. Primary carbocations: One alkyl group is attached to the carbon having a positive charge.
2. Secondary carbocations: In these carbocations, two alkyl groups are attached to a carbon atom with a positive charge.
3. Tertiary carbocations: The carbon atom having a positive charge is attached to three alkyl groups.

For example: ${\text{CH}}_{\text{3}}\text{-}\stackrel{\text{+}}{\text{C}}{\text{H}}_{\text{2}}$

This is an example of a primary carbocation.

The carbocation given in the question is tertiary because the carbon atom having a positive charge is attached to three alkyl groups, i.e., methyl $\left({\text{CH}}_{\text{3}}\right)$ groups.

The carbon atom having a positive charge has a steric number (S) = 3 calculated by using the formula

Steric number = number of bond pairs + number of lone pairs

Therefore, the hybridization of the carbocation is ${\text{Sp}}^{\text{3}}$, and its shape will be trigonal planar. As in tertiary carbocation, the carbon atom having a positive charge will have a partial positive character. Other carbon atoms of the alkyl group will have a partial negative charge; therefore, polarity difference will be there, but as it has a trigonal planar shape, the compound is symmetrical. Therefore, tertiary carbocation has no net dipole moment or zero dipole moment.