From its formula, one might expect CO to be quite polar, but its dipole moment is low (0.11 D).

(a) Draw the Lewis structure for CO.

(b) Calculate the formal charges.

(c) Based on your answers to parts (a) and (b), explain why the dipole moment is so low

It represents the valence electrons of the atom around the atom in the form of dots and shows the bonds in sigma and pi bond form.

Carbon has one lone pair represented in the form of dots, and oxygen has one pair represented in dots, and both form a triple bond with each other to complete the octet. They have a total of 10 electors because carbon has a valency of 4 and oxygen has a valency of 6, as shown:

Lewis structure of CO

It is defined as the charge on an atom when atoms are in bonding form .it can be calculated as:

$\mathrm{FC}=\frac{1}{2}\left(\mathrm{V}-\mathrm{N}-\frac{\mathrm{B}}{2}\right)$

Where

V=valence electrons

N= non-bonded electrons

B= boded electrons

But in CO, the charge is zero because carbon has one lone pair with a triple bond, so 4-5=-1 and oxygen has one lone team with a triple bond, and valency is so 6-5=1, so,

C=+1 and O=-1

Net CO=+1-1

CO=0

Zero formal charge

The dipole moment of CO is low because the maximum number of bonding electrons is occupied by the sigma and pi bond; therefore, oxygen can’t get a negative charge due to a lack of lone pairs, and the carbon monoxide has a triple bond, and oxygen is more electronegative. Still, it is a polar molecule, so that dipole moment is low.