Use formal charge arguments to explain why \(\mathrm{CO}\) has a much smaller dipole moment than would be expected on the basis of electronegativity.

First, we need to determine the Lewis structure of CO. Carbon has 4 valence electrons and oxygen has 6 valence electrons, giving a total of 10 valence electrons in the CO molecule. The most stable Lewis structure for CO is a triple bond between the carbon and oxygen atoms, with one lone pair on the carbon atom and two lone pairs on the oxygen atom. Here's an example of what this structure looks like: ``` :C≡O: ```

Next, we will calculate the formal charges of the carbon and oxygen atoms in the Lewis structure. Formal charge is calculated using the formula: Formal charge = (number of valence electrons on the free atom) - (number of non-bonding electrons) - (½ × number of bonding electrons) For carbon, the formal charge is: Formal charge of C = (4) - (2) - (½ × 6) = 0 For oxygen, the formal charge is: Formal charge of O = (6) - (4) - (½ × 6) = 0 Both atoms have a formal charge of 0.

Since both the carbon and oxygen atoms have a formal charge of 0, there is no charge separation between the atoms in the CO molecule, despite their different electronegativities. The equal sharing of electrons in the triple bond between carbon and oxygen results in a less polar bond than would be expected based on electronegativity alone. As a result, the dipole moment of CO is much smaller than expected, based on the electronegativities of carbon and oxygen. This means that the CO molecule is less polar, leading to a smaller overall electric dipole moment. In conclusion, the smaller dipole moment in CO can be explained by the formal charge arguments, which reveal that there is no charge separation between the carbon and oxygen atoms in the molecule due to the equal sharing of electrons in the triple bond. Although oxygen is more electronegative than carbon, the structure of the molecule leads to a smaller dipole moment than would be expected based on their electronegativities.