Question

Where, approximately, is the negative pole on each of these molecules?

Which molecule should have higher dipole moment, and why?

Answer 1

Concepts and reason

Bonded atoms in a molecule have unequal electronegative values wherein the electrons present in the atom shift towards the most electronegative site and is represented as polarity. The atom in the molecule carries charge according to its polarity. Both positive and negative charges are present in the molecule.

Fundamentals

The atoms in the molecule show positive and negative poles. The positive pole represents that the atom is the least electronegative and the negative pole represents that the atom is the most electronegative.

Example:

In Hydrogen bromide $\left( {{\rm{HBr}}} \right)$ , bromine $\left( {{\rm{Br}}} \right)$ is more electronegative than hydrogen. In this molecule, the positive pole is hydrogen and the negative pole is bromine. The electrons in the bonded atoms move to a more electronegative site.

Consider the structure of compound as shown below:

In this molecule, one oxygen, one carbon and two fluorine atoms are present. In general, oxygen atom is less electronegative than fluorine atom therefore, C-O bond is less polar than C-F bond. Therefore, the negative pole will not present towards the ${\rm{O}}$ atom.

Since, both the fluorine atoms are pulling the electrons from C equally in a common direction thus, the resultant negative pole will be present I between the two F atoms.

Consider the structure of compound as shown below:

This molecule contains one oxygen atom, one carbon atom, one hydrogen atom and one fluorine atom. In general, oxygen atom is less electronegative than fluorine atom therefore, C-O bond is less polar than C-F bond.

But both the polar bonds are not present in opposite direction thus they won’t cancel out each other effect instead both oxygen and fluorine atom are pulling the electrons from the carbon atom in a common direction thus, the resultant negative pole will be present between oxygen and F atom.

Consider the resultant dipole moments in each compound as shown below:

Therefore, ${\rm{COFH}}$ shows more dipole moment as polar bonds in ${\rm{CO}}{{\rm{F}}_{\rm{2}}}$ nearly cancel out each other effect.

Ans:

The negative pole on these molecules will be present:

The negative pole on these molecules will be present:

The molecule that shows a higher dipole moment: