Inferring Charge Distribution

Central goal: Understand how to infer charge distribution in a molecule based on its atomic composition, connectivity, and three dimensional geometry

Electron Distribution

Electronegativities of different elemental atoms

The electronegativity of different atoms in a molecule influence the distribution of electrons in the electron cloud

Partial Charges

Electrons will spend more time in regions around atoms that are more electronegative, resulting in partial charges in a molecule

Partial charge = small electric charge resulting from uneven distribution of valance electrons in a molecule

Partial charges are represented by a lowercase delta, followed by either a negative or positive sign δ- = partial negative δ+ = partial positive

We write the partial negative/positive charges next to the corresponding regions in the molecule Partial charge =/= formal charge!

Bond Dipoles

Differences in electronegativity (EN) results in bond dipoles and different classifications of bonds

Polar covalent bond = Covalent bond in which the participating atoms have an EN difference of 0.4 or greater (involve partial positive and negative charges) Nonpolar covalent bond = Covalent bond in which the participating atoms have an EN difference less than 0.4

Tells us about the polarity of a bond. The greater then EN difference, the more polar the bond

Ionic bonds have EN differences greater than 1.8, but are a metal + nonmetal and therefore don’t get these classifications

EN of O = 3.5 EN of H = 2.1

3.5-2.1 = 1.4

Bond Polarity

Draw a dipole vector along a bond starting at the partially positive and ending at the partially negative region of the bond

EN of C = 2.5 EN of H = 2.1 EN of F = 4.0

For C-F: 4.0-2.5 = 1.5

For C-H: 2.5-2.1= 0.4

A longer arrow = a more polar bond (use this when you have bonds of different polarities in the same molecule)

EN of O = 3.5 EN of H = 2.1

3.5-2.1 = 1.4

We represent the polarity of polar bonds using arrows

net dipole moment

Summing the vectors results in one of two scenarios:

1. The dipoles cancel and the net dipole moment is zero (non-polar molecule)

2. The diploes sum to give a net dipole in one direction (polar molecule)

This will be important to predicting interactions between molecules in Unit 3!

We sum the vectors drawn on each bond to determine the net dipole moment

Scenario #1: A nonpolar molecule

Each bond is polar, but because their dipoles are of equal magnitude (same EN difference) and pointing in opposite directions, they cancel out and our molecule is non-polar

Scenario #2: A nonpolar molecule

If, in the molecular geometry of a compound, the bond dipoles are evenly spaced, they will cancel! (even if the angles are hard to visualize)

Scenario #3: A polar molecule

Net dipole moment

Add the vectors “head to tail” to get the net dipole moment

In water, we have two bond dipoles. When summed, the total dipole of the molecule is pointing up

Hydrocarbons

EN differences between C and H is barely polar CH bonds are evenly distributed Overall, hydrocarbons are nonpolar

Hydrocarbons (molecules made up of only hydrogen and carbon) are a special case

EN of C = 2.5 EN of H = 2.1

For C-H: 2.5-2.1= 0.4

Infared Radiation Relations

Molecules must have a polar bond in order to be IR light active

Which of the following molecules are IR active?

Bond polarity allows us to predict which molecules will interact with IR radiation

A note: often structures are drawn with lone pairs implied. On an exam you will be expected to include lone pairs when asked to draw Lewis structures, but I will specify this in the question.