# Polarity

## Polarity: Wherein All Things Are Not Created Equal

### Introduction

In our discussion of valence bond theory we learned that covalent bonds are formed when two atoms share one or more pairs of electrons. We now want to examine the nature of that sharing in greater detail. First, we will consider the question of whether two atoms share a pair of electrons equally, i.e. is the electron density in the inter-nuclear region symmetrically distributed or not? Bonds in which the electron density is symmetrically distributed between the nuclei are called non-polar bonds.

### Non-Polar Bonds

Any bond between two identical atoms is non-polar since the electronegativities of the two atoms is identical. The simplest examples are the diatomic molecules such as $$\ce{H2}$$, $$\ce{N2}$$, and $$\ce{F2}$$. The $$\ce{C-C}$$ bond in ethane, $$\ce{H3C-CH3}$$, is also non-polar.

### Polar Bonds

Any bond between two non-identical atoms is by definition polar. The difference in electronegativity of the two bonded atoms means that the electron density is unsymmetrically distributed between them. The bond in $$\ce{HF}$$ is polar. So are the $$\ce{C-H}$$ bonds in $$\ce{CH4}$$.

Figure 1 presents color-coded schematic representations of polar and non-polar bonds.

Figure 1: Imagining Chemical Bonds

The dots in the figure represent the positions of the nuclei. The colored shapes define areas of high electron density, i.e. a covalent bond. In the top panel the left-hand shape indicates that the electron density is symmetrically distributed between the two nuclei. This could represent the situation in a molecule of dihydrogen, $$\ce{H2}$$, for example. The other image implies that the electron density is higher closer to one nucleus than the other. This would be the case for a molecule like hydrogen fluoride, $$\ce{HF}$$, where the electron density lies closer to the more electronegative fluorine atom. In the lower panel, the color gradations correspond to variations in electron density; red represents a region of higher electron density than blue. Note, again, the symmetrical nature of the electron distribution in the left-hand image. The picture on the right suggests that the electron density is higher near one nucleus than the other.

One convention chemists use to depict the electron distribution in organic molecules involves electrostatic potential maps, MEPs. The MEPs of dihydrogen and hydrogen fluoride are shown in Figure 2. By convention, regions of relatively high electron density are colored red, while those of relatively low electron density are blue. Variations in electron density are indicated by variations between these two color extremes. Note the similarity between the models in Figure 2 and the images in the lower panel of Figure 1.

 $$\ce{H-H}$$ $$\ce{F-H}$$

Figure 2: Mapping Electron Density in Organic Molecules

The red region in the model of $$\ce{HF}$$ indicates that the electron density is highest around the fluorine atom. This makes sense because the $$\ce{F}$$ atom has three non-bonding pairs of electrons around it. The deep blue color around the H atom is more telling; this hydrogen is electron deficient because the more electronegative $$\ce{F}$$ atom has pulled the electrons in the $$\ce{H-F}$$ bond away from the $$\ce{H}$$ atom.

The distribution of electron density in a chemical bond depends on the difference in electronegativies of the atoms that share the electrons that comprise that bond. Another convention chemists use to represent the uneven distribution of electron density between two nuclei as shown in structure 1 for the specific case of hydrogen fluoride.

Here the symbols δ- and δ+ are read as "partial minus" and "partial plus". This is the symbolism used to describe a bond dipole. Here's what this convention means: if you compare $$\ce{H-F}$$ to $$\ce{F-F}$$, the electron density around the fluorine atom in $$\ce{H-F}$$ is higher than it is in $$\ce{F-F}$$; if you compare $$\ce{H-F}$$ to $$\ce{H-H}$$, the electron density around the hydrogen atom in $$\ce{H-F}$$ is lower than it is in $$\ce{H-H}$$. The models of $$\ce{H2}$$, $$\ce{HF}$$, and $$\ce{F2}$$ shown in Figure 3 should make this comparison more obvious. Note that the model of $$\ce{HF}$$ is rotated by 180o relative to its orientation in Figure 2.

 $$\ce{H2}$$ $$\ce{HF}$$ $$\ce{F2}$$

Figure 3: Bond Dipoles:Polar and Non-Polar Molecules

Both $$\ce{C-C}$$ bonds in propane $$\ce{CH3CH2CH3}$$ are polar. This is because the terminal carbon atoms and the central carbon are not identical. There are two unique types of carbon atom in propane. The terminal carbon atoms are both bonded to three hydrogen atoms and the central carbon atom. The central carbon, however, is bonded to two hydrogen atoms and two carbon atoms. An alternative way to think about this is to identify the groups that are attached to each carbon in propane. The left-hand carbon is attached to three hydrogen atoms and a $$\ce{CH2CH3}$$ (ethyl) group. The central carbon is attached to two hydrogen atoms and two $$\ce{CH3}$$ (methyl) groups. Since the right-hand carbon is attached to three hydrogen atoms and an ethyl group, it is identical to the left-hand carbon. Looking at propane in this way allows us to introduce the idea of group electronegativities.

### Group Electronegativities

In the same way that the electronegativity of an atom is a measure of the tendency of that atom to attract electrons, group electronegativity is a measure of the tendency of a polyatomic group to attract electrons. Propane, $$\ce{CH3CH2CH3}$$, contains two $$\ce{CH3}$$ groups and one $$\ce{CH2}$$ group (methylene group). Since these two groups are not identical, they have different group electronegativities. Even though the $$\ce{H3C-CH2}$$ bond is a bond between two carbon atoms, the carbons are not identical because they have different atoms attached to them. As far as the pair of electrons that the two carbons share goes, they experience a different Coulombic attraction from the $$\ce{CH3}$$ group than they do from the $$\ce{CH2}$$ group. One way to investigate group electronegativities experimentally involves nuclear magnetic resonance (NMR) spectroscopy. If you want to know more...

Exercise 1 If the four groups attached to the left-hand carbon in $$\ce{CH3CH3}$$ are $$\ce{H}$$, $$\ce{H}$$, $$\ce{H}$$ and $$\ce{CH3}$$, what are the four groups attached to the left-hand carbon in $$\ce{CH3CH2OH}$$? Note-You must type a space between each of the four groups.

Exercise 2 What are the four groups attached to the central carbon in $$\ce{CH3CH(Cl)CH2F}$$? Note-You must type a space between each of the four groups.

Exercise 3 Are the three $$\ce{CH2}$$ groups in pentane, $$\ce{CH3CH2CH2CH2CH3}$$, identical? Yes No

### Polar and Non-Polar Molecules

Our discussion to this point has focused on the polarity of individual bonds. Now we want to turn our attention to entire molecules. We'll start with methane, $$\ce{CH4}$$. As we have seen, the $$\ce{C-H}$$ bonds in methane are polar. However, a molecule of methane is non-polar. Specifically, the dipole moment of methane is zero. A dipole moment of zero means that the "center of negative charge" in the molecule corresponds to the "center of positive charge". In the case of methane, the "center of positive charge" and the "center of negative charge" are focused on the carbon atom. Think of the "center of charge", whether positive or negative, in the same way that you think of the "center of mass". From that perspective, a molecule with a dipole moment of zero is like a perfectly balanced see-saw.

Exercise 4 Select those structures which have a dipole moment of zero. In order to answer the question correctly you must select all of the compounds that have a dipole moment of zero and none of those that do not. When you select a correct option, it will be highlighted in pink.

 $$\ce{CCl4}$$ $$\ce{CHCl3}$$ $$\ce{CH2Cl2}$$ $$\ce{CH3CH3}$$ $$\ce{CH3CH2CH3}$$

An alternative way to determine the polarity of a molecule is to assess the interaction of all the bond dipoles in the molecule. A dipole is a vector quantity, and methane has a dipole moment of zero because the vector sum of the individual bond dipoles equals zero. Figure 2 illustrates the idea for a molecule of methane where one $$\ce{C-H}$$ bond is aligned along the +z axis of a set of Cartesian coordinates. The arrows represent the magnitude of the z component of the bond dipoles. The magnitude of the z component of each of the $$\ce{C-H}$$ bond dipoles shown in color is 1/3 that of the bond dipole that is aligned along the +z axis. The sign of the z component of each of the colored bond dipoles is negative.

Figure 2: A Vector Analysis of the Dipole Moment of Methane

This analysis is based upon the assumption that methane is a tetrahedral molecule. In a way, that assumption puts the cart before the horse. The measurement of dipole moments provided chemists with experimental evidence that allowed them to deduce molecular shapes. Consider the case of dichlorodifluoromethane, $$\ce{CF2Cl2}$$, a component of freon gas. This molecule has a dipole moment of 0.51 Debye. While this fact does not prove that dichlorodifluoromethane is tetrahedral, it does distinguish between square planar structures 2 and 3 .

Exercise 5 Which structure is consistent with the experimental fact that $$\ce{CF2Cl2}$$ has a dipole moment of 0.51 Debye?

2 3

The value of understanding the concept of polarity derives from the insights it provides into many physical and chemical properties of molecules. Let's consider one example, the boiling points of liquids.

### Boiling Points

Coulomb's law tells us that as the distance between opposite charges decreases, the attractive force between them increases. If those charges are molecular dipoles, then intermolecular attractions cause the molecules to aggregate. When the aggregates become large enough, droplets of liquid are formed. In the gas phase, the distance between molecules ( r in Coulomb's law) is very large in comparison to the size of the molecules themselves. The intermolecular attractions are essentially zero. In order to go from a state where there are strong intermolecular interactions to one in which they are essentially zero, it is necessary to add energy to the system. The boiling point of a liquid is a measure of the amount of energy that is required. In other words, the boiling point of a liquid provides an indication of the strength of the forces that hold one molecule next to another.

Exercise 6 Is the relationship between the polarity of a molecule and its boiling point a direct or an inverse correlation?

direct inverse

The boiling point of dihydrogen is -251oC. The boiling point of hydrogen fluoride is +20oC, more than 270o higher!! Clearly the intermolecular forces holding $$\ce{HF}$$ molecules together are much greater than those between HH molecules. Figure 3 provides a pictorial rationalization of the intermolecular forces that hold hydrogen fluoride molecules together in the liquid state.

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Figure 3: Intermolecular Attractions in Liquid HF

The dashed red lines in the figure represent the intermolecular Coulombic attractions between the positive end of a bond dipole in one $$\ce{HF}$$ molecule and the negative end of a bond dipole in another $$\ce{HF}$$ molecule. Note that the separations between molecules are not much greater than the size of the molecules themselves. The dashed red lines do not represent covalent bonds. The interactions they depict are called secondary bonding interactions.

We'll revisit the idea of polarity when we discuss the classification of solvents.

### Contributors

• Otis Rothenberger (Illinois State University) and Thomas Newton University of Southern Maine)