explain how dipole moments depend on both molecular shape and bond polarity. predict whether a molecule will possess a dipole moment from the molecular formula or structure. use the presence or absence of a dipole moment as an aid to deducing the structure of a given compound.

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Study Notes

You must be able to combine your knowledge of molecular shapes and bond polarities to determine whether or not a given compound will have a dipole moment. Conversely, the presence or absence of a dipole moment may also give an important clue to a compound’s structure. BCl3, for example, has no dipole moment, while NH3 does. This suggests that in BCl3 the chlorines around boron are in a trigonal planar arrangement, while the hydrogens around nitrogen in NH3 have a less symmetrical arrangement - trigonal pyramidal.

Remember that the C-Hbond is assumed to be non-polar.

### Molecular Dipole Moments

In molecules containing more than one polar bond, the molecular dipole moment is just the vector combination of what can be regarded as individual "bond dipole moments". Mathematically, dipole moments are vectors; they possess both a magnitude and a direction. The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO2, a linear molecule (Figure $$\PageIndex{1a}$$). Each C–O bond in CO2 is polar, yet experiments show that the CO2 molecule has no dipole moment. Because the two C–O bond dipoles in CO2 are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO2 molecule has no net dipole moment even though it has a substantial separation of charge. In contrast, the H2O molecule is not linear (Figure $$\PageIndex{1b}$$); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H2O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H2O to hydrogen-bond to other polarized or charged species, including other water molecules. Figure $$\PageIndex{1}$$: How Individual Bond Dipole Moments Are Added Together to Give an Overall Molecular Dipole Moment for Two Triatomic Molecules with Different Structures. (a) In CO2, the C–O bond dipoles are equal in magnitude but oriented in opposite directions (at 180°). Their vector sum is zero, so CO2 therefore has no net dipole. (b) In H2O, the O–H bond dipoles are also equal in magnitude, but they are oriented at 104.5° to each other. Hence the vector sum is not zero, and H2O has a net dipole moment.

The following is asimplified equation for a simple separated two-charge system that is present indiatomic molecules or when considering a bond dipole within a molecule.

\< \mu_{diatomic} = Q \times r \label{1a}\>

This bond dipole,µ (Greek mu) is interpreted as the dipole from a charge separation over a distance $$r$$ between the partial charges $$Q^+$$ and $$Q^-$$ (or the more commonly used terms $$δ^+$$ - $$δ^-$$); the orientation of the dipole is along the axis of the bond. The units on dipole moments are typically debyes (D) where one debye is equal to 3.336 x 1030 coulomb meters (C · m) in SI units. Consider a simple system of a single electron and proton separated by a fix distance. The unit charge on an electron is 1.60 X 1019 C and the proton & electron are 100 pm apart (about the length of a typical covalent bond), the dipole moment is calculated as:

\<\begin{align} \mu &= Qr \nonumber \\<4pt> &= (1.60 \times 10^{-19}\, C)(1.00 \times 10^{-10} \,m) \nonumber \\<4pt> &= 1.60 \times 10^{-29} \,C \cdot m \label{2} \end{align}\>

\<\begin{align} \mu &= (1.60 \times 10^{-29}\, C \cdot m) \left(\dfrac{1 \;D}{3.336 \times 10^{-30} \, C \cdot m} \right) \nonumber \\<4pt> &= 4.80\; D \label{3} \end{align}\>

$$4.80\; D$$ is a key reference value and represents a pure charge of +1 and -1 separated by 100 pm. However, if the charge separation were increased then the dipole moment increases (linearly):

If the proton and electron were separated by 120 pm:

\<\mu = \dfrac{120}{100}(4.80\;D) = 5.76\, D \label{4a}\>

If the proton and electron were separated by 150 pm:

\<\mu = \dfrac{150}{100}(4.80 \; D) = 7.20\, D \label{4b}\>

If the proton and electron were separated by 200 pm:

\<\mu = \dfrac{200}{100}(4.80 \; D) = 9.60 \,D \label{4c}\>

Example $$\PageIndex{1}$$: Water

The water molecule in Figure $$\PageIndex{1}$$ can be used to determine the direction and magnitude of the dipole moment. From the electronegativities of oxygen and hydrogen, the difference is 1.2e for each of the hydrogen-oxygen bonds. Next, because the oxygen is the more electronegative atom, it exerts a greater pull on the shared electrons; it also has two lone pairs of electrons. From this, it can be concluded that the dipole moment points from between the two hydrogen atoms toward the oxygen atom. Using the equation above, the dipole moment is calculated to be 1.85 D by multiplying the distance between the oxygen and hydrogen atoms by the charge difference between them and then finding the components of each that point in the direction of the net dipole moment (the angle of the molecule is 104.5˚).

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The bond moment of O-H bond =1.5 D, so the net dipole moment is

\<\mu=2(1.5) \cos \left(\dfrac{104.5˚}{2}\right)=1.84\; D \nonumber\> api/deki/files/66332/dipole_CHCl.png?revision=1&size=bestfit&width=522&height=136" />

Consider $$CCl_4$$, (left panel in figure below), which as a molecule is not polar - in the sense that it doesn"t have an end (or a side) which is slightly negative and one which is slightly positive. The whole of the outside of the molecule is somewhat negative, but there is no overall separation of charge from top to bottom, or from left to right. In contrast, $$CHCl_3$$ is a polar molecule (right panel in figure above). However, although a molecule like CHCl3 has a tetrahedral geometry, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments do not cancel one another, and the result is a molecule which has a dipole moment. The hydrogen at the top of the molecule is less electronegative than carbon and so is slightly positive. This means that the molecule now has a slightly positive "top" and a slightly negative "bottom", and so is overall a polar molecule. 