Molecular dipole

As any charge distribution, a molecule may possess a permanent electric dipole, a vector known as the molecular dipole. The size of this dipole, the molecular dipole moment, gives an indication of the polarity of the molecule, that is, the amount of charge separation in a molecule.

In chemistry, polarity is usually explained by the presence of electronegative and/or electropositive atoms in the molecule. An electronegative atom attracts electrons (becomes negative) and an electropositive atom donates electrons (becomes positive). These concepts are semi-quantitative and different measures for electronegativity and electropositivity of elements are in use, leading to different absolute values of molecular dipoles. However, all measures predict the same direction of the dipole. For physicists the direction of the dipole vector is from negative to positive charge and for many chemists it is the opposite, namely from positive to negative charge.

Molecular dipole moments may be obtained experimentally; the main techniques are microwave spectroscopy and measurements of dielectric constants. Dipole moments can be computed reliably for smaller molecules (say up to 10 second-period atoms) by means of quantum chemical methods.

Selection rule

Whether a molecule has one or more non-zero permanent dipole components depends on the symmetry of the molecule and on the symmetry species of the molecular state under consideration. Symmetry arguments can be used to explain the absence or presence of components of a molecular dipole (symmetry gives a "selection rule"), but such arguments cannot easily predict the magnitudes of non-vanishing dipoles.

Usually one considers molecules in their ground (lowest energy) electronic state and this state is almost always totally symmetric, i.e., invariant under all symmetry operations. Provided a molecule is in a totally symmetric state, it can be shown that a dipole component vanishes when it is not totally symmetric (not invariant under all symmetry operations).

This rule can be proved formally, but can also be understood intuitively. By definition a symmetry operation changes a molecule to a conformation that is indistinguishable from the original conformation. If a dipole component would change under a symmetry operation, it would give a handle on distinguishing the old from the new conformation, so that the two conformations would be distinguishable. This is a contradiction and, hence, either a dipole component is zero or it is invariant (does not change).

More technically: the symmetry operations of a molecule that is rigid—i.e., the nuclei are clamped in space, but the electrons "move" in the quantum mechanical sense of the word—form a group, the point group of the molecule. This point group has irreducible representations among which is the totally symmetric one, commonly denoted by A₁. Most molecular ground states transform as A₁ (alternatively expressed as: "the symmetry species of the ground state is A₁"). Only the components of the dipole that transform according to A₁ are non-vanishing. This is true not only for molecules in an A₁ state, but holds for molecules in any non-degenerate state (a state that transforms according to a one-dimensional irreducible representation of the point group).

Molecules with an inversion center, such as ethene (C₂H₄) and sulfur hexafluoride (SF₆), do not have a permanent dipole moment because all three components of the dipole change sign under inversion (are non-invariant).

Note that the presence of an inversion center is not necessary for the vanishing of permanent dipoles, for instance, methane (CH₄) is a tetrahedrally shaped molecule without an inversion center and yet has no permanent dipole. The point group of methane is the tetrahedral group T_d. The components of the dipole span a 3-dimensional irreducible representation of this group, so that none of the three components is invariant under the operations of T_d; all three components vanish when methane is in a non-degenerate electronic state.

Bond dipoles

There is a pedestrian way of predicting the vanishing of dipole moments, one that does not require knowledge of point groups and symmetry operations. One assigns to each bonded pair of atoms a bond dipole and one adds these bond dipoles vectorially to a total dipole moment of the molecule. Equal bonds have equal dipoles. In this way one predicts, for instance, that the linear molecule O-C-O has no dipole because the two C-O bond dipoles are antiparallel and cancel each other. (By the symmetry argument it follows from the presence of an inversion center that that O-C-O has no dipole.) As another example we mention that in the case of methane (CH₄), a tetrahedrally shaped molecule, the four C-H bond dipoles add up to zero. It is apparent that in the addition of the four bond dipoles it must be used that these vectors point from the center of the regular tetrahedral molecule to its vertices. Hence the symmetry of methane is used implicitly in this argument.

A clear advantage of the bond dipole method is that, given a table of bond dipoles (and the assumption that these are transferable between different molecules) a fair estimate of the total molecular dipole is obtained by the vector addition of the bond dipoles.

Units and order of magnitude

An electric dipole moment has the dimension charge times length. The SI unit of dipole is accordingly coulomb times meter. However, this unit is unwieldly large and therefore hardly used in chemistry and molecular physics. The Gaussian unit of debye (D) is most widely applied. It is 10⁻¹⁰ esu times ångstrom. An ångstrom is 10⁻⁸ cm = 10⁻¹⁰ m. An esu (electrostatic unit of charge, now called statcoulomb) is C/(10⋅c) ≈ 3.335 640 95⋅10⁻¹⁰ C, where C is coulomb and c is speed of light. Hence,

$1\;\mathrm {D} =10^{-10}\;{\frac {10^{-10}}{10c}}\;\mathrm {C\,m} \approx 3.335\,640\,95\cdot 10^{-30}\;\;\mathrm {C\,m} .$

In quantum chemistry and molecular physics, a common unit of dipole moment is the atomic unit: the charge e of an electron times the bohr radius a₀. Since 1 e = 1.602 176 487 ⋅ 10⁻¹⁹ C and a₀ = 0.529 177 208 59 ⋅ 10⁻¹⁰ m, it follows that

$1\;\mathrm {au} =8.4783528106\cdot 10^{-30}\;\mathrm {C\cdot m} =2.54174623039\;\mathrm {D}$

The debye is of such magnitude that most molecules have dipole moments on the order of 1 to 10 D. For instance, the water molecule has an electric dipole moment of 1.85 D and HCl has 1.09 D. In both cases the direction of the dipole is determined by the fact that the hydrogen atom(s) is(are) slightly positive. In the case of the bent molecule H₂O the dipole bisects the H-O-H angle and in the case of HCl the dipole points from Cl to H (in the physics convention).

Quantum mechanical theory

A few remarks are made about the quantum mechanical background of the molecular dipole. Readers without knowledge of quantum mechanics are advised to skip this section.

Dipole operator

As an observable property, a dipole is represented by a Hermitian operator on the state space of the molecule. This operator is obtained from the corresponding classical expression by reinterpreting the spatial coordinates as multiplicative operators. Classically a molecule is a charge distribution consisting of N point charges q_k at positions r_k (electrons and nuclei),

$\rho (\mathbf {r} )=\sum _{k=1}^{N}q_{k}\,\delta (\mathbf {r} -\mathbf {r} _{k}),$

where δ is the Dirac delta function. Classically, the dipole is the first moment of this charge distribution,

${\boldsymbol {\mu }}\equiv \int \mathbf {r} \,\rho (\mathbf {r} )\;\mathrm {d} \mathbf {r} =\sum _{k=1}^{N}q_{k}\,\mathbf {r} _{k}.$

The rightmost expression is the quantum mechanical dipole operator, where the sum is over electrons and nuclei.

Interaction with electric field

When the molecule is brought into an electric field F it obtains an energy with operator

$V=\mathbf {F} \cdot {\boldsymbol {\mu }}=\mathbf {F} \cdot \left(\sum _{k=1}^{N}\,q_{k}\;\mathbf {r} _{k}\right),$

which is added to the usual field-free Hamiltonian H₀ of the molecule.

Definition observed dipole

If one solves the Schrödinger equation with total Hamiltonian H = H₀ + V one obtains a total energy E(F) and a corresponding wave function Ψ(F), both depending on F. We assume that the state Ψ is non-degenerate. The observed dipole is given by the first derivative (it is a so-called first-order property, a second-order property is a second derivative, third derivative gives a third-order property, and so on),

${\boldsymbol {\mu }}_{\mathrm {obs} }\;{\stackrel {\mathrm {def} }{=}}\;\left({\frac {\partial E(\mathbf {F} )}{\partial \mathbf {F} }}\right)_{\mathbf {F} =\mathbf {0} }.$

Hellmann-Feynman theorem

When the Hellmann-Feynman theorem holds a much simplified expression for the observed dipole may be derived. Use that

$E(\mathbf {F} )=\langle \Psi (\mathbf {F} )\;|\;H_{0}+V\;|\;\Psi (\mathbf {F} )\rangle \equiv \langle \Psi (\mathbf {F} )\;|\;H\;|\;\Psi (\mathbf {F} )\rangle$

and we get

${\begin{aligned}{\boldsymbol {\mu }}_{\mathrm {obs} }&=\left({\frac {\partial \langle \Psi (\mathbf {F} )\;|\;H\;|\;\Psi (\mathbf {F} )\rangle }{\partial \mathbf {F} }}\right)_{\mathbf {F} =\mathbf {0} }\\&={\Big \langle }{\frac {\partial \Psi (\mathbf {F} )}{\partial \mathbf {F} }}\;|\;H\;|\;\Psi (\mathbf {F} ){\Big \rangle }_{\mathbf {F} =\mathbf {0} }+{\Big \langle }\Psi (\mathbf {F} )\;|\;H\;|\;{\frac {\partial \Psi (\mathbf {F} )}{\partial \mathbf {F} }}{\Big \rangle }_{\mathbf {F} =\mathbf {0} }+\langle \Psi (\mathbf {0} )\;|\;{\boldsymbol {\mu }}\;|\;\Psi (\mathbf {0} )\rangle ,\end{aligned}}$

where we used that

${\frac {\partial H}{\partial \mathbf {F} }}={\frac {\partial V}{\partial \mathbf {F} }}={\boldsymbol {\mu }}.$

The Hellmann-Feynman theorem (that holds in any case if Ψ is an exact eigenfunction of H) states,

${\Big \langle }{\frac {\partial \Psi (\mathbf {F} )}{\partial \mathbf {F} }}\;|\;H\;|\;\Psi (\mathbf {F} ){\Big \rangle }+{\Big \langle }\Psi (\mathbf {F} )\;|\;H\;|\;{\frac {\partial \Psi (\mathbf {F} )}{\partial \mathbf {F} }}{\Big \rangle }=0.$

When the theorem holds the observed dipole is given as an expectation value

${\boldsymbol {\mu }}_{\mathrm {obs} }=\langle \;\Psi (\mathbf {0} )\;|\;{\boldsymbol {\mu }}\;|\;\Psi (\mathbf {0} )\;\rangle .$

The wave function Ψ(0) does not depend on F. It is an eigenfunction of the field-free molecular Hamiltonian H₀. This function is not perturbed by V, and is referred to as zeroth-order in V, so that the observed dipole is the first derivative of the first-order energy, whenever the Hellmann-Feynman theorem holds,

${\boldsymbol {\mu }}_{\mathrm {obs} }={\frac {\partial E^{(1)}}{\partial \mathbf {F} }}={\frac {\partial \langle \Psi ^{(0)}|V|\Psi ^{(0)}\rangle }{\partial \mathbf {F} }}=\langle \,\Psi ^{(0)}\,|\,{\boldsymbol {\mu }}\,|\,\Psi ^{(0)}\,\rangle \quad {\hbox{with}}\quad \Psi ^{(0)}\equiv \Psi (\mathbf {0} ).$

There are computational methods yielding wave functions that do not obey the Hellmann-Feynman theorem. In that case the first derivative of the total energy does not give the same result as the first derivative of the first-order energy. It is generally found that the derivative of the total energy (computed in the presence of a field) is the more reliable route to μ. It is also a more expensive route because the computation of μ is usually done by numerical differentiation. This requires the computation of the total energy for several different values of F, while the route via the first-order energy only requires the computation of the expectation value of μ.

Symmetry

In order to indicate how symmetry can predict the vanishing of a molecular dipole, we consider the first-order formula and space inversion (parity) Π. Parity is a unitary operator. Assume

$\Pi \Psi ^{(0)}=\pm \Psi ^{(0)}$

and we know that

$\Pi \,{\boldsymbol {\mu }}\Pi ^{\dagger }=-{\boldsymbol {\mu }}\quad {\hbox{with}}\quad \Pi ^{\dagger }\,\Pi =1.$

Now

${\begin{aligned}{\boldsymbol {\mu }}_{\mathrm {obs} }&=\langle \Psi ^{(0)}\;|\;\Pi ^{\dagger }\,\Pi \,{\boldsymbol {\mu }}\,\Pi ^{\dagger }\,\Pi \;|\;\Psi ^{(0)}\rangle \\&=\langle \Pi \,\Psi ^{(0)}\;|\;-{\boldsymbol {\mu }}\;|\;\Pi \,\Psi ^{(0)}\rangle \\&=-(\pm )^{2}\langle \Psi ^{(0)}|{\boldsymbol {\mu }}|\Psi ^{(0)}\rangle =-{\boldsymbol {\mu }}_{\mathrm {obs} },\end{aligned}}$

where we used that the adjoint of Π, moved to the bra, becomes Π. And finally

${\boldsymbol {\mu }}_{\mathrm {obs} }=-{\boldsymbol {\mu }}_{\mathrm {obs} }\;\Longrightarrow \;{\boldsymbol {\mu }}_{\mathrm {obs} }=0.$

This argument can generalized to arbitrary point groups and leads to the rule: the product of the symmetry species (irreducible representations) of bra, ket, and operator (component of dipole) must contain the totally symmetric representation in order that the expectation value of the operator does not vanish.