To begin smoothly, we will describe the simplest uncharged molecule: the hydrogen. It is composed of one proton with a positive charge e and one electron of opposite charge –e that revolves around the proton at a distance r.

In quantum mechanics, the system is described by the equation of Schrödinger

Ψ is a wave function and Ĥ is the Hamiltonian, composed of one kinetic term and of one potential term.

The potential term is the potential of attraction between two opposite charges in the void (ϵ₀ is the void permittivity).

The kinetic term describes the movement of the electron in orbit around the nucleus, where ћ=h/2π and the gradient ∇ is

in Cartesian coordinates. It is interesting to change for spherical coordinates. The expression looks more complex but allows to separate some expressions.

Then

This expression seems more complex than the previous one, but there will be one big advantage. The equation of Schrödinger becomes

This equation, developed, gives

As a result, the wave function can now be separated into one radial equation and one angular equation.

The solution of the equations are directly related to the quantic number l. As a reminder, the quantic numbers n, l and m give the electronic configuration of the atoms with l between zero and n (O ≤ l ≤ n) and m between –l and l (–l ≤ m ≤ l). For instance if n=2, l can have a value of 1 or 0, giving 3 possible values for m: -1, 0 (twice), 1.  Fermi told us that we may put 2 electrons of opposite spin on each orbital. For the atoms, the orbitals are noted s, p, d, f, g,… for l=0, 1, 2, 3, 4,… preceded by the quantic number n.

We will first try to solve the radial equation. To do that we multiply everything by ћ²/2m_er² and change the variable from R(r) to P(r)=rR(r).

There are three terms in the equation: the radial kinetic energy, the Coulomb potential and the centrifugal term. The equation is similar to the initial Hamiltonian but there is an additional centrifuge term. The potential term is now composed of a Coulomb term (potential of attraction) and of the centrifugal term.

If we consider the effective potential, we see that the centrifuge term repulses the electrons from the nucleus. The s orbitals (l=0) have no centrifugal term and their electrons are thus close to the nucleus. The other orbitals involve a positive centrifugal term moving the electrons away from the nucleus.

The energy E_n of the orbitals depends on n and not on l or m.

It is proportional to 1/n². The orbitals are thus closer and closer of each other when n increases. The energy is negative and tends to zero for n→∞, i.e. the ionisation. The energy is also proportional to the square of the atomic number. The energy is thus different for H or He⁺, or any atom with one single electron despite the similar structure.

The solution for the wave function is

The wave function R_nl(r) will thus depend on the quantic numbers n, l, the atomic number and the radius. In the case of s orbitals, the wave function is not equal to zero at r=0. Yet the density of probability is r².R_nl(r)² and there is thus no electron at the nucleus.

As the density of probability is the square of something, the areas with a negative value of R_nl(r) can possess electrons. The points where the density of probability equals zero are called nodes and their number is equal to n-l-1. Note that there is thus no fixed radius for an electron but for its largest probability of presence and that even if the energy of the orbitals are equal, the electrons are not at the same distance from the nucleus. For monoelectronic atoms, the orbitals 3s and 3p are thus at the same energy. They are degenerated. It is the case for all the orbitals with a same value of n.

Now, let’s take a look at the angular wave function Y.

Where P_l^∣m∣(cos ) is the associated functions of Legendre. The solution here does not depend on n but on l and m. A few solutions are given below:

The first angular wave function Y₀⁰ gives a sphere as it has no dependence on the angles and ϕ. The other functions have different shapes that depends on the angles.

To obtain atomic orbitals we just have to combine the angular and the radial wave functions. The orbitals s are spherical. There are 3 orbitals p because there are 3 possible values for m: -1, 0 or 1. The orbitals are identical but point in different directions. Actually it is a bit more complicated: the angular wave functions for m=-1 and 1 have an imaginary term that we get rid of by a change of coordinates, back to the Cartesian ones. We can do this change with a combination of the orbitals. We can draw the orbitals with the following technique:

First we take the functions that depend on one angle and draw their value as a function of the angle. For the orbital p_z the functions are f()=cos and F(ϕ)=1. While F=1 for any angle, f goes from 1 to -1. The orbital is given by the multiplication of the two functions. F=constant means that the orbital is identical for any value of ϕ. The axis z is thus an axis of symmetry. is the angle in the z direction. When =0 (point a), we are on the axis z and the intensity is 1. When >0, we are not on the axis. The intensity is smaller than one so we draw a point at a smaller distance from the centre than the previous point. The intensity keeps decreasing to reach zero at =π/2 (point g), the intersection with the plane x=0, y=0. In fact, we just drew one sphere above the plane x=0, y=0. The same is done at the other side of the plane (π/2< <π). This part of the orbital has a negative sign because the function is negative (cos <0). We can repeat this method for the other directions.

Operators can be applied to the wave functions and respect the equation of Schrödinger. The operator inversion Î is an operator such as, if a central symmetry can be found,

For instance, the orbital s have a centre of symmetry. We say that this state is even. If we apply the inversion operator to this orbital s, the sign of the orbital is still unchanged at the opposite side of the atom (i.e. obtained by central symmetry).

The orbital p is odd because if we look at one point in the positive part of the orbital and apply the symmetry, we are now in the negative lobe of the orbital. The d orbitals are even. If fact, we have

The operator Î can also be applied to electrons to obtain the opposite spin (up à down).

Angular moments

The angular moment L is the cross product of two vectors r and p. The cross product or vector product is a binary operation on two vectors and is denoted by the symbol ×. Given two linearly independent vectors a and b, the cross product, a × b, is a vector c that is perpendicular to both and therefore normal to the plane containing them (two linear vectors always share a plane).

The intensity of the new vector is a × b = ∣a∣∣b∣ sin n (with n the unity vector pointing in the good direction) and the direction of the cross product is given by the rule of the right hand: the index stands points in the direction of the first vector and the middle finger points in the direction of the second vector. If you put your thumb perpendicular to the two other fingers (as to show your approval), it points the direction of the cross product. It can be necessary to rotate your hand strangely to get the good directions, for instance to obtain the b x a product from above.

In quantum mechanics, p is an operator

We have thus

Note that L_z does not depend on and there is thus a symmetry around the z axis for this operator. If we apply the operator Lz on the angular wave function, we obtain the wave function multiplied by ћm:

It means that the angular wave function is a proper function of the operator Lz. It is also the case with the operator L².

We can thus simultaneously determine the proper values of L² (=ћ²l(l+1)), characteristic of the angular moment L, and of L_z (=ћm), the projection of L on the z axis, for fixed values of the quantic numbers.

The proper value of an operator is the integral of the wave function multiplied by the operator acting on the wave function:

The length of the vector L and of its projections on the axes.

For instance, for l=1, we can draw L and its projection as a function of m. The length of L doesn’t depend on m but the projection on the axis z does.

L² and L_z commute: two operators that commute have the same wave function. The commutativity means that

It is indeed the case: [L²,L_z]=0 and it is not the case between L_x, L_y and L_z:

Note that L_x and L_y also commute with L², like L_z does, but their proper values <L_x> and <L_y> equal zero.

The same kind of properties is true for the spin of the electrons with the operators S² and S_x, S_y, S_z.

To describe the whole system, we have thus a series of equations containing the 5 quantic numbers n, l, m_l, s and m_s. They form a CSCO (complete set of commuting observables).

Matrix representation

In an orthonormal base of functions {Ψ₁, Ψ₂,…Ψ_N}, the operator  has a matrix representation (Â) which is a matrix composed of operators Â_ij such as

If Ψ₁=Ψ₂, the result would have been the proper value of Ψ₁ if Ψ₁ is a proper function of Â. It is usual to use the bra-ket notation of Dirac:

The part before the operator is called the bra and the part after the operator is the ket.

If Ψ₁ is not a proper function of  (ÂΨ≠aΨ), then we can find a linear combination (combili) of wave functions such as

or written as matrices

Let’s take an example with the operators L_z and L² and the angular functions Y_m^l that we used before for l=1. The complete base of functions is Y₁^m={Y₁¹, Y₁⁰, Y₁^-1}. The equations are

The values of a_ij can easily be found

The equation for Y₁¹ has no dependence on Y₁^-1 or Y₁⁰. As a result, aij=0 for i≠j and aii=2ћ².

The same goes for Y₀¹ and Y₁¹

Then we have that the matrix representation (L²) of the operator L²

This matrix is diagonal, i.e. the components of the matrix that are not on the diagonal are all equal to zero. (L_z) is also diagonal

The rule is that if the operators are operators of functions that are proper functions of this operator, then the matrix is diagonal. All the operators of the CSCO have a diagonal matrix representation on the base of their common proper functions. Note that it means that L_x and L_y are not part of the CSCO: they don’t commute with L_z and their matrix representation are

It is a direct consequence of the Heisenberg’s uncertainty principle: we cannot know the exact position and speed of a particle at the same time. As a result, L² cannot commute with all of the L_x, L_y and L_z.

We can define the operators of climbing L₊ and of descent L_– from L_x and L_y.

For our example with l=1, they are

Their role is to move from one orbital to another one.

As the last line shows it, it is impossible to move out of the system. There is no Y₁^-2 orbital.

The presence of a second electron induces a term of repulsion between electrons.

This term is positive so it increases the energy of the orbitals. The rest of the equation is similar to the Hamiltonian of the hydrogen. We can compare the energy of the orbitals with those two models.

The repulsion is large in the 1s orbital wherein the electrons are close to each other (the radius being small). For instance, the energy of the orbital 1s² of the Helium is at -2.9au (atomic units). If we calculate the energy without the repulsion term, i.e. as a monoelectronic atom (Ĥ= Ĥ₁ + Ĥ₂), we find -4au (E₁=E₂=-Z²/2n²=2au). Another consequence is that the orbitals s and p don’t have the same energy anymore. The configurations 1s¹2s¹ has an energy of -2.17au and the configuration 1s¹2p¹ an energy of -2.13au.

We can separate the Hamiltonian into one monoelectronic Hamiltonian term and the term of repulsion between the electrons.

ĥ(i) is the monoelectronic Hamiltonian of one electron.  We sum up the terms for all the electrons to obtain the Hamiltonian. To resume, the repulsion spreads the degenerated orbitals from the monoatomic model.

The repulsion can be large enough to have an orbital of level n+1 of lower energy than an orbital of n. It is the case for orbitals d and f and it is why we use the following technique to fill the orbitals with electrons:

The 4s orbital is thus filled before the 3d orbital.

The operators that commute are simply the sum of the operators applied to the electrons independently (but perturbed by the repulsions). We write these operators with a lower case to denote them from the global operators (Ĥ→ĥ, Ŝ→ŝ, …).

The electrons are indistinguishable. It means that if we exchange an electron of an atom with another electron of the atom on the same orbital or on a different orbital, the atom remains unchanged. We can thus introduce a permutation operator P_ij that commutes with the Hamiltonian Ĥ and the proper value of which is ±1. This last affirmation is true because we get back to the initial conditions if we apply the permutation operator P_ij twice. The identity operator Ê is the operator with which nothing changes: ÊΨ=Ψ.

The value of p is imposed by the type of particle. Bosons have a integer spin and p_boson=1. Fermions have half-integer spins and p_fermion=-1. Electrons are thus fermions. Some nuclei are bosons and some are fermions. In an atom with N electrons, N! permutations of electrons are possible. For instance, the lithium has 3 electrons with spherical coordinates (r₁, σ ₁), (r₂, σ₂) and (r₃, σ ₃) (considering the orbitals 1s and 2s) that can be permuted between 3 spots a, b and c. There are 6 possible permutations 6=3! = 3x2x1): P₁₂, P₁₃, P₂₃, (P₁₃,P₁₂), (P₁₃,P₂₃) and Ê.

We don’t know which electron is in which spot (a, b or c). The wave function Ψ is one of the functions

To write the wave function we use the determinant of Slater which is a combination of the possible states.

with a, b, c, …, z are the spin orbitals.

In mathematics, a determinant is read in diagonal. The elements of the diagonals (from top left to bottom right) are multiplied and the diagonals are summed up. Then we deduce the diagonals in the other way (from top right to bottom left). In the case of the lithium we have

The determinant of Slater has a few properties:

• if we exchange two lines or two columns, the sign of the determinant changes.
• if two columns are identical, the determinant equals 0.

It can be translated by the law of Fermi: 2 electrons must differ by at least one quantic number.

Let’s apply this to the helium and a few of its excited states. He has 2 electrons that are distributed on the 1s orbital in its ground state. There can be two electrons on this orbitals if they have different spins (1/2 and -1/2). In the notation of Slater, the spin is indicated by the presence of a line above the orbital if the spin=-1/2.

For the excited state wherein one electrons jumped to the orbital 2s, there are 4 possible states depending on the spin of the electrons:

For instance,

If instead the electron jumped to the 2p orbital, there are 12 possible states:

For instance,

Coupling of Russel-Saunders

This coupling is also called the L-S coupling. We consider it for multi-electron atoms with weak spin-orbit couplings. In this case, the orbital angular moments of the individual electrons add to form a resultant orbital angular momentum L and the same is true for the spin moments to form a spin angular moment S. L and S combine to form the total angular moment J.

J=L+S

The resultant of two vectors is their sum. If we consider two operators Ĵ₁ and Ĵ₂ (which can be L and S), their resultant Ĵ = Ĵ₁+ Ĵ₂.

Ĵ_z is simply equal to the sum of Ĵ_1z and Ĵ_2z and Ĵ² is the development of the square of the sum

We can look at the coupling between the electrons in He. The electrons have a spin s=1/2 and m_s=±1/2. Let’s say that m_s=1/2 is the state α and that m_s=-1/2 is the state β. The coupling can be αα, αβ, βα or ββ coupling. The coupled M_s=m_s1+m_s2 can thus be equal to 1, 0, 0 or -1. We write these values in a table. I insist on the fact that M_s=0 is degenerated.

To know the resulting states of the coupling between the electrons, we take the largest S=s₁+s₂ and look for its M_s in the results of the coupling. It is S_max=1 with M_s=-1, 0 and 1. We discard these values of M_s from the 4 couplings that we obtained earlier. If there are one or more remaining M_s in the table, we do the same for S=S_max-1, etc. Here, there is one last M_s=0 that comes from S=0. The 4 couplings correspond thus to a singlet state (S=0 and M_s=0) and 1 triplet state (S=1 and M_s=1, 0 or -1). S goes from ∣s₁+s₂∣ to ∣s₁-s₂∣ by step of 1. The multiplicity is equal to 2S+1. The difference between the states is always by step of ΔS=1 and ΔM_s=1. The states can be represented this way:

Note that the vectors for M_s=1 or M_s=-1 are not two times the length of the vectors for ms1 and ms2. The vertical component is. Note too that the resulting vector for M_s=0 may have a nonzero length (S=1).

If 3 electrons are coupled, we proceed step by step. First we do the coupling between two electrons. As previously we obtain two different states, S₁₂=0 (singlet) and S₁₂=1 (triplet). There is no need yet to determine Ms. The last electron is coupled next to the two possible states S₁₂. The coupling with the singlet S₁₂=0⊗s₃=1/2 gives S₁₂₃=1/2. It is a doublet (M_s=1/2 or M_s=-1/2, remember that ΔM_s=1 between ∣S₁₂+s₃∣ to ∣S₁₂-s₃∣). The coupling S₁₂=1⊗s₃=1/2 gives a total of S₁₂₃=3/2 (1+1/2), leading to two states S₁₂₃=3/2 and S₁₂₃=1/2 (ΔS=1). It means that there is a quadruplet (2(3/2)+1=2) and a doublet (2(1/2)+1=4). In total, there are 2ⁿ=8 possible states, n being the number of electrons.

In the case of degenerated orbitals as the orbitals p or d, the amount of configurations is larger (see the part about the determinant of Slater) but the method is identical and is also applied to the quantic number l. Let’s take a few examples: the electronic configurations (1s)¹(2s)¹, (1s)¹(2p)¹ and (2p)¹(3p)¹.

Configuration (1s)¹(2s)¹

The degeneration is 4: the two electron can have a positive or a negative spin (α or β). We have thus 2×2=4 degenerated configurations:

There is only one value for the orbital number L=0: l_1s=0⊗l_2s=0. As a result, M_L=0. As we have seen above, the coupling between two electrons s_1S=1/2⊗s_2s=1/2→S=1, 0 gives one singlet (S=0, degeneration =1) and one triplet (S=1, deg=3).

We write the electronic term depending on L and S. L gives a capital letter S, P, D, F the same way n gives s, p, d, f. The electronic degeneration is written as an exponent before the orbital letter. The electronic terms for the configuration (1s)¹(2s)¹ are thus ¹S, ³S. The degeneration of the terms is the multiplication of the degeneration of L and S and we sum up the degeneration of the electronic terms: 1.1 (for ¹S) +1.3 (for ³S) = 4. We have thus a concordance between the degeneration of the terms and the degeneration of configuration.

Configuration (1s)¹(2p)¹

The degeneration is 12 in this case: both electrons can have a positive or negative spin (α or β) but the 2p electron can be on 3 different orbitals (2p_–, 2p₀ or 2p₊). We have thus (1.2).(3.2)=12 degenerated configurations (written earlier in the chapter).

The coupling l_1s=0⊗l_2p=1 gives L=1 (M_L=-1, 0, 1). There is no L=0 here: if we look at the details of the coupling.

We find the M_L’s of L=1 but once they are discarded, there is no other M_L. There is thus no L=0. An easier way to know it is that L goes from ∣l₁=l₂∣ (=1) to ∣l₁-l₂∣ (=1-0=1).

As L=1, the electronic terms will be ^xP with x=1 and 3 from the spin. As always, the coupling between two electrons gives a triplet and a singlet. The electronic terms are thus ¹P, ³P with a degeneration (3.1)+(3.3)=12.

Configuration (2p)¹(3p)¹

The degeneration is 36 ((3.2).(3.2)): 3 orbitals with 2 possible spins for each electron. The coupling l_2p=1⊗l_3p=1 gives L=2 (M_L=-2, -1, 0, 1, 2), L=1 (M_L=-1, 0, 1) and L=0 (M_L=0). The coupling of spins gives one singlet and one triplet as usual. We have thus the electronic terms

¹S, ¹P, ¹D, ³S, ³P, ³D that can also be written as ¹(S, P, D), ³(S, P, D) or even ^1,3(S, P, D)

The degeneration are respectively (1.1), (3.1), (5.1), (1.3), (3.3) and (5.3) for a total of 36.

If we want to couple a third electron, we obtain the electronic terms by coupling separately the electronic terms if the electrons are not equivalent. For instance, if we add a 3d electron (²D) to the 2p3p electrons, we proceed this way:

One can see that some electronic terms are obtained several times (²F for instance). Those terms are not identical and have not the same energy because they come from different couplings.

In the case of equivalent electrons, the amount of states is more limited. For instance we had 36 possibilities for the 2p3p configuration. This amount drops to 15 for the 2p² configuration. This amount is equal to X!/(X-N)!N! where X is the degeneration of the state and N the number of electrons. In our case, 2p², N=2 and X=6 for the orbital 2p: there are 3 orbitals (2p₊, 2p₀ and 2p_–) and 2 spins. There are thus 6.5/2=15 possibilities. To determine the electronic terms of this coupling, we will make a list of the 15 possibilities and identify each one by the couple (M_L,M_S). To do so we draw a table with a column for each one of the 6 states and we put pairs of electrons together. In a seventh column, we write down the (M_L, M_S) of the couple of electrons.

Once it is done, we seek for the emergents. An emergent is one couple of projections that is unique. We take the one with the largest values, i.e. (2,0) in this case and we determine the electronic term from which the pair comes. M_L=2 means that L=2 and M_S=0 means that S=0. The corresponding electronic term is ¹D that has a degeneration of 5. As a consequence we will find 5 pairs of projections in the table that come from the electronic term ¹D.

We can remove them from the list and look for another emergent: (1,1). The corresponding electronic term is ³P with a degeneration of 9. From the 15 possibilities, 14 correspond to 2 electronic terms (³P and ¹D). The last emergent is (0,0), what corresponds to the electronic term ¹S. The electronic terms of the configuration 2p² are thus ¹D ³P ¹S. In comparison with inequivalent electrons (2p3p), the ³D, ¹P and ³S terms disappeared. Also note that several pairs were several times in the list and we randomly selected which one corresponds to which electronic term. For instance, we don’t know which (1,0) comes from ¹D or from ³P.

In terms of energy, the rules of Hund tell us that the term with the highest M_S is the most stable. If there are identical M_S projections, then the highest M_L is the most stable. For instance, the ³P coupling of 2p² has a lower energy than ¹D itself lower than ¹S.

Here come two properties of the L-S coupling:

• Complete orbitals are all ¹S:

As a result, they do not modify the other orbitals. Note that even if they do not perturb the other orbitals, they do contribute to the energy of the electronic states.

• A “hole” (the absence of electron) in one orbital gives the opposite signs of M_L and M_S than the presence of this electron. For instance

But the sign doesn’t change the state. As a result, the carbon (2p²) and the oxygen (2p⁴) have the same state ²P.

Spin-orbit coupling

The magnetic moment induced by the spin of the electron interacts with the magnetic field induced by the electric current resulting from its orbital movement. This relativist effect induces a further separation of the states and an additional term in the Hamiltonian.

The coupling introduces a new observable J=L+S

J goes from L+S to ∣L-S∣ by step of 1. Considering this, [Ĥ,L²] and [Ĥ,Ŝ²]≠0 but [Ĥ,Ĵ²]=0.

The relativist effects grow with the atomic number of the atom: the larger the atomic number, the heavier the nucleus. Its charge increases as well meaning that the electrons have to revolve faster to compensate.

Energy of the states

The energy of the states can be found from the equation of Schrödinger

From that, we can write

The energy is thus the average value of Ĥ. To determine it, we start from a linear combination of determinants of Slater (M_L,M_S). An emergent determinant of Slater always gives a proper function. From this point, we can apply operators L₊, L_–, Ŝ₊ or Ŝ_– to increase/decrease the value of M_L or M_S and find their energy.

We do it for each state that has at least an emergent. If there is no emergent, we can apply the rules of orthogonality: 2 wave functions that differ by at least one proper value of operators that commute together are orthogonal. Eventually, the proper functions are normalised.

With this method, the energies of the states can be found. One simple example says more than a thousand words. The coupling of two electrons gives a singlet (S=0, MS=0) and a triplet (S=1, M_S=1, 0, -1). We will place the values of M_S in boxes for S=1 and S=0, as shown below.

An emergent is found for M_S=1 (S=1) so we can find the wave function Ψ_S=1,Ms=1 (the state αα). We also have the wave function of Ψ_1,-1 (the state ββ) because it is an emergent too.

To obtain the wave function of the state M_S=0, we apply the operator of descent Ŝ_– on Ψ_1,1 (we could have applied Ŝ₊ on Ψ_1,-1 instead).

One electron in the α state is thus now in the β state. We need to norm the function. To do so, we must have the norm N equal to

Applied here, we have

The orbitals are orthonormal, meaning that the integrals equal 1 if the two terms of the integrant are identical and zero if they are different: ∫αα=1, ∫ββ=1, ∫αβ=0, ∫βα=0. The cross terms of the square (in blue) are thus equal to zero and the others equal 1.

As a result,

To obtain the wave function of the singlet, we can use the property of orthogonality between Ψ_1,0 and Ψ_0,0. Posing β(1)α(2)=x and α(1)β(2)=y,

Meaning that, once normed,

The same method can be applied on more complex cases. For instance with two 2p electrons:

The Hamiltonian quickly becomes monstrously difficult when several atoms and electrons are considered. To illustrate this point, the equations of the Hamiltonians for H, H₂⁺ and H₂ are showed below:

For a molecule with M nuclei of atomic number Z₁, Z₂, Z₃, …, Z_M and n electrons, the global expression is

The Hamiltonian can be decomposed into 5 terms, two from the kinetics of the electrons and nuclei and 3 for the interactions between the electron-electron, nucleus-nucleus and nucleus-proton.

As soon as there are several nuclei, there is no central field anymore and we can’t develop a spherical symmetry to build the CSCO. To simplify the problem, we use the Born-Oppenheimer approximation. The electrons have a mass that is way smaller than the one of nuclei (around 1800times smaller than the mass of a proton) and they move way faster. The approximation is to consider that the electrons are adapting themselves instantly to the movements of the nuclei and that we can consider the nuclei as immobile to determine the movements of the electrons.

As a result, we get one equation of Schrödinger for the electrons in the field of the fixed nuclei (T_N=0).

The coordinates of the nuclei are a parameter and not a variable anymore in this equation. We can solve the equations for any position of the nuclei. Then we solve the Schrödinger’s equation for the nuclei.

We treat a single electronic state at a time.  It gives the surface of potential energy. We talk about a surface of potential energy but the dimension depends on the quantity of nuclei. The surface of potential energy depends on 3M-6 intern coordinates) (3M-5 for linear molecules) where M is the amount of nuclei. The -6 comes from the fact that the 3 degrees of translation and of rotation do not impact the potential energy. The global wave function of the molecule is the combination of the wave functions for the electrons and the nuclei. We can plot the potential of BO simply by the addition of the electronic energies as a function of the distance between the nuclei with the nucleic potential.

  

It forms the potential of Born-Oppenheimer that can show a minimum and the state is binding, or no minimum and the state is not binding: the most stable distance between the nuclei is infinite.

During a reaction, the distances between the nuclei vary to form a new liaison and we can plot isocurves of potential energy as a function of the internucleic distances but the approximation that the nuclei are immobile becomes bad as their movements are an important parameter of the problem. It is the case during the cleavage or the formation of a liaison.

Something that was neglected in the approximation of BO is that the different states can interact together and that the couplings between these states can be important. We isolated one state and neglected the couplings it can have with the other states. The exact solution takes them into account. The difference is especially important if two states are close in term of energy or if there is a crossing of the states of energy as a function of the distance between the nuclei (see below an example for the diazomethane, the parallel lines are for various angles of approach). In these cases the coupling terms are not negligible anymore and the approximation.

Because of the particular geometries of some molecules, the CSCO may be different. Instead of the CSCO that we had with the atom, we want to determine the CSCOH: the complete set of operators commuting with Ĥ. It is thus a set larger than the CSCO because the operator don’t have to commute between them. The CSOCH can be subdivided into groups (SCO) of similar operators but they do not necessarily commute between each other.

When two operators of the CSOCH don’t commute together, it implies a degeneration of the states.

In the case of atoms, when we have a spherical symmetry (it has been shown recently that some atoms are slightly deformed and have an elliptic shape (and then a quadrupole moment) or a pear shape (and then an octupole moment)), we have the usual CSCO. We can build 3 SCO:

This last group is present for any system of electrons: the spin of the electrons is independent of the geometry of the molecule. The small letters correspond to operators of orbitals. In the case of a linear molecule, we lose the spherical symmetry but there is still a cylindrical symmetry. Instead of an infinity of axis of rotation passing by the nucleus, we have now only one axis of rotation on the axis of the two (or more) nuclei. This difference induces a modification of the SCO’s: we lost the L² symmetry. As reminder, L²=L_x² + L_y² + L_z². L_Z is still in the SCO but L_y and L_x are not anymore commuting with H. We have now

In the case of nonlinear molecules, we also lose the operator L_Z.

 

We have other symmetry operators that we have to add to the CSOCH (showed by the ? above):

• a bilateral symmetry σ_v: there is an infinity of planes of symmetry passing by the axis of the molecule.

• a centre of inversion Î if the linear molecule is centro-symmetric (same atoms at each side of the centre of the molecule, ex: CO₂, N₂, C₂H₂).

Those operators are necessary to describe completely the molecule: the operators of symmetry characterise the spatial behaviour of the system. As a result, the ensemble of all the operators that commute with H define the state of the system from the spatial point of view. If we forget some operators in the CSOCH, one part of the quantic information is lost. We say that the ensemble of the operations of symmetry form a mathematic group.

Theory of groups

 G={a, b, c, …} forms a mathematic group with regards to one law (*) if

• * is intern and defined everywhere: a*b=c a, b, c ϵ G
• * is associative: (a*b)*c=a*(b*c)
• ∃ e neutral (e ϵ G) : a*e=e*a=a
• Reversibility: ∀ a, ∃ x=a^-1 ϵ G : a*a^-1=e.

For a group of symmetry, a*b means that we apply the operation b first then the operation a.

The order h of a group of symmetry G is the amount of operations it contains. A group of symmetry can be continuous (order h of G is infinite, i.e. a symmetry of revolution) or finished (h is finite), commutative (or abelian) if a*b=b*a ∀ a and b ϵ G, or non-abelian (what leads to a degeneration).

The representation of a group is a set of n by n matrices (n being the dimension of the representation) {D(a), D(b),…} associated to the elements {a, b, …} of G such as

• the matrix product is associated to the law (*)
• the matrix unity is associated to the neutral e.

The representation is said to be reducible if the matrices can be diagonalised and irreducible if it is not the case. Any reducible representation D of G can be expressed as a linear combination of the irreducible representations D_i of G.

We can thus search for the operators of symmetry of a molecule that let the Hamiltonian unchanged. It is the interactions electron-nucleus that impose this symmetry.

There are 5 operations of symmetry:

• identity: E – no displacement
• inversion: I – central symmetry or centre of inversion

• reflexion: σ_v(vertical) σ_h(horizontal), σ_d(diedre) – planar or bilateral symmetry

• proper rotation: C_n – rotation of 2π/n rad around the axis

• improper rotation: S_n – commutative product of a rotation of 2π/n around the axis with a reflexion in the plane perpendicular to the axis.

To apply several consecutive rotations we add an exponent x to C_n or S_n. It means that we apply the rotation 2π/n x times. x goes from 0 to n for Cn and up to 2n for Sn. The Some elements are equivalent. For instance C³₃=E

The groups are named following this picture:

Now lets come back quickly on the properties of the groups.

Faire pareil: d’abord expliquer avec les opérations sur la base 3 puis dire que pour l’explication il était plus simple de montrer avec les axes x,y,z mais qu’en réalité, en utilisant  d’autres coordonnées on arrive à une représentation de dimension 4 qui donne lieu à une ligne supplémentaire.

Let’s take a look on how we build the representation of the group of symmetry C_2V, i.e. the group of shouldered molecules as H₂O, NO₂ but also as CH₂O. If we draw H₂0 in the Cartesian coordinates such as

If we apply an operation of symmetry on the molecule, for instance σ_v(xz), i.e. the reflexion in the plane xz, the molecule did not change but if we follow one of the hydrogen atoms, its coordinate y changed of sign ((x,y,z)à(x,-y,z)). We can translate this into a matrix of dimension 3 that we apply on the coordinates x, y and z:

Such a matrix can be found for the other operators of the group

These matrices commute together and the group is intern and defined everywhere. For instance

Each column of the representation of the group is the value on the diagonal of the matrix for the corresponding base (here the coordinates x, y, z)

Those values are associated with the quantic numbers and the parity. 1 means symmetric and -1 antisymmetric. The character of one matrix associated to the operation a is the sum of the elements on its diagonal and is noted χ(a).

Let’s come back to something we said earlier: any reducible representation D of G can be expressed as a linear combination of the irreducible representations D_i of G.

The coefficients can be determined from the characters of the matrices associated to the operations of this group because of the properties of orthogonality.

Where χ ̅(a) is the value of the irreducible representation associated to one base. For instance, in the base A₁ we have χ ̅(E)=1, χ ̅(C₂(z))=-1, χ ̅(σ_v(xz))=1, χ ̅(σ_v(yz))=-1

Applied to the C_2v group, it gives

We will see next that there is in fact a fourth line (A₂) in the table and why it is necessary to correctly describe molecule.

And thus

A group is normed if its order h, equal to the amount of operations, is also equal to the sum of the squares of χ ̅(a), and this for each irreducible representation, i.e.

For C_2v, h=4=1²+1²+1²+1² (in the case of the operation E). For the operation C₂(z) we find the same value: h=4=1²+1²+(-1)²+(-1)². Two representations k and l are orthogonal if

For instance, we obtain for A₁ and B₂

The group C_3v describes molecules such as NH₃. It contains 6 operations: E, two C₃ and three σ_v. We can represent it as

The irreducible representation E is degenerated: it contains a value different from -1,0 or 1. Yet the group is still normed (don’t forget there are two C₃ and three σ_v):

The representations are orthogonal (for instance between A₁ and E):

Application of the theory of groups

Each orbital of the atoms of one molecule is a basic function that impacts the geometry of the molecule. We need all of those functions to describe correctly and completely the molecule. As a result, a molecule such as NO₂ is described by one representation of base 15 that can be reduced into 10 complete bases (5 of dimension 1 for the nitrogen and 5 of dimension 2 for the oxygen) because the two oxygen atoms are equivalent.

We have seen previously that the NO₂ molecule belongs to the group of symmetry C_2v. If one representations of N can be transformed by one operator of this group into itself in absolute value, it means that this representation is an irreducible representation of the group for the atom N. It is indeed the case for NO₂:

• all of the orbitals of N are irreducible: the orbitals s are spherical so the operation of symmetry changes nothing while the orbitals p change of sign for some operations of symmetry.

• the orbitals of the oxygens are transformed into themselves or into an orbital of the other oxygen, in absolute value.

We can thus find a matrix D(R) such as when it is applied to the base we find the linear combination that belongs to the group.To build a representation of the group C_2v, we have to build one matrix of dimension n for each operation of symmetry of the group, i.e. 4 matrices. Those matrices define the transformation with regard to the four operations of symmetry of one base of n functions. The base has to be complete, i.e. the action of one operation on one of the functions of the base has to give a linear combination of functions of the base. In mathematical words,

The set {D(R), D(R’),…} is a representation of the group G.

The trace T(R) is the sum of the diagonal members of the matrix D(R) (here it is a+d). The set {T(R), T(R’),…} characterises the representation of the group D.

From a few irreducible representations of one group, we can find the other representations of the group. We proceed that way:

3 representations can be found from the nitrogen (see the table behind).  One representation (A₁) is when none of the orbital changes during any operation of symmetry. This representation works for the orbitals 1s, 2s or 2p_z. A second representation (B₁) stands for the orbital 2p_x for which C₂(z) and σ_v(yz) change the sign of the function. The third representation (B₂) is when C₂(z) and σ_v(xz) change the sign of the function, i.e. in the case of the orbital 2p_y.

We can write the representations in a table

These three representations are not the only ones that apply to the NO₂ molecule. We must also describe the orbitals of the atoms of oxygen to give a complete description of the molecule. In this case, it is a bit more complicated because the operations of symmetry can transform one oxygen into itself or into the other oxygen, in absolute value. As a result, the representation is not a series of numbers but a series of 2×2 matrices. If the orbitals remains unchanged by an operation, the matrix corresponding to this operation of symmetry is

If the atoms are interchanged, the matrix is

and if the sign changes it is

For instance, D(E) of the 1s orbitals is (1 0 sur 0 1) because

The operator σ_v(xz) interchanges the atoms of oxygen: and D(σ_v(xz))=(0 1 sur 1 0) because

And we can do that for the other operations as well. That being done, the table of the group C_2v is completed:

However, there is still a problem: the representations that are matrices are not irreducible. But we can find from which linear combination of irreducible representations they come from the traces of the matrices.

One can see that we can combine the representation A₁≡{1 1 1 1} with the representation B₂≡{1 -1 -1 1}. The representation corresponding to the base (1s₁, 1s₂) is thus A₁ + B₂. We can repeat this process for the other bases of the group. It works fine for 3 bases but not for the base (2p_x1, 2p_x2). There is thus an unknown representation A₂ in the group so that we can build the matrices of the base (2p_x1, 2p_x2). Obviously the combination will involve the representation B₁ to obtain a trace of -2 for the operation σ_v(yz). The representation A₂ is thus A₂≡{1 1 -1 -1}.

If you remember well, earlier in the course we discussed on the operators of symmetry of the water and we already talked about a fourth representation in the group C_2v. Now you know why we needed this fourth representation. It is required to build the CSCO that would not be intern without A₂.

Amongst the 15 atomic orbitals of NO₂, we have 7A₁ + 1A₂ + 2B₁ + 5B₂:

7A₁: 1s, 2s, 2p, (1s₁,1s₂), (2s₁, 2s₂), (2p_y1, 2p_y2), (2p_z1, 2p_z2)

1A₂: (2p_x1, 2p_x2)

2B₁: 2p_x, (2p_x1, 2p_x2)

5B₂: 2p_y, (1s₁,1s₂), (2s₁, 2s₂), (2p_y1, 2p_y2), (2p_z1, 2p_z2)

To build the representation A₁, we will thus need a linear combination of the 7 orbitals.

Method of projection

The projection allows to determine which orbitals take part to which representation. In a general way, the operator of projection P⁽ⁱ⁾ for  non-degenerated representations is given by

with i the corresponding representation, χ the trace of the operator and h the norm. For instance, applied to NO₂,

To obtain π liaisons, the p orbitals have to be oriented correctly. On the other hand, antibonding liaisons are the result of p orbitals that are (also) correctly oriented but with opposed signs. We know that there are 3 unoccupied orbitals in NO₂ (15 orbitals and 23 electrons). These orbitals are some of the antibonding orbitals and are the highest in energy.

For degenerated representations, we can still do the projection using the diagonal elements of the matrix D(i). The formula changes a bit:

with n_i the degree of degeneration.

Application to NH₃

NH₃ belongs to the C_3v group.

Obviously the presence of a 2 in the irreducible representation E indicates that this IR is degenerated. The orbitals to consider are

The orbitals s of N belong to A₁ because they are spherical and the 2p_z also does because it is on the axis of rotation. The rotations that we can apply on NH₃ are of 120°, so our system of coordinates is not very convenient here. We can however obtain the matrix representation of {2p_x, 2p_y}

The matrix representation of {2p_x, 2p_y} is thus the irreducible representation E: the traces of the matrices correspond to the elements of the IR E in the table of C_3v.

We repeat the process for the three hydrogen atoms

It doesn’t correspond to any IR of the table C_3v but we can find a linear combination of them:

There are two projections for the operation E as it is degenerated twice. The atomic orbitals adapted to the C₃ symmetry are, for the hydrogen atoms

We already verified the orthogonality of C_3v previously

Proper functions

The proper functions of Ĥ (polyelectronic states) and of ĥ (molecular orbitals) that have a proper value E in common form a base for the irreducible representation of G.

Non-degenerated case:

We can thus reduce the molecular representation of dimension n into n matrix representations of dimension 1.

Degenerated case (n times):

Several proper functions have the same energy. As a result, the reduction leads to several representations amongst which some are not of dimension 1.

Direct Product of two representations

The direct product between two representations of dimensions n and m give a n.m representation.  It allows to determine the symmetry of the product of two or more representations, i.e. in case of coupling between orbitals. For instance, the direct product of one representation D⁽¹⁾ of base f (dimension n₁=3)  with one representation D⁽²⁾ of base g (dimension n₂=2) gives a representation D of base fg (dimension n=n₁.n₂=6)

The character of the new representation will simply be the product of the characters of the old representations (no need to do all the stuff).

We can thus determine the symmetry of the product of several functions.

The direct product with one 1D representation gives an irreducible representation. If the representation is reducible, it is a linear combination of the representations of the group.

Most of the molecules have a fundamental state that is an A₁ representation because the octet rule is respected. I understand that the direct product and its interest are vague for now but we will see them with the example of NH₃.

 Example of NH₃

NH₃ belongs to the group of symmetry C_3v.

The direct product of two E representations gives in the group C_3v gives

This product corresponds to a linear combination of the IR of the group.

The coefficients can be determined from the relation

The molecular orbitals are called in function of their irreducible representation, written in small letters. The fundamental state of NH₃ is

The MO 1a₁² results from a coupling between two A₁ states: A₁xA₁=A₁. It is the same for the other a₁² molecular orbitals. For the 1e⁴, we have the coupling of 4 E representations. The E representation is of dimension 2 and the coupling gives a degeneration of 16 (2⁴). However, the principle of Pauli has to be respected as well, i.e. the only possibility is that the 4 electrons are distributed amongst the two orbitals of same energy with opposite spins. As the layer is full, the representation is A₁.

If one electron of the HOMO (orbital 1e⁴) is excited to the orbital 2e (LUMO) we obtain two ²E states (1e³ and 2e¹ give the same state by the symmetry hole/particle).

Cases of the spherical and cylindrical molecules (or atoms) – group of symmetry K_h

I don’t know the exact reason, but for those two cases the representations have specific names. For atoms, the representations have names identical to the atomic symmetry. The orbitals s belong to the group S (dim 1), the orbitals p belong to the group P (dim 3), etc. We can put an index to the group that translates the parity in this group. The index is g (gerade from German) if the state is symmetric and u (ungerade) if the state is antisymmetric.

If we apply the inversion operator on the nitrogen, we have

The coupling between two identical indexes gives gerade and between two different indexes gives ungerade:

The groups for linear molecules that are not centrosymmetric are called differently

The distinction between σ⁺ and σ^– is the direction of the rotation (clockwise or anticlockwise).

For centrosymmetric molecules, we note the distinction between ungerade and gerade

The electrons revolving on an orbital generate an angular moment.

M_L is the quantic number associated to the projection of L on the internuclear axis.

The projection is degenerated because it can either be in the positive values of the z axis or in the negative ones. The projection of L can thus give M_L or –M_L. We can define a new quantic number L=∣M_L∣.

The fact that there are two projections explains the dimension 2 or the orbitals π, δ, … of linear molecules. The length of the projection of L grow if  L gets larger but the degeneration is always 2. The only exception is Σ for which M_L=0. In the atoms we had the possibility to choose the orientation but we cannot do that with molecules.

As a result, the operator σ_v doesn’t commute with L_z:

except for L=0 that gives the states Σ⁺ and Σ ^–. This distinction + or – is not present for the states Π, Δ, …

The fact that σ_v doesn’t commute with L_z induces the degeneration of the states.

The inversion operator Î still commutes with Ĥ, L_z and σ_v in the case of centrosymmetric molecules.

As a resume, the operator σ_v is separated from the other operators of the CSCO of linear molecules because it does not commute with L_z anymore.

Application to H₂

Let’s take a look at the possible electronic configurations of the molecule H₂.

 

The unexcited state of H₂ has two equivalent electrons on the ground orbital 1σ_g:

This state is binding: between the nuclei, the probability of presence of electrons is positive. In antibonding states, there are some points between the nuclei where the probability to find electrons is zero.

The first excited state is the configuration 1σ_g1σ_u. In this configuration the electrons are not equivalents and the degree of degeneration is 4 (2×2):

To determine the states, we take the emergent (M_L=0,M_S=1). It corresponds to the triplet ³Σ _u ⁺ (the pairs (0,1), (0,0) and (0,-1)). The second emergent (0,0) corresponds to the singlet state ¹Σ _u ⁺(the pair (0,0)).

To determine the energy of the states, we proceed as for the atoms with the determinants of Slater

In the case of an emergent such as (0,1) we have a proper function and thus the energy can be determined. From this function we find the other functions with operators of rise/descent S₊ and S_– and with the principle of orthogonality we find the energy of the singlet state.

The dashed curves are correspond to unstable states because there is no minimum of energy: the atoms are get more stable as they get away one from each other.

Application to O₂

The same method can be applied to O₂. Its fundamental configuration is

As usual, we only consider the highest occupied molecular orbitals (HOMO). Those are the 2 1π_g orbitals with 2 electrons to place, represented above by the circles: they can be on the same orbital or separated. There is thus a degeneration of 6:

The first emergent is (2,0). That corresponds to the group 1Δ_g. Remember that the degeneration for linear molecules is 2 except for M_L=0. In the atomic case, we had a degeneration of 2L+1 (i.e. M_L, M_L-1, …, 0, …, –M_L-1, -M_L) but here we just have M_L and –M_L. The next emergent is (0,1), a triplet ³Σ_g. Finally there is a singlet ¹Σ_g. To know if it is Σ⁺ or Σ^–, we have to apply the operator σ_v. In this case we have ³Σ_g^– and ¹Σ_g⁺.

Ozone is obtained by the excitation of one molecule of O₂ at its fundamental state into two oxygen atoms in the ³P state. In this state, they react with another molecule of O₂ and a catalyst to produce the ozone.

This theory says that each molecular orbital Φ_a is described by a linear combination of atomic orbitals {χ} centred on the M nuclei of the molecule.

The molecular orbitals have the symmetry of one of the irreducible representations of the group G. This symmetry is taken into account in the LCAO coefficients. Some are null (some symmetries are not used) and some are equals in absolute value: only functions of same symmetry interact together to form molecular orbitals.

Given a MO Φ_a ∈ D⁽ⁱ⁾ of G and the AO {χ}, we can adapt the atomic orbitals to the symmetry of D⁽ⁱ⁾: {χ}à{χ⁽ⁱ⁾}. Then

The advantage of doing this is that the number n⁽ⁱ⁾ of functions χ⁽ⁱ⁾ in the last expression is smaller than (or equal to) the number n of functions χ in the atomic orbitals. Let’s apply the LCAO theory to an example in which the orthonormal base is composed of two function {Ψ₁, Ψ₂}, i.e. a case where two states are in interaction with each other.

The secular determinant is

As the base is orthonormal, S is a delta of Dirac: S₁₂=S₂₁=0, S₁₁=S₂₂=1 and H₁₂=H₂₁. Posing that the two states H11 and H22 are equidistant from the zero energy, they are separated in energy by 2H:

We can represent the problem as follow:

The secular determinant is thus reduced to

and

If we set c₁=1, then c₂=(H+E)/H₁₂. The coefficients must still be normed.

Two cases can be considered:

• H=0 with H₁₂<0.

Two degenerated states interact with each other. The result of this interaction is that the states are repelling from each other. One is stabilised and the other one is destabilised by ΔE=H₁₂. We will obtain a bonding state and an antibonding state. H₁₂ is thus a measure of the interaction between the states. The bigger it is, the bigger is the separation between the resulting states. H was the distance between the states that interact together.

Posing that c₁=1, then c₂=-1. The coefficients must still be normed: c₁=1/√2 and c₂=-1/√2. As a result, the wave function of the state of energy E=-H₁₂ is

We can do the same for the second state (E=H₁₂) and obtain

• H≠0 and H>>H₁₂

The states that are interacting together have not the same energy. For instance, let’s consider H=2 and H₁₂=-1/2.

The interaction between the two states separated the states but just by a bit. The states did not mix a lot together.

The mixing between two states is inversely proportional to the difference of energy between the states.

 

If H₁₂=0, there is no interaction between the states. It is the case only when the states don’t share any symmetry, i.e. H₁₂ can be different from zero only if Ψ₁ and Ψ₂ have common proper values with all the operators that commute with Ĥ. As a result, a triplet does not interact with a singlet even if the energies of those states are similar.

Rule of non-cancellation of an integral:

An integral is not equal to zero if the integrand is invariant with regards to all the operation of the group G, i.e. if the reduction of the direct product contains the totally symmetric irreducible representation D⁽¹⁾. In other words, the integral is different from zero if the integrand is totally symmetric.

Cases of H₂⁺ and H₂

The orbitals 1s of the two atoms are interacting together to form molecular orbitals σ_g and σ_u. σ_g is a binding orbital resulting from a constructive interference:

Here we considered that the base is not orthonormal. It is why the norm is 1/√(2(1-S)) and not 1/√2. S is the deviation to the orthonormal base

In an orthonormal base, S=0. The antibonding state σ_u is resulting from a destructive interference

The interactions can be seen this way:

When the electron is in a molecular orbital such as the attraction it produces on both nuclei brings the nuclei closer to each other, the effect is binding.

In the figure above, the electron is between the two nuclei. The attraction is produces on the nuclei is represented by the black arrows. The green arrows are the projection of the attraction on the axis passing by the two nuclei. The nuclei move thus in the direction of the other nucleus and remain thus together because of the presence of the electron.

If the position of the electron leads to a separation of the nuclei, then the effect is antibonding. The interaction of the electron-nucleus is positive but the intensities and directions are such as the nucleus-nucleus distance increases.

In the picture above, both nuclei are attracted by the electron but they move in the same direction with different speeds. The nucleus of the right moves faster than the nucleus of the left and the nuclei move thus away from each other.

The function 1s can be expressed as an exponentially decreasing function centred on the nucleus.

We can superimpose the functions 1s of two hydrogen atoms. In the case of σ_g, the functions add together:

We can do the same for σ_u.

If we put those function to the square, we obtain the probability of presence of electrons.

 In the second case, there is a place between the nuclei where no electron can be found. No liaison can thus be done between the nuclei.

Interaction between 2p orbitals

All the 2p orbitals don’t interact the same way. The 2p_z orbitals interact together to give the σ orbitals. This time, it is not the sum of the atomic orbitals that give the molecular orbital of lower energy and the binding orbital σ_g.

The other 2p orbitals (2p_x, 2p_y) lead to the π orbitals.

Note that the energy of the orbitals depends on the atoms. They all decrease in energy with Z but not with the same speed (see the figure below).

The energy of the π_u orbitals is almost constant while σ_g 2p_x decreases quickly with Z. σ_g 2p_x falls under π_u at O₂.

The energy of the liaison between the two atoms increases up to N₂ and decreases after because electrons are placed in antibonding orbitals.

The Walsh diagram

This diagram relates the energies of molecular orbitals of a molecule as a function of the angle that separates the liaisons. It helps to visualise the stability of the liaisons with regards to the symmetry of the molecular orbitals. The following figure shows the Walsh diagram for AH₂.

On the left one sees the linear molecules. As we go towards the right, the angle between the two liaisons goes towards the right angle, i.e. towards a bent conformation. As the bond angle is distorted, the energy for each of the orbitals can be followed along the lines, allowing a quick approximation of molecular energy as a function of conformation. As we move towards the top, the energy of the liaisons increases. Note that the 1π_u orbitals are degenerated for an angle of 180° but separate if we change the conformation of the molecule.

For one molecule, we count the number of electrons of valence. For instance BeH₂ has 6 electrons of valence (4 for Be and 1 for each H). We place 2 electrons by line, starting from the bottom (note that the 1a₁ line, binding the σ_g orbitals, is not plotted. The last electrons are on the 1b₂ line. One can see that the most stable angle for this molecule is 180°. For BH₂ and CH₂, the molecules are bent. If one electron is excited, then the conformation of the molecule can change.

Method of Hückel

This method limits the LCAO method to the π electrons. The reason is that a lot of physicochemical properties of the molecules can be explained by the π-π* orbitals. In the method of Hartree-Fock, the secular determinant was

We have to solve the determinant for all the orbitals of the molecule, what can quickly become complicated. If we apply the method of Hückel on C₂H₄ for instance, there is only one π liaison in the molecule and thus only one secular determinant to solve. Considering two degenerated levels of energy α, the secular determinant is

α=H₁₁=H₂₂ is the energy of the perpendicular to the plane atomic orbitals of the carbons and beta is the energy of resonance/interaction.

The solution found with this method (we won’t do it here) is close to the one obtained with the Hartree-Fock method that consider all of the electrons.

This method can be extended to other systems with π electrons if we pose that

If we consider the butadiene, we consider it as the interaction of two π systems

With the increase of π electrons, there are more binding states and one can see that, looking from the bottom to the top, the organisation of the orbitals follows a simple rule: the number of times that the signs are reversed increases by one at each orbital. One talk about the “wavenumber” of the orbitals. Indeed, on the lowest energy state, all the orbitals are aligned. There is no change of orientation of the orbital. On the second lowest state, the two orientations of the orbitals are present but they are grouped. There is only one change of orientation. The third level has 2 changes of orientations, there are 3 changes on the fourth lowest state, 4 on the fifth, etc… If we look at the orbitals as a wave, the wavenumber is indeed increasing with the energy of the state.

When the amount of π liaisons increases,

• the separation in energy between the states of same type (binding or antibonding) decreases and for a large amount of liaisons (in polymer for instance), we talk about a band of valence for the block of binding states and about a band of conduction for the block of antibonding states, separated by a gap.
• the amount of states increases in both bands,
• the separation in energy between the band of valence and of conduction decreases.

We have seen quite a lot of new stuff up to now. We described monoelectronic and polyelectronic atoms and developed the description to molecules through the approximation of Born-Oppenheimer, the theory of groups and the CSOC. All of this teaches us how orbitals are and how they change during a reaction. Yet, we did not find the energies of the orbitals and can’t say which one is more stable than the others. The general equation is, as seen at the beginning of the course,

From the wave function we want to determine the energy of the orbitals but we can’t solve the equation exactly except for systems with one electron. For other species, we can only find an approached solution. To obtain it, we use the theory of perturbations or the method of variations.

Theory of perturbations

We can apply this theory to states that are independent of the time, i.e. stationary states. We can approximate the Hamiltonian and the energy of one state if this state is the result of a small perturbation λ of a state the solutions of which are known (Ĥ⁽⁰⁾, E⁰).

The solutions are then expressed as a series of correction to the model at order zero.

The corrections are calculated from the solutions at the order 0. For instance, the correction of order 1 are

The corrections usually decrease in intensity with their order but they will always approach the approximation from the exact solution. Note that we can go beneath the exact solution.

Method of variations

To find the exact energy, we use a trial wavefunction. This function has a known form but will probably not give the exact energy but will give us a superior born for the exact energy. Next, we modify the parameters of the trial wavefunction to obtain a better approximation of the energy. Unlike the method of perturbation, we can’t go beneath the exact energy with the method of variations. Each time we modify the parameters, we obtain a solution of lower energy and we approach the exact value of the energy. For instance if we chose a wavefunction with two parameters α and β. We can try to improve the values of the parameters to obtain the best possible approximation.

We reach an optimal estimation when the energy does not vary anymore when we change the parameters, i.e. when dE/dparameters=0.

Application to the helium

We will apply those two methods to estimate the energy of the helium. The exact solution is E=-2.903u.a. and the complete Hamiltonian is

One model the solution of which is known is the hydrogenous model:

The solution of this model for 1 electron is

with Z the atomic mass and the quantic number n. As there are two electrons in the helium, we simply multiply the energy by 2 to obtain a first approximation from the hydrogenous model:

This approximation is far from the exact solution and overestimates the stability of the atom because we did not take the repulsion between electrons into account (the +1/r₁₂ term).

The theory of the perturbation will start from this model and introduce the repulsion term as a pertubation of the model:

This estimation is way better than the hydrogenous model alone.

The method of variations is slightly different as we choose a wave function Ψ that depends on a few parameters that we may vary. As for the theory of perturbations, we select the hydrogenous wave function

As there are two electrons in the helium, we use a combination of two wave functions:

The parameter that we will vary is the effective charge Z=Z_eff. One part of the charge of the nucleus is indeed hidden to one electron by the second electron.

From the chosen wave function, we find an equation for the energy that depends on Z

We optimise the parameter to obtain the best approximation that we can, such as dE/dZ=0. Solving this, we find

We apply this particular value of Z to the equation for the energy to find

This result is close to the exact solution.

Principle of linear variation: linear combination of atomic orbitals (LCAO)

In this method, we assume that Ж can be expressed as a linear combination of a base of functions {Ψ₁, Ψ₂, …}

where c_i is the variational parameter associated to the wave function Ψ_i. These coefficients are adjusted to approximate the exact energy. For more simplicity, let’s take an example in which Ж is a linear combination of two wave functions Ψ₁ and Ψ₂.

It is corresponding to the case of H₂:

The Hamiltonian is

Next we multiply this equation by Ψ₁* and integer it over tau:

We can do the same with Ψ₂*:

Now, we introduce H_ij and S_ij such as

The previous integrals can be written

or

This system of two equations can be written as a 2×2 secular determinant

In a general way, it forms a n by n secular determinant if there are n wave functions (or particles)

The n solutions of the secular determinant give the n most stable states of energy of the system. If we inject one of the energies E=E_q in the system, we obtain the optimised coefficients {c_iq} related to the state q.

If the base is orthonormal, then

And then

If the base is complete, the linear combination corresponds to the exact solution due to the theorem of superposition (Quantum superposition is a fundamental principle of quantum mechanics. It states that, much like waves in classical physics, any two (or more) quantum states can be added together (“superposed”) and the result will be another valid quantum state; and conversely, that every quantum state can be represented as a sum of two or more other distinct states). The more the base is complete, the more one tends to the exact solution.

Method of Hartee-Fock

This method is an application of the variational method in which the trial wave function is a normalised determinant of Slater Ж = ∣ Φ ₁ Φ ₂ Φ ₃… Φ _N ∣. We try to minimise the energy of the orbitals (∂E/∂Φ_i=0) and to keep them orthonormal. We obtain a system of N coupled equations of Hartree-Fock that describe the movement of one electron in the averaged field u_b created by the other electrons, such as

where ε_a is the HF energy of the molecular orbital Ψ_a. If we introduce the operator of Fock

we can rewrite the system of N equations as

or on one line:

This equation only depends on the position and the movement of one electron, the effect of the other electrons being averaged. As the averaged field is determined by the orbitals that we are looking for, we have to solve the system of equations by iterations, starting from a set of orbitals of trial.

The convergence is guaranteed by the variational principle. In practice, we develop each orbital as a linear combination of Gaussian atomic orbitals centred on the nuclei. Then we iterate.

In the Born-Oppenheimer approximation, we froze the position of the nuclei to find the electronic energy. The position of the nuclei was considered as a parameter that can be modified and we were able to construct the Lenard-Jones potential for the liaisons or the surface (or hypersurface) of potential energy for molecules with more than one liaison. We will now discuss in further details about the vibration and rotation modes of molecules ant thus incorporate the movements of nuclei in the model.

The first step is to choose the set of coordinates in which we will work. A molecule with M nuclei has 3M coordinates: (x_i,y_i,z_i) for each of the M atoms.

The laboratory axes system (LAS) is the first set of coordinates that we use. In this system we determine the position of the centre of mass of the molecule. Those 3 coordinates allow us to determine the translation of the molecule but not the rotation or the vibration modes: the centre of mass doesn’t move because of those modes.

To determine those modes, we need two other sets of coordinates: the LAS’ that is fixed to the centre of mass of the molecule and consequently is independent of the translation and the MAS: molecular axes system that is fixed to the molecule and turns with it. 2 angles 0≤θ≤π and 0≤Φ≤2π are used to locate the linear molecules and a third one 0≤χ≤2π is needed for the nonlinear molecules. Those 3 angles are called the angles of Euler and are

 

From the 3M coordinates, we used 5 (linear molecules) or 6 (non-linear molecules). The rest correspond to the modes of vibration of the molecule. In a diatomic molecule, there will be only one mode of vibration: M=2 and 5 coordinates are used to locate it.

To resume, the coordinates of one molecule are given by the vector

with 3M dimensions. The translation of the centre of mass is determined in the LAS

The condition here reflects the facts that the centre of mass does not move in the LAS’ referent. The MAS rotates with the molecule. To move into this referent, we apply the matrix S to the vector R_A’.

The matrix S defines the orientation of the axes (x’,y’,z’) of the LAS’ from the coordinates (x,y,z) of the LAS:

Small vibrations around the equilibrium R_A^eq are given by the vector dA

with the conditions of Eckart that

The first condition reflects that the centre of mass does not move because of the vibration and the second condition that there is no rotation in the MAS.

The kinetic energy of the molecule is in this notation

with M_A the mass of the nucleus A and Ṙ_A=dR_A/dt. Replacing Ṙ_A by its expression

We obtain (the red terms equal zero)

If Eckart is respected, then the interaction term Trot/vib can be neglected. We can thus approximate that the energies of rotation and of vibration are separable. The order of magnitude is indeed different: the vibration is found into the infrared while the rotation is observed in the microwave range.

Vibration

For a diatomic molecule, the oscillation is characterised by a force of recall

We find that

The energy of vibration can thus be approximated to a constant. The kinetic energy T_vib is always positive and is thus equal to

From the quantum mechanics, we know that

The separation in energy between the levels characterised by a number of nodes v=0, 1, 2, … There is thus regular a ΔG_V=ῡ.

There is however a deviation to the harmonicity that we found. If we develop the potential as a series of Taylor, we obtain

The two first terms are simply equal to zero (V₀=0 and the potential is at a minimum at the equilibrium). The third term is the harmonic result that we just obtained but further terms express the deviation to the harmonicity. The potential of Morse gives an empiric formula that fits correctly the real potential.

The anharmonicity modifies slightly the energy of the states

In the case of polyatomic molecules, we can consider that all the nuclei oscillate in phase, giving a base of 3M-6(5) independent movements. The result doesn’t noticeably differ from the diatomic molecule in this case.

Rotation

The rotation can be considered as a rigid rotation: the difference of frequency and of energy between the rotation and the vibrations is huge enough to make this approximation. The angular speed ω is thus identical for all the nuclei of the molecule.

The kinetic energy due to the rotation for a diatomic molecule is thus

with I the moment of inertia that is common to the two atoms.

μ is the reduced mass of the molecule. The angular moment J is given by

Its absolute value is

For a diatomic molecule, we get

The angular moment and the kinetic energy are thus directly bound:

We can go from the classical mechanics to the quantum mechanics by the application of the Hamiltonian on a wave function.

The multiplicity is g_J=2J+1. A small correction has to be added due to the centrifugal distortion, correction which is normally very small and negligible except when the rotation is very fast:

The spectrum of the rotation is thus composed of bands regularly spaced.

The IR spectroscopy has a different goal than the UV/visible. It is one of the strongest methods to determine the structure of organic compounds. An IR spectrum is similar to the finger print of one molecule and the matching of peaks can tell us if a molecule is in the sample or not. The spectrum showed below is the one of the 1-hexene in black and of the cis-2-hexene in orange. The difference of position of the double liaison leads to a huge difference in the IR spectrum.

The UV/visible spectroscopy involves electronic transitions while the IR spectroscopy involves transitions between vibrational states. As for the UV/visible, we observe a series of bands, and not rays, because of the sublevel rotational states. There are two main modes of vibration: the elongation and the bending.

A liaison between two atoms has not a constant length. The atoms vibrate around their mass centre with a frequency that is characteristic of the pair of atoms. It is the elongation mode: a vibration in the axis of the liaison. However, the atoms are often bound to more than one other atom, modifying the frequency of the vibration and giving place to additional elongation processes.

The bending is a vibration out of the axis of the liaison. It leads to a variation of the angles between liaisons.

For a molecule of n atoms, the amount of degrees of liberty of the molecule is equal to the sum of the degrees of liberty of each atom. Each atom has 3 degrees of liberty corresponding to the 3 Cartesian coordinates required to describe its position in the molecule. The molecule has thus 3n degrees of liberty. Amongst the 3n, 3 are used to describe the translation modes and 3 are used to describe its rotation (2 if the molecule is linear). There are thus 3n-6 (or 3n-5) modes of vibration for each molecule. For instance, the water has 3 modes of vibration: 2 modes of elongation and 1 mode of bending.

One mode of elongation is symmetric (the centre of mass is displaced) and one is asymmetric. Each mode shows a specific vibration, with a given wavelength. The wavelengths of the elongation modes are very close to each other and the bending mode has a smaller wavenumber. The fact that the elongation modes and the bending modes are distant in terms of wavelength and that the bending modes have smaller wavelengths are generalities that we can find for any liaison.

There are 3 atoms in the carbon dioxide but 4 modes of vibrations as the CO₂ is linear (3n-5).

There is one symmetric and one asymmetric modes of elongation and two equivalent modes of bending (one in the x-y plane and one in the y-z).  The bending modes have the same wavelength at 666cm^-1. We say that those vibration modes are degenerated twice. Moreover, the symmetric elongation is inactive in IR because this mode of vibration does not induce any variation of the dipolar moment of the molecule.

In bigger molecules, it is rare to observe the exact number of modes of vibration (3n-6) because some modes come from combination from two vibrations or more, or are harmonics from strong modes of vibration. Those two effects increase the number of bands and other effects decrease this number

• bands that are too weak to be observed
• bands that are to close and that merge together
• degenerated bands
• inactive modes of vibration
• modes with wavenumbers out of the analysed range, usually between 4000 and 400cm^-1.

It is possible to approximate the frequency of vibration of a given liaison with the law of Hooke that considers the liaison as a simple harmonic oscillator between two masses M₁ and M₂.

c is the speed of light and f is a constant that reflects the strength of the liaison, the value of which is approximately 5.10⁵dyne/cm (1dyne=10⁵N=10⁵kg m s^-2) for simple liaisons and two and three times this value for double and triple liaisons. f increases from left to right in the Mendeleev table. For instance, the law of Hooke gives a wavelength of 3040cm^-1 for the C-H liaison. If we take a look to the vibration modes of CH₂ inside a carbon chain, we find 6 modes of vibration (2 elongation modes and 4 bending modes).

Remember that the 3n-6 rule only applies to complete molecules. The elongation modes that we observe have frequencies that are a bit lower than the ones obtained by the law of Hooke (2926 and 2853cm^-1 vs 3040cm^-1) because of the environment of the C-H liaison that is not taken into account by Hooke.

From the law of Hooke, it is obvious that heavy atoms vibrate at smaller frequencies. We can barely separate a IR spectrum in several regions depending on the pair of atoms involved in the liaison and the type of liaison:

3800-2700cm-1: C-H, O-H, N-H	1900-1500cm-1: C=C, C=O, C=N, N=O
2300-2000cm-1: C≡C, C≡N	1300-800cm-1: C-C, C-O, C-N

On the figure we can thus already deduce that there is no triple liaison (grey region) in our molecule (the 1-hexene from early), that there are some double liaisons (green region) and some simple liaisons (blue and red regions). However we cannot yet determine which peak corresponds to which vibration.

Coupled interactions

When two oscillators share a common atom, they rarely act as simple oscillators except if their vibrations modes are very different. The coupling between two modes of vibration produce two new modes of vibration with frequencies greater and smaller than the one in absence of interaction.

For instance, in the CO₂, the asymmetric elongation (2350cm^-1) and the symmetric  elongation (1340cm^-1) are the result of a coupling between the two C=O oscillations, one shortening when the other one elongates.

From what we have seen previously, one C=O should vibrate between 1900-1500 cm^-1 but it is clearly not the case here. The coupling between the two C=O vibrations has for effect to displace the vibration towards larger frequencies (only the asymmetric is visible). The coupling becomes negligible when one or more carbons separate the liaisons. Two carbonyls separated by one or more carbons would show an absorption around 1725cm^-1.

So the conditions to observe a coupling are that the vibrations share a common atom and have similar frequencies. However interactions are possible between fundamental vibrations and harmonic vibrations/vibrations of combination.  Such coupling is called a resonance of Fermi. Still with the CO₂, the symmetric elongation that is not active in IR but that can be observed in the Raman spectrum at 1340cm^-1. In fact, there are two bands at 1286cm^-1 and 1388cm^-1 because of the coupling of the vibration with the harmonic of the bending modes (666cm-1). The first harmonic is thus at 1332cm^-1 and interacts with the symmetric elongation mode at 1340cm^-1.

Hydrogen bonds

Hydrogen bonds decrease the frequencies of the involved liaisons and usually enlarge and enhance their bands. The importance of this effect depends on the strength of the H bond. A H bond is strong when it is in the exact direction of the lone pair of the electronegative atom. The strength of the bond also depends of the distance between the atoms, of the atom and if a cycle can be formed by the H bond.

The decrease of frequency varies from 300 to more than 500cm^-1 if the H bond is intermolecular and from less than 100cm^-1 to more than 300cm^-1 if the bond is intramolecular. H bonds are often present between a molecule and the solvent. It is thus important to indicate the solvent and the concentration of the compound on a spectrum. The presence of water in the sample has a visible effect on spectra.

Analysis of a spectrum

An IR spectrum shows the transmittance as a function of the wavelength/wavenumber. The spectrum is to be read from top to bottom with bands dropping low in the spectrum.

An accurate analysis of an IR spectrum is not conceivable. The couplings, the absence of modes of vibration and the width of some bands make it difficult to determine the exact nature of the bands. However, as it has been said before, one IR spectrum is like the finger print of one compound and if one spectrum superimpose with the spectrum of a known molecule (with the same solvent, concentration and setup), the job is done. In other situations, an IR spectrum is used in combination with MS, UV and NMR spectra.

The spectrum, usually between 4000 and 400cm^-1 can be split in two important regions to analyse: one between 4000 and 1300cm^-1, called region of the functional groups, and one under 900cm^-1 that shows the bending modes.

The elongation modes of functional groups with OH, NH, C=O can be found in the region of the functional groups. If this region is empty, we can assume that the molecule does not wear any of them. In rare situations, a weak and very wide band can be mistaken for an absence of band. Yet, be careful with low intensity bands that can also be harmonics or combination bands.

Characteristic absorbances

It would be too long to describe all the bands that can be observed on a IR spectrum. The goal here is more to give the very characteristic bands that can be easily spotted on the spectrum to have a general idea of what is in the molecule and what is not. In combination with a mass spectrum, it will be possible to determine the correct structure of the analysed molecule.

High frequencies vibrations

The region of the spectrum above 2850cm^-1 is rich in information’s that can easily be studied. In this region we find the elongation bands of Y-H, Y=C, O, N. The bands are usually intense and are not hidden even if another band is nearby. The position of one band depends on the direct environment of the liaison. When necessary, we write this liaison (simple, double or triple) to demark =C-H, i.e. a hydrogen bond to a carbon with a double liaison, from –C-H for instance, i.e. a hydrogen bond to a carbon with a simple liaison.

A first hint on the nature of the compound is found in the vicinity of 3000cm^-1 where the C-H liaisons absorb. The difference of frequency between aromatic =C-H and aliphatic -C-H is small but is characteristic:

Methylic C-H are found as 2 bands at 2962cm^-1 and 2872cm^-1 and methylenic C-H are found a bit lower (2926 and 2853cm^-1). =C-H are found slightly above 3000cm^-1. It is often synonym of an aromatic but is can also simply be an alkene (or conjugated double liaisons). On the spectrum of the hexene, the peak at 3080cm^-1 is characteristic of the double liaison and the peaks for the –C-H are hidden in the large cluster. Three small peaks are however visible around the values cited above for Methylic and methylenic –C-H.

It is thus easy to guess the presence of an aromatic cycle from the –C-H bands but it is not easy to determine if the =C-H absorbance band come from an aromatic or an alkene: both alkene =C-H and aromatic =C-H show an intense band in the low frequencies region, respectively between 1000-650cm^-1 and 900-675cm^-1. The aromatics (and heteroaromatics) will also show 3 peaks around 1600cm^-1, 1500-1400cm^-1 and 1300-1000cm^-1 (note that conjugated alkene also show that kind of bands). Still in the spectrum of the hexene, the peak between 1300 and 1000cm^-1 is not present and the bending peak is lower than 675cm^-1.

The C-H bond of aldehydes vibrates at smaller wavenumbers, between 2830 and 2695cm^-1 because the carbonyl takes electrons from the C-H and this liaison is thus longer, making it vibrate slower. The peak is not very intense and can be doubled because of a resonance of Fermi with the first harmonic of the bending mode at 1390cm^-1.

≡C-H bond to triple liaisons are absorbing at frequencies above 3100cm^-1, between 3333cm^-1 (this one is easy to remember) and 3267cm^-1 but are not the single species absorbing above 3100cm^-1. We can also find –O-H and N-H elongation bands as well as H bonds. The ≡C-H bands are usually thinner than the bands for –O-H and –N-H and are intense.

-O-H elongation bands

The –O-H bands are large and intense, and very large in the case of carboxylic acids. In fact, those large bands are the result of the H bonds and the formation of dimers when the concentration of alcohol/acid are much diluted.

Free –O-H give small and thin bands between 3650 and 3584cm^-1 for alcohols and around 3550cm^-1 for acids but are easily erased by the large bands of  H:O-H that are spreading over hundreds of cm^-1 are way more intense. Those large bands should be centred between 3550 and 3200cm^-1 (for the benzyl alcohol it is at 3320cm^-1) for alcohols and 3000cm^-1 for acids.

-N-H elongation bands

The high frequencies bands for aliphatic amines are quite weak between 3400 and 3250cm^-1. There are two bands in the case of primary amines and 1 band in the case of secondary amine. The N-H elongation modes are slightly displaced towards larger frequencies if the amine is aromatic. The spectrum is different in the case of amine salts. Ammonium ions give a large and intense band of absorption between 3300 and 3030cm^-1. The frequency decreases with the degree of the amine: primary amine: 3000-2800cm^-1, secondary amines: 3000-2700cm^-1, tertiary amine: 2700-2250cm^-1).

In absence of H bond, the primary amide shows two bands around 3520 and 3400cm^-1 while secondary amides show one band of absorbance between 3500 and 3400cm^-1. In more concentrated solutions or in solids, H bonds displace the bands towards smaller frequencies, respectively around 3350 and 3180cm^-1 for primary amides and multiple bands between 3330 and 3060cm^-1 for secondary amides. The presence of multiple bands instead of one single band is explained by the formation of a dimmer of amides.

Middle range frequencies

Between 2850 and 900cm^-1, we can find bands characteristic of elongation vibrations between two atoms other than H and some bending vibrations of Y-H that can confirm the observations from the high frequencies region of the IR spectrum. We will focus on the elongation bands.

The S-H liaisons produce a small band of absorption between 2600 and 2550cm^-1. This band is lonely in this region of the spectrum but in diluted solutions the band can be to small to be detected.

C≡C shows a small band of absorbance between 2260 and 2100cm^-1. This region is dedicated to triple liaisons (C≡N shows a band between 2260 and 2240cm^-1) but harmonics and combination bands can be found in this region too. The bands of C=C are or middle or weak intensity between 1667 and 1640cm^-1 for alkenes that are not conjugated. Those alkenes are found between 1650 and 1600cm^-1. The cumulated alkenes (c=c=c) absorb between 2000 and 1900cm^-1.

Elongation vibrations of C=O

The absence of absorbance between 1870 and 1540cm^-1 is synonym of the absence of carbonyl. In this region, the position of the peak depend on the conjugation of the carbonyl and of the substituents (which substituent and its mass) and the possible H bonds.

An aliphatic ketone is found at 1715cm^-1. Modifications of the environment of the ketone can induce either an increase of a decrease of the wavenumber of the peak that depends on the predominance of the inductive or of the resonance effect. The inductive effect reduces the length of the C=O liaison, leading to the increase of its frequency of vibration. A resonance increases the length of the C=O liaison, displacing the peak towards the C-O peak that is at a lower frequency. For instance, Cl has a predominant inductive effect and the peak of the C=O is at 1815-1785cm^-1. The carbonyl of a primary amide vibrates at 1695-1650cm^-1 because the resonance effect is predominant. There are two bands for primary amides and only one for the other ones. Carbonyls conjugated with alkenes have peaks of absorption at smaller wavenumbers than the aliphatic ketone between 1685 and 1666cm^-1. An aldehyde absorbs a bit higher than the ketone (1740-1720cm^-1). An acid absorbs strongly (way more intensely than a ketone) around 1760cm^-1 but H bonds in dimmers of acids can displace the peak towards smaller frequencies (1720-1706cm^-1). Esters absorb at 1750-1735cm^-1. An acyl halide shows a strong absorption between 1815 and 1785cm^-1.

H bonds decrease the frequency of vibration of the carbonyls. The effect is small for intermolecular H bonds (a dozen of cm^-1) but can be important for intramolecular H bonds.

Elongation vibrations of C-O

In the middle range spectrum, between 1300 and 900cm^-1 the spectrum is usually complicated. It is the “digital print” of the compound where we can find the bands of alcohols C-C-O that should be accompanied with O-H bands at large frequencies.

The C-O liaison of alcohols shows an intense band due to the vibrations of elongation between 1260 and 1000cm^-1. This mode of vibration is usually coupled with the next carbon, as an asymmetric elongation C-C-O. The C-O-C of ethers and epoxides systems can also be found in this region of the spectrum, between 1150 and 1086cm^-1 (for its asymmetric elongation, the symmetric being very weak because it does not modify the dipolar moment) as an intense band or several bands if the nearby carbons are ramified. If one side is an aryl group, we observe one intense band (asymmetric elongation) at 1275-1200cm^-1 and one intense band (symmetric elongation) at 1075-1020cm^-1). Finally, the vinyl ethers show similar bands (1225-1200cm^-1 and 1075-1020cm^-1).

Acids give two bands of in unequal intensities, in addition of the one of C=O, coming from an interaction between the elongation mode of C-O and the bending mode of O-H. We talk about the bands of C-O-H. The most intense band between 1315 and 1280cm^-1 is called the elongation band C-O. The second band, between 1440 and 1395cm^-1 is the band called the band of bending O-H. This band falls in the same region than the shearing mode of CH₂. The carboxylate also gives two bands of unequal intensities, the most intense one between 1650 and 1550cm^-1 and the other one around 1400cm^-1.

A particularity of the esters is their intense band of absorption at the place of the elongation mode of the ketones. This band is not the band of C=O, that is displaced towards larger frequencies but one of the two bands of C-O. The C=O band is indeed between 1750 and 1735cm^-1 and 20cm^-1 lower if the carbonyl is conjugated. The C-O vibration is coupled with both sides of the liaison, leading to two bands of absorption. The most intense one is C-C(=O)-O giving a peak between 1300 and 1000cm^-1. Esters of aromatic acids absorb strongly between 1310 and 1250cm^-1. The other band between 1111 and 1031cm^-1 comes from the coupling O-C-C. Its frequency depends on the substituent of the ester. Esters of primary alcohols are at the lowest frequencies (1064-1031cm^-1) while esters of secondary alcohols and of aromatics absorb around 1100cm^-1and 1111cm^-1 respectively.

Elongation vibrations of NO

Nitro groups have a symmetric and an asymmetric elongation mode of vibration. The asymmetric mode gives an intense band between 1661 and 1499cm^-1. The symmetric mode gives a band between 1389 and 1259cm^-1. The wavenumbers of both bands depend on the substituent of the nitro group. In the case of an alkane, the bands should be observed around 1550 and 1372cm^-1 respectively. An electronegative group on the alpha carbon increases the frequency while a conjugation decreases the frequency.

Nitrates show 2 intense bands of elongation for N=O (one asymmetric and one symmetric) between 1300-1255cm^-1 and 1660-1625cm^-1. Nitrites have two bands as well, one for the cis (1625-1610cm^-1) and one for the trans isomer (1680-1650cm^-1).

Nitroso compounds show an absorption band between 1585 and 1539cm^-1.

Elongation modes of C=S

The elongation modes of C-S are found in the low frequencies range and the C=S slightly above this region, between 1250 and 1020cm^-1. This band is less intense than the one of a C=O because it is less polar.

Elongation modes of S=O

The S=O liaisons produce intense bands in general. The S=O liaison of sulfoxides vibrates between 1070 and 1030cm^-1. Sulfones show two bands of absorption between 1350-1300cm^-1 and 1160-1120cm^-1. The bands are displaced towards larger frequencies in the case of a sulfonyl chloride (1410-1380cm^-1 and 1204-1177cm^-1) or of a sulfonamide (1370-1335cm^-1 and 1170-1155cm^-1).

Bending in the middle range region

Several liaisons with elongation modes of high frequency, typically Y-H liaisons, have bending modes that fall in the same region than other elongation modes (Y-Z). They complicate the reading of the spectrum but can also confirm peaks of the high frequencies region.

C-H bending modes

The bending modes of the methylenic C-H liaisons have been seen at the beginning of this chapter. One mode is in the low frequencies region at 720cm^-1 and the three other are in the middle range region. The shearing mode has an almost fixed wavenumber of 1465cm^-1 but the two last modes (nod and torsion) give peaks between 1350 and 1150cm^-1. In a cycle, those frequencies are slightly decreased. Methylic C-H have two bending modes: one where the three H oscillate in phase and one wherein one H is not in phase. The symmetric mode (in phase) gives a peak at 1375cm^-1 and the asymmetric mode at 1450cm^-1, almost at the same frequency than the shearing mode of a methylenic C-H. =C-H of alkenes absorb at 1415cm^-1 due to a shearing mode. Aromatics have a mode of bending out of the plane that give intense peaks in the low frequencies region and a bending mode in the plane give peaks between 1300 and 1000cm^-1.

Bending O-H

Peaks of alcohols are not very characteristic and are in the region between 1420 and 1330cm^-1. If there is a proton on the carbon wearing the –OH (i.e. primary and secondary alcohols), there is a coupling that gives two bands around 1420 and 1330cm^-1. Tertiary alcohols give only one peak. In the case of the acids, there is an interaction between the elongation mode of C-O and the bending mode of O-H (discussed in the C-O elongation modes section). The C-O-H bending mode should be found between 1440cm^-1 and 1395cm^-1.

Bending N-H

Primary amines have a bending mode that gives a peak of average to high intensity between 1650cm^-1 and 1580cm^-1. The bending mode of secondary amines is hardly detectable except for aromatic secondary amines that absorb around 1515cm^-1. In the case of amides, the N-H bending mode should be observed by a band between 1655 and 1590cm^-1 with an intensity way smaller than the intensity of the C=O peak. As C=O absorbs in the same region, the peaks of the elongation C=O and of the bending N-H can merge.

Low frequencies region

In this region we find mostly the bending modes but some elongation modes are also present. In my opinion, this region is less interesting than the high and middle frequencies regions and is pretty hard to read because several modes have large ranges of possible frequencies that overlap with other modes. The most characteristic peaks are the peaks for dimmers and for aromatics. Between 900 and 650cm^-1, a large and intense band is characteristic of a dimmer of acid, amide or amine. In the same region, the absence of intense peaks indicates that the molecule is not an aromatic.

Bending modes of CH

The peak of the swinging mode of methylenic -C-H at 720cm^-1 has a low intensity or is not visible. A very intense band (usually the most intense of the spectrum) is visible between 1000 and 650cm^-1 if an alkene is present in the molecule, coming from a bending mode out of the plane. It is usually this kind of bending out of the plane that gives intense bands in the low frequencies region. A very intense peak can thus be observed for aromatics between 900 and 650cm^-1. The alkynes give a thick and intense peak at lower frequencies, between 700 and 610cm^-1.

Other bending modes

The bending mode of alcohols gives a thick peak between 769cm^-1 and 650cm^-1 in addition of the bending modes of the middle frequencies region. Dimers of acids give a very broad band (not as large as the band in high frequencies that the acids give but it is characteristic) of average intensity centred near 920cm^-1. Amides also give a thick band of average intensity but between 800 and 666cm^-1, again due to a bending out of the plane. This bending is stronger and at higher frequencies for amines (909-666cm^-1).

The bending out of the plane of the amines and amides generate one average/intense band respectively between 909-666cm^-1 and 800-666cm^-1. Nitro groups seem to show a bending mode between 763 and 690cm^-1.

Elongation modes in the low frequencies region

A band of elongation of N-O appears between 870-833cm^-1 from nitro groups and between 850-750cm^-1 from nitrites. C-S absorbs between 700 and 600cm^-1.

Finally, halogens have their elongations modes in the low frequencies. In an aliphatic chain, C-Cl absorbs between 850 and 550cm^-1, C-Br between 690 and 515cm^-1, C-I between 600 and 500cm^-1 and C-F between 1400 and 730cm^-1. The band of C-F is strong and large. If the substituent is an aromatic, the bands are displaced towards larger frequencies, above 1000cm^-1 in the case of chlorobenzenes.