The Heitler-London model

There are two main models to approach the mathematical representation of the molecular bond:

1. Heitler-London, or valence bond model;
2. MO-LCAO, that stands for Molecular Orbital from Linear Combination of Atomic Orbitals.

In this section we're going to illustrate the first of this two models. So, for simplicity, we consider the molecule ${\displaystyle H_{2}}$ and we just try to find a way to describe the wave function which represents the two electrons orbiting around the bounded nuclei. Firstly we consider that both electrons, taken singularly, are described by the wave function corresponding to the ground state of Hidrogen (1s orbital). If the two nuclei were far enough, the superposition would be neglegible and the two electrons could be described by a total wave fuction given by the multiplication of the two single wave functions for atom ${\displaystyle A}$ and atom ${\displaystyle B}$:

$${\displaystyle \psi \left({\vec {x}}_{1};{\vec {x}}_{2}\right)=u_{A}\left({\vec {x}}_{1}\right)u_{B}\left({\vec {x}}_{2}\right)}$$

where ${\displaystyle {\vec {x}}_{1}}$ and ${\displaystyle {\vec {x}}_{2}}$ represent the positions of the two electrons. In this case we can consider that the global Hamiltonian is separable, because the terms involving superposition between the effects of the two nuclei vanish when the distance between them is big. If the two nuclei are close to each other it is necessary to antisymmetrize the wave function (electrons are fermions). In order to do this we have to takein account all the possible electrons configuration:

It is fundamental to observe now that not all four states represented by the image above are eigenstates of the total spin operator ${\displaystyle S=S_{1}+S_{2}}$. In particular, only the two states on the left (${\displaystyle \alpha \left(1\right)\alpha \left(2\right)}$ and ${\displaystyle \beta \left(1\right)\beta \left(2\right)}$) are eigenstates of ${\displaystyle S}$. To obtain a coplete system of eigenstates of ${\displaystyle S}$ we have to cosider, a part from the first two states, linear combinations between the two states on the right:

$${\displaystyle \chi _{\pm }={\frac {\alpha \left(1\right)\beta \left(2\right)\pm \beta \left(1\right)\alpha \left(2\right)}{\sqrt {2}}}}$$

.

${\displaystyle \chi _{-}}$ is called a state of singlet (${\displaystyle S=0}$), and from now we will denote it with ${\displaystyle \chi _{sing}}$, whereas the other three are states of triplet (${\displaystyle S=1}$) and will be all denoted by ${\displaystyle \chi _{trip}}$. Of course we have to match the spin component of the wave function with the correct (symmetrical or antisymmetrical) spatial one, in order to obtain a totally antisymmetrical wave function. We denote:

$${\displaystyle \psi _{\pm }=\left(u_{A}\left({\vec {x}}_{1}\right)u_{B}\left({\vec {x}}_{2}\right)\pm u_{A}\left({\vec {x}}_{2}\right)u_{B}\left({\vec {x}}_{1}\right)\right)}$$

We get four antyssimetrical states:

• ${\displaystyle Singlet:\psi _{+}\chi _{sing}}$

• ${\displaystyle Triplet:\psi _{-}\chi _{trip}}$

It is possible to write the complete electrons Hamiltonian:

$${\displaystyle {\hat {H}}_{e}={\frac {{\vec {p}}_{1}^{2}}{2m_{e}}}-{\frac {e^{2}}{4\pi \varepsilon _{0}\left|{\vec {x}}_{1}-{\vec {R}}_{A}\right|}}+{\frac {{\vec {p}}_{2}^{2}}{2m_{e}}}-{\frac {e^{2}}{4\pi \varepsilon _{0}\left|{\vec {x}}_{2}-{\vec {R}}_{B}\right|}}-{\frac {e^{2}}{4\pi \varepsilon _{0}\left|{\vec {x}}_{1}-{\vec {R}}_{B}\right|}}-{\frac {e^{2}}{4\pi \varepsilon _{0}\left|{\vec {x}}_{2}-{\vec {R}}_{A}\right|}}+{\frac {e^{2}}{4\pi \varepsilon _{0}\left|{\vec {x}}_{1}-{\vec {x}}_{2}\right|}}+{\frac {e^{2}}{4\pi \varepsilon _{0}\left|{\vec {R}}_{A}-{\vec {R}}_{B}\right|}}}$$

The problem now is to normalize ${\displaystyle \psi _{sing}}$ and ${\displaystyle \psi _{trip}}$. In fact, when superposition occurs, the quantity ${\displaystyle \left\langle u_{a}|u_{b}\right\rangle }$ is not equal to ${\displaystyle 0}$, because the two states ${\displaystyle u_{a}}$ and ${\displaystyle u_{b}}$ are not orthogonal. We define the overlap ${\displaystyle S}$ as:

$${\displaystyle S=\left\langle u_{a}|u_{b}\right\rangle }$$

and the normalization coefficient that will appear in the correct expression of ${\displaystyle \psi _{sing}}$ and ${\displaystyle \psi _{trip}}$ will be ${\displaystyle {\frac {1}{\sqrt {2\left(1\pm S^{2}\right)}}}}$. It's easy to check that, if the overlap is null (${\displaystyle S=0}$, distant atoms), the normalization coefficient is ${\displaystyle {\frac {1}{\sqrt {2}}}}$, which is consistent with the orthonormality of ${\displaystyle u_{a}}$ and ${\displaystyle u_{b}}$. Taking in consideration th Hamiltonian we wrote above, we calculate:

$${\displaystyle \left\langle \psi _{\pm }\left|{\hat {H}}\right|\psi _{\pm }\right\rangle =2E_{1s}+2{\frac {J_{AB}\pm SK_{AB}}{1\pm S^{2}}}+{\frac {e^{2}}{4\pi \varepsilon _{0}\left|{\vec {R}}_{A}-{\vec {R}}_{B}\right|}}+{\frac {J_{ee}\pm K_{ee}}{1\pm S^{2}}}}$$

where ${\displaystyle E_{1s}}$ represents the energy of the two atomic orbitals. ${\displaystyle J_{AB}}$ and ${\displaystyle K_{AB}}$ come from the interaction between electrons and nuclei. ${\displaystyle J_{AB}}$ is called direct integral and represents the classical Coulomb potential energy:

$${\displaystyle J_{AB}={\frac {-e^{2}}{4\pi \varepsilon _{0}}}\int d^{3}{\vec {x}}{\frac {\left|u_{A}\left({\vec {x}}\right)\right|^{2}}{\left|{\vec {x}}-{\vec {R}}_{B}\right|}}}$$

${\displaystyle K_{AB}}$ is called switch integral and has no classical interpretation. It's defined as:

$${\displaystyle K_{AB}={\frac {-e^{2}}{4\pi \varepsilon _{0}}}\int d^{3}{\vec {x}}{\frac {u_{A}^{*}\left({\vec {x}}\right)u_{B}\left({\vec {x}}\right)}{\left|{\vec {x}}-{\vec {R}}_{B}\right|}}}$$

It can be demonstrated that ${\displaystyle J_{AB}<0}$ and ${\displaystyle K_{AB}<0}$, moreover if we switch the labels ${\displaystyle A}$ and ${\displaystyle B}$ in the integrals, their values don't change. This is the reason of the factor ${\displaystyle 2}$ in the expression of the energy. The term ${\displaystyle {\frac {e^{2}}{4\pi \varepsilon _{0}\left|{\vec {R}}_{A}-{\vec {R}}_{B}\right|}}}$ comes from the interaction between the two nuclei. Eventually the last term of the energy is due to the repulsive interaction between the two electrons. ${\displaystyle J_{ee}}$ and ${\displaystyle K_{ee}}$ are defined in analogy with ${\displaystyle J_{AB}<0}$ and ${\displaystyle K_{AB}<0}$:

$${\displaystyle J_{ee}={\frac {e^{2}}{4\pi \varepsilon _{0}}}\int d^{3}{\vec {x}}_{1}d^{3}{\vec {x}}_{2}{\frac {\left|u_{A}\left({\vec {x}}_{1}\right)\right|^{2}\left|u_{B}\left({\vec {x}}_{2}\right)\right|^{2}}{\left|{\vec {x}}_{1}-{\vec {x}}_{2}\right|}}}$$

$${\displaystyle K_{ee}={\frac {e^{2}}{4\pi \varepsilon _{0}}}\int d^{3}{\vec {x}}_{1}d^{3}{\vec {x}}_{2}{\frac {u_{A}^{*}\left({\vec {x}}_{1}\right)u_{A}\left({\vec {x}}_{2}\right)u_{B}\left({\vec {x}}_{2}\right)u_{B}^{*}\left({\vec {x}}_{1}\right)}{\left|{\vec {x}}_{1}-{\vec {x}}_{2}\right|}}}$$

with ${\displaystyle J_{ee}>0}$ and ${\displaystyle K_{ee}>0}$. The minimum energy can be obtained taking the ${\displaystyle +}$ between ${\displaystyle J_{AB}}$ and ${\displaystyle K_{AB}}$. So the minimum energy state is the singlet state. We will denote ${\displaystyle E_{+}}$ the energy of the singlet state and ${\displaystyle E_{-}}$ the energy of the triplet state. It can be shown that it may happen, for certain values of interatomic distance that ${\displaystyle E_{+}<2E_{1s}}$, which means that the energy of the two atoms bounded in this state is lower than the energy they would have if they were completely separated. This is the reason why we call this state bonding state. On the other hand ${\displaystyle E_{-}>2E_{1s}}$, so the triplet state is called antibonding state. This behaviour can be explained by talking about the charge density for the elctrons. Considering the square module of the wave function as a propability density it's possible to write:

$${\displaystyle {\frac {\rho \left({\vec {x}}\right)}{\left(-e\right)}}=\int d^{3}{\vec {x}}_{1}\left|\psi _{\pm }\left({\vec {x}}_{1};{\vec {x}}\right)\right|^{2}+\int d^{3}{\vec {x}}_{2}\left|\psi _{\pm }\left({\vec {x}};{\vec {x}}_{2}\right)\right|^{2}}$$

Because of the simmetry of ${\displaystyle \psi _{\pm }}$ for the exchange between ${\displaystyle 1}$ and ${\displaystyle 2}$ indexes, the two integrals have the same value. This leads to:

$${\displaystyle {\frac {\rho \left({\vec {x}}\right)}{\left(-e\right)}}=2\int d^{3}{\vec {x}}_{1}\left|\psi _{\pm }\left({\vec {x}}_{1};{\vec {x}}\right)\right|^{2}={\frac {\left|u_{A}\left({\vec {x}}\right)\right|^{2}+\left|u_{B}\left({\vec {x}}\right)\right|^{2}\pm 2u_{A}\left({\vec {x}}\right)u_{B}\left({\vec {x}}\right)}{1\pm S^{2}}}}$$

where ${\displaystyle +}$ sign is for the singlet state and ${\displaystyle -}$ for the triplet. For the former state, in the middle of the molecule there is an increase of charge density, related to an higher probability to find the electrons between the two nuclei and to a stronger bond. The latter state, on the contrary, implicates a decrese of charge density and also of probability to find the electrons in the middle of the nuclei.

This model predicts a distance between bound nuclei of ${\displaystyle 1\,\mathrm {\AA} {}}$ when the experimental value is ${\displaystyle 0,74\,\mathrm {\AA} {}}$. For the binding energy the theoretical values is ${\displaystyle 3,14\,eV}$ and the experimental one is ${\displaystyle 4,75\,eV}$. These results are indicative that the Heitler-London model has some intrinsic limitits. In order to overcome these problems it's useful to change point of view and consider the Molecular Orbital model.

reference:

https://en.wikitolearn.org/Course:Physics_of_Atoms_and_Molecules/Molecules/The_Heitler-London_model