   PHY.K02UF Molecular and Solid State Physics

## Covalent bond potential

An important contribution to a covalent bond is the confinement energy. This can best be explained by considering a one dimensional example. The Schrödinger equation for a particle in a one dimensional potential is,

$$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} +V(x)\psi(x) = E\psi(x).$$

For the case of an infinite square well, the potential energy term $V(x)=0$ and only the kinetic energy term remains. The solutions are sinusoidal with a wavelength that is $\lambda_n = 2L/n$ where $n = 1,2,3,\cdots$ is an integer. The energy associated with state $n$ is,

$$E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} = \frac{\hbar^2k_n^2}{2m}.$$

Here $k_n= \frac{2\pi}{\lambda_n}$ is the wave number of the wave function. The momentum is related to the wave number by the de Broglie relation, $p = \hbar k$ so the energy can also be written as $E = \frac{\hbar^2k^2}{2m} = \frac{p^2}{2m} = \frac{mv^2}{2}$, where $v$ is the velocity. This shows that the confinement energy is the kinetic energy of the particle. This makes sense since we set the potential energy term to zero. You can adjust the slider to change the width of the potential well and see how that changes the energies of the two lowest square well states.

 Your browser does not support the canvas element. $E_2=\frac{\hbar^2k_2^2}{2m}=$ 150 [eV]$E_1=\frac{\hbar^2k_1^2}{2m}=$ 37.6 [eV] $L=$ 1 [Å] $\lambda_1 = 2L =$ 2 [Å]    $k_1 = \frac{2\pi}{\lambda_1}=$ 3.14 [Å-1] $\lambda_2 = L =$ 1 [Å]     $k_2 = \frac{2\pi}{\lambda_2}=$ 6.28 [Å-1]

Some elementary textbooks try to explain a covalent bond by the electrostatic interaction of the electrons being concentrated between the positively charged nuclei. While there is a electrostatic component to a covalent bond, any concentration of the electron wavefunction will increase the confinement energy. The dominant contribution to a covalent bond is the reduction in confinement energy as the electrons spread out over both atoms in the bond.

The quantum mechanically calculated bond potentials of H2 for the lowest energy singlet and the lowest energy triplet are shown below. In the simplest molecular orbital description, the singlet would be $| \psi_+\uparrow ,\,\psi_+\downarrow \rangle$ and the triplet would be $|\psi_+\uparrow ,\,\psi_-\uparrow \rangle$.

H2
U(R) [eV]

$U_{0}=7.263$ eV
$R_{E}=0.74$ Å
$a=1.610$ Å-1
$U_{\infty}=3.629$ eV

R [Å]

Notice that the triplet state has the lower energy for large separations. The singlet state energy was fit with a Morse potential:

$$U(R)= U_0\left(e^{-2a(R-R_E)}-2e^{-a(R-R_E)}\right) +U_{\infty},$$

and the energy of the triplet state was fit by a simple exponential $U(R) = U_t\exp(-\beta R)$, where $U_t = 35.14$ eV and $\beta = 2.36$ Å-1. The dissociation energy from the minimum of the singlet to the singlet-triplet crossing point is 4.3 eV.