LCAO and stability of H2 molecule

Linear Combination of Atomic Orbitals (LCAO) is a method of finding molecular orbitals by adding or subtracting the atomic orbitals of the atoms that form the molecule:

$$\psi_{MO} = c_A\psi_A + c_B\psi_B$$

For two identical atoms $c_A = \pm c_B$, which gives two molecular orbitals, one bonding and one antibonding.

Molecular orbitals of H₂

Each H atom has one electron in a 1s orbital, $\psi_{1s}(r_A)$ and $\psi_{1s}(r_B)$. When the atoms come close, these overlap and form

$$\psi_\sigma = \psi_{1s}(r_A) + \psi_{1s}(r_B) \qquad \text{(bonding)}$$ $$\psi_{\sigma^*} = \psi_{1s}(r_A) - \psi_{1s}(r_B) \qquad \text{(antibonding)}$$

   psi_A + psi_B  (sigma)        psi_A - psi_B  (sigma*)
       .-.   .-.                    .-.
      /   \_/   \                  /   \
  ___/    ^      \___        _____/     \     ______
     A  charge    B               A      \   / B
        builds up                         '-'
        between A, B            node at mid-point

• Bonding orbital: the two waves add between the nuclei, so $|\psi_\sigma|^2$ is large there. This shared negative charge attracts both protons, so the energy $E_\sigma$ is lower than the energy of an isolated 1s electron.
• Antibonding orbital: the waves cancel at the mid-point; there is no shared charge, the protons repel, so $E_{\sigma^*}$ is higher than the 1s energy.

Filling the orbitals

H₂ has two electrons. By the Pauli principle both can go into the lowest orbital $\sigma$ with opposite spins; $\sigma^*$ remains empty.

 Energy
   |              ____  sigma*  (empty)
   |  1s  _^_   /      \   _^_  1s
   |  H_A      \        /       H_B
   |             _^v_   sigma (2 e, opposite spins)

Energy versus separation

The total energy of the molecule as a function of nuclear separation $R$:

 E
 ^
 |\  sigma* : always above zero -> repulsive
 | \__
 |    ''---...________
0+-----------------------------------> R
 |  \             ___...---
 |   \       _.-''  sigma : bonding
 |    \   _-'
 |     '-'  <- minimum at R0 = 0.074 nm
 |            depth = 4.5 eV

• At large $R$ the atoms do not interact, $E = 0$ (two separate H atoms).
• As $R$ decreases, electrons in $\sigma$ lower the energy.
• At very small $R$ the proton–proton repulsion dominates and the energy rises.
• The minimum at $R_0 = 0.074$ nm is the bond length. The molecule's energy here is about 4.5 eV lower than that of two separate atoms (bond energy).

Conclusion

Since the molecule, with both electrons in the bonding orbital, has less total energy than the separated atoms, and every system moves to its lowest energy state, the formation of the H₂ molecule is energetically favourable and H₂ is stable. Bond order $= \frac{1}{2}(2 - 0) = 1$ (a single covalent bond). Energy of 4.5 eV must be supplied to break it.