Two helium atoms: bonding and antibonding orbitals

When two helium atoms are brought very close, their 1s orbitals overlap, but no stable He₂ molecule forms; the atoms repel each other. Molecular orbital (LCAO) theory explains why.

Molecular orbitals formed

Each He atom has the configuration $1s^2$ (two electrons). Overlap of the two 1s orbitals gives:

$$\psi_\sigma = \psi_{1s}(A) + \psi_{1s}(B) \qquad \text{(bonding, lower energy)}$$ $$\psi_{\sigma^*} = \psi_{1s}(A) - \psi_{1s}(B) \qquad \text{(antibonding, higher energy)}$$

Filling the orbitals

There are 4 electrons in total. Each MO can hold only 2 electrons with opposite spins (Pauli exclusion principle):

• 2 electrons go into the bonding orbital $\sigma$;
• the other 2 are forced into the antibonding orbital $\sigma^*$.

 Energy
   |           sigma*   _^v_   (2 e, raises E)
   |                   /    \
   |  1s  _^v_  ------        ------  _^v_  1s
   |   He(A)           \    /          He(B)
   |           sigma    _^v_   (2 e, lowers E)

Result

• The rise in energy of the antibonding orbital is larger than the fall of the bonding orbital (the antibonding level is pushed up more than the bonding level is pulled down).
• So with both orbitals full, the total energy of He–He is higher than that of two separate He atoms.
• Bond order $= \frac{1}{2}(N_b - N_a) = \frac{1}{2}(2 - 2) = 0$: no bond.

 E (He-He)
  ^
  |\
  | \     total energy: purely repulsive,
  |  \    no minimum
  |   '.
  |     '-.__
 0+-----------'--------------------> R

As the separation decreases further, the electron clouds overlap more, the antibonding electrons put charge outside the internuclear region, and the nuclei repel strongly. Hence the two He atoms push each other apart.

Conclusion

Helium does not form He₂; it exists as a monatomic gas. Only very weak Van der Waals forces act between He atoms, which is why helium liquefies only at about 4.2 K. In contrast, H₂ is stable because its 2 electrons fill only the bonding orbital.