Bonding orbitals and instability of H3

Bonding and antibonding molecular orbitals

When two atomic orbitals $\psi_A$ and $\psi_B$ overlap, they combine (LCAO) into two molecular orbitals:

Each molecular orbital can hold at most two electrons, with opposite spins (Pauli exclusion principle).

Why H₃ is not stable

H₂: Each H atom has one 1s electron. The two 1s orbitals form $\sigma$ and $\sigma^*$. Both electrons go into the bonding orbital $\sigma$ with opposite spins, and $\sigma^*$ stays empty. The energy falls by about 4.5 eV, so H₂ is stable.

Adding a third H atom: The third electron cannot enter $\sigma$, because $\sigma$ is already full with two electrons of opposite spin (Pauli principle). It must go into the higher-energy antibonding orbital $\sigma^*$.

 Energy
   |          sigma*  _^_   <- 3rd electron forced here
   |                 /    \
   |  1s  __  ---           ---  __ 1s
   |                \      /
   |          sigma  _^v_   <- full (2 electrons)
   |
   |  H2 + H:  bonding gain is cancelled by the
   |           energy cost of the sigma* electron

• The electron in $\sigma^*$ puts its charge outside the region between the nuclei, so it causes repulsion and raises the energy.
• The energy gained from the bonding pair is largely cancelled by the energy cost of the antibonding electron, so the total energy of H₃ is not lower than that of H₂ + H.
• A system always goes to the state of lowest energy, so H₃ breaks up into a stable H₂ molecule and a free H atom.

(In a full three-orbital treatment of linear H₃, the third electron goes into a non-bonding/antibonding level above the bonding one, and the result is the same: H₃ has no energy minimum and is not a stable molecule.)

Same idea for He₂

He has two 1s electrons, so He₂ would have 2 electrons in $\sigma$ and 2 in $\sigma^*$. Bond order $= \frac{1}{2}(2 - 2) = 0$, so He₂ does not form. In general, a molecule is stable only if bonding electrons outnumber antibonding electrons enough to lower the total energy.