# For the $\mathrm{H}_2$ molecule which spin states correspond to bonding? and which spin states correspond to anti-bonding?

Do I understand correctly?

In the case of the $\mathrm{H}_2$ molecule, the overall state is $S = \Psi(r)\chi(s)$.....(Spatial and spin aspects)

• $\Psi(r)$ is inherently anti-symmetric under exchange since we're dealing with Fermions (electrons in this case)

• In order for there to be bonding S must be symmetric under exchange to experience an "attractive exchange force"

• Therefore $\chi(s)$ must be anti-symmetric under exchange in order to make S symmetric under exchange

ConclusionAnd for these reasons the singlet state (which is anti-symmetric) is bonding, and triplet state (which is symmetric) is anti-bonding

Correct conclusion but incorrect argument. The full wave function - not only its spatial part - must be antisymmetric. The spatial part can be symmetric if the spin part is antisymmetric.

In fact in the case of $H_2$ the bonding is for a symmetric spatial wavefunction (see here). Indeed “singlet” usually refers to spin-singlet, which is $S=0$ and antisymmetric, thus forcing the spatial part to be symmetric under permutation.

• Thank you for your response, let me see if I understand correctly 1) S = Psi(r)Chi(s) must be anti-symmetric for fermions 2) Psi(r) being symmetric leads to bonding. Psi(r) being anti-symmetric leads to anti-bonding. 3) Chi(s) can be either symmetric or anti-symmetric Conclusion 1: Chi(s) being symmetric leads to anti-bonding because it forces Psi(r) to be anti-symmetric in order to satisfy 1) – Omar Azami Aug 5 '18 at 20:20
• Conclusion 2: Chi(s) being anti-symmetric leads to bonding because it forces Psi(r) to be symmetric in order to satisfy 1) Conclusion 3: Triplet states of H2 are symmetric and are therefore anti-bonding. Conclusion 4: Singlet states of H2 are anti-symmetric and are therefore bonding. – Omar Azami Aug 5 '18 at 20:30