# 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

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.