 Solution

The spin isomers of hydrogen illustrate how nuclear spin statistics constrain the product of electronic, vibrational, rotational and nuclear-spin exchange symmetries. In the ground electronic and vibrational states of these homonuclear molecules, the first two factors are symmetric. Exchanging the nuclei reverses the internuclear direction, so a rotational state has parity $(-1{)}^J$.

For ${}^1{\mathrm{H}}_2$, the protons are spin-one-half fermions and the total nuclear-exchange wavefunction must be antisymmetric. The antisymmetric spin singlet pairs with even $J$ (para-hydrogen, one spin state); the symmetric spin triplet pairs with odd $J$ (ortho-hydrogen, three spin states). For ${}^2{\mathrm{H}}_2$, spin-one deuterons are bosons, so the total must be symmetric. Nuclear spins $I=0,2$ supply six symmetric states paired with even $J$ (ortho-deuterium); $I=1$ supplies three antisymmetric states paired with odd $J$ (para-deuterium). The high-temperature ortho/para statistical ratios are $3:1$ for hydrogen and $2:1$ for deuterium, while low-temperature equilibrium populations also depend on rotational energies and whether spin conversion can occur.