= 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 \b[<even> $J$ (para-hydrogen, one <spin> state)]; the symmetric <spin> triplet pairs with \b[<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 \b[<even> $J$ (ortho-deuterium)]; $I=1$ supplies three antisymmetric states paired with \b[<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.
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