= Solution
A <planar rigid rotor> has angular eigenfunctions $e^{im\phi}$ and single-valuedness imposes $m=0,\pm1,\pm2,\ldots$. Its <energy> depends on $m^2$, so if the level is labelled by $J=|m|$, then
$$
\boxed{J=0,1,2,\ldots;\qquad g_0=1,\quad g_J=2\ (J\ge1).}
$$
The two states at $J>0$ represent opposite directions of <angular momentum>. If the signed <quantum number> itself is called $J$, each signed state is counted once instead; one must not both sum over both signs and multiply by two again.
For an unrestricted three-dimensional <linear rigid rotor>, $J=0,1,2,\ldots$ has $M=-J,\ldots,J$ and <state degeneracy> $2J+1$, with $E_J=\hbar^2J(J+1)/(2I)$. Planar confinement removes one rotational coordinate and changes both the spectrum and degeneracy. These are the spatial counts before <nuclear spin statistics>. For homonuclear <hydrogen>, the allowed even/odd spatial states must also be paired with the appropriate <parahydrogen> or <orthohydrogen> nuclear states, as considered explicitly in A8.
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