= Solution
For <hyperfine coupling for a spin-one nucleus>, let the <electron spin> be $s=1/2$ and the <nuclear spin> $I=1$. First couple $\mathbf j=\mathbf L+\mathbf s$, giving $j=L\pm1/2$, then $\mathbf F=\mathbf j+\mathbf I$, giving $F=|j-I|,|j-I|+1,\ldots,j+I$.
|| $L$
|| $j$
|| Allowed $F$
| $1$
| $1/2$
| $1/2,3/2$
| $1$
| $3/2$
| $1/2,3/2,5/2$
| $2$
| $3/2$
| $1/2,3/2,5/2$
| $2$
| $5/2$
| $3/2,5/2,7/2$
For each $F$, the projection is $m_F=-F,-F+1,\ldots,F$; electron projections similarly satisfy $m_j=-j,\ldots,j$. Summing $2F+1$ gives 18 states for $L=1$ and 30 for $L=2$, agreeing with $(2L+1)(2s+1)(2I+1)$.
\b[Notation qualification:] if the printed leading “2” denotes the principal <quantum number>, a hydrogenic $2d$ bound state does not exist because $L\leq n-1$; a physical $d$ state needs $n\geq3$. If a doublet term symbol $^{2}P$ or $^{2}D$ was intended, its leading 2 instead means $2s+1$, and the angular-<momentum> table applies without fixing $n$. The answers above use the explicitly specified $L$ values, while retaining this ambiguity.<Quantum Mechanics by David Tong>[Three-dimensional angular momentum and hydrogenic quantum-number restrictions]{ref}
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