(a)$\sum_{J=0,1,2 \ldots}(2 J+1) e^{-\beta h c B J(J+1)}$
(b)$\sum_{J=1,3,5 \ldots}(2 J+1) e^{-\beta h c B J(J+1)}$
(c)$\sum_{J=0,2,4 \ldots}(2 J+1) e^{-\beta h c B J(J+1)}$
(d)$\frac{1}{4}\left[\sum_{J=0,2,4 \ldots}(2 J+1) e^{-\beta h c B J(J+1)}+3 \sum_{J=1,3,5 \ldots}(2 J+1) e^{-\beta h c B J(J+1)}\right]$
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
For H₂ each nucleus has I = ½, giving (2I+1)² = 4 nuclear-spin states: 1 antisymmetric (para, combines with even J) and 3 symmetric (ortho, combine with odd J).Hence $q_{\text{rot,nuc}}=\sum_{J=0,2,4\ldots}(2J+1)e^{-\beta hcBJ(J+1)}+3\sum_{J=1,3,5\ldots}(2J+1)e^{-\beta hcBJ(J+1)}$; dividing by the 4 nuclear-spin states gives the rotational partition function