Suppose we are looking at the valence electron of a Hydrogenic atom for which $I = 3/2,\ J = 1/2$ (and hence $L = 0$). Our Hamiltonian for this setting is given by: $$ H = \frac{p^2}{2m} + V_\text{ECP} + \eta LS + hA_\text{hfs} I J $$ where $V_\text{ECP}$ is the effective core potential.
We start by writing our state in the coupled basis (i.e., $F = I + J = I + S$ because $L = 0$) and by using the Clebsch-Gordan coefficients we can transform it to the uncoupled basis \begin{align*} |F=1, m_F=0\rangle &= \sqrt{\frac{1}{2}}|I=3/2, S=1/2, m_I=1/2, m_S=-1/2\rangle \\ &- \sqrt{\frac{1}{2}}|I=3/2, S=1/2, m_I=-1/2, m_S=1/2\rangle \end{align*}
Since the main part of the energy comes from the first two terms of the Hamiltonian and these are independent of $I, S$, they cancel out and the energy of this state becomes much smaller than, for example, the energy of $|F=2, m_F=2\rangle$.
The difference in energy shouldn't be that big, of course, so where is my reasoning error?
