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For an axial spin Hamiltonian ${\mathcal{H}}_D=D{S}_z^2$ with $D/k_B=-25$ K, the minima are $m_S=\pm 5/2$. The highest levels along the reversal ladder are the $m_S=\pm 1/2$ doublet. Henceabout 1.25 kJ mol${}^{-1}$ or 12.9 meV per molecule. Using $|D|S^2$ would incorrectly put a half-integer spin at a nonexistent $m_S=0$ level. Transverse anisotropy and quantum tunnelling of magnetization can create faster relaxation pathways, so this axial ladder barrier need not equal an experimentally fitted effective barrier.