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An equilibrium bond length requires $B_e$, not merely the ground-vibrational-state constant. Extrapolating the linear vibration-rotation interaction law givesUsing integer isotope masses, the reduced mass isFor two nuclei $I_e=\mu r_e^2$, soUsing $B_0$ instead gives $r_0\simeq 177.1$ pm, an effective vibrational separation. The small difference is precisely why the weaker vibrational progressions matter when the target is equilibrium geometry. Higher terms in $B_v$ and the rigid-rotor approximation limit this estimate's precision.

The two extra progressions are rotational transitions within thermally populated $v=1$ and $v=2$ vibrational levels. Isotopic purity rules out attributing them to different isotope masses. Their rotational constants areThe Boltzmann distribution makes each higher vibrational manifold less populated, so its rotational spectrum is weaker. A Morse oscillator explores larger separations as it is excited: the bond's vibrationally averaged moment of inertia increases and the rotational spacing decreases. More precisely, $B_v$ involves an average of the inverse square separation; it is not simply the inverse square of the mean separation.

The observed decrement is constant to the given precision:This fits the leading vibration-rotation interaction law $B_v=B_e-{\alpha }_e(v+\frac{1}{2})$.