[bmu123]

Lines near the band origin in the high-resolution infrared spectrum of ¹²C¹⁶O are observed at 2131.58, 2135.50, 2139.38, 2147.04, 2150.81 and
2154.55 cm^-1. Assign J_initial and J_final quantum numbers to each transition
and, hence, determine the rotational constant of this molecule in both the v = 1 and the v = 0 vibrational levels. Determine the band origin.

OK so, in the P branch:
2131.58  J=2  J=3
2135.50  J=1  J=2
2139.38  J=0  J=1

ΔE=hν₀+hB[J(J+1)-J'(J'+1)]
ΔE=hν₀+hB[J(J-1)-J(J+1)]
ΔE=hν₀-2hBJ

R branch:
2147.04  J=1  J=0
2150.81  J=2  J=1
2154.55  J=3  J=2

R branch: ΔE=hν₀+hB[J(J+1)−J′(J′+1)]
ΔE =hν₀ +hB [(J+1)(J+2)-J(J+1)]
ΔE =hν₀+2hB(J+1)

These are the equations I've found but I'm guessing that they're not right as I don't have v₀, but I can't do the rotational constant from the line spacing because they're all different? Just need a point in the right direction, not the answer, thanks.

[Corribus]

Those equations assume the rotational constant for the two vibrational states are equivalent. In some cases this assumption is ok, but by the wording of the question clearly it isn't here.

Have you seen this website?

http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Rotational_Spectroscopy/Rovibrational_Spectroscopy

(There is a procedure there, near the bottom under "real spectra", that can be used to get the correct answers using your input values.)

[bmu123]

Ah yes I remember doing this before, didn't think about that! Thanks so much