2 added detail edited Mar 18 '17 at 10:35 porphyrin 19.8k11 gold badge3535 silver badges6060 bronze badges I'm assuming you are measuring the rotationally well resolved band at $$\approx 2349 \pu{cm^{-1}}$$ and you are correct. The rotational band structure for the antisymmetric vibration centred at $$\approx 2349 \pu{cm^{-1}}$$ (also called parallel $$v_3$$ band) has alternate lines missing, which are those having odd J. $$\ce{CO2}$$ is a linear symmetric molecule ($$D_{\infty h}$$ point group) but nuclear spins on $$^{12}\ce{C}$$ and $$^{16}\ce{O}$$ are both zero and means that antisymmetric rotational lines are missing. The ground vibrational state is $$\Sigma^{+}_g$$ resulting in the odd levels being absent. The effect is due to the symmetry properties of the wavefunction, and what happens on exchange of coordinates. For fermions ( multiples of spin 1/2 , eg. electrons, protons,) the wavefunction must be antisymmetric but symmetric for Bosons (integer spin 0,1,2 etc) which is the case for $$\ce{CO2}$$. Look for 'statistical weight', 'linear molecules', and 'nuclear spin' in a good spectroscopy book, Herzberg 'Infra Red and Raman Spectra' for example. I'm assuming you are measuring the rotationally well resolved band at $$\approx 2349 \pu{cm^{-1}}$$ and you are correct. The rotational band structure for the antisymmetric vibration centred at $$\approx 2349 \pu{cm^{-1}}$$ (also called parallel $$v_3$$ band) has alternate lines missing, which are those having odd J. $$\ce{CO2}$$ is a linear symmetric molecule ($$D_{\infty h}$$ point group) but nuclear spins on $$^{12}\ce{C}$$ and $$^{16}\ce{O}$$ are both zero and means that antisymmetric rotational lines are missing. The ground vibrational state is $$\Sigma^{+}_g$$ resulting in the odd levels being absent. I'm assuming you are measuring the rotationally well resolved band at $$\approx 2349 \pu{cm^{-1}}$$ and you are correct. The rotational band structure for the antisymmetric vibration centred at $$\approx 2349 \pu{cm^{-1}}$$ (also called parallel $$v_3$$ band) has alternate lines missing, which are those having odd J. $$\ce{CO2}$$ is a linear symmetric molecule ($$D_{\infty h}$$ point group) but nuclear spins on $$^{12}\ce{C}$$ and $$^{16}\ce{O}$$ are both zero and means that antisymmetric rotational lines are missing. The ground vibrational state is $$\Sigma^{+}_g$$ resulting in the odd levels being absent. The effect is due to the symmetry properties of the wavefunction, and what happens on exchange of coordinates. For fermions ( multiples of spin 1/2 , eg. electrons, protons,) the wavefunction must be antisymmetric but symmetric for Bosons (integer spin 0,1,2 etc) which is the case for $$\ce{CO2}$$. Look for 'statistical weight', 'linear molecules', and 'nuclear spin' in a good spectroscopy book, Herzberg 'Infra Red and Raman Spectra' for example. 1 answered Mar 18 '17 at 8:56 porphyrin 19.8k11 gold badge3535 silver badges6060 bronze badges I'm assuming you are measuring the rotationally well resolved band at $$\approx 2349 \pu{cm^{-1}}$$ and you are correct. The rotational band structure for the antisymmetric vibration centred at $$\approx 2349 \pu{cm^{-1}}$$ (also called parallel $$v_3$$ band) has alternate lines missing, which are those having odd J. $$\ce{CO2}$$ is a linear symmetric molecule ($$D_{\infty h}$$ point group) but nuclear spins on $$^{12}\ce{C}$$ and $$^{16}\ce{O}$$ are both zero and means that antisymmetric rotational lines are missing. The ground vibrational state is $$\Sigma^{+}_g$$ resulting in the odd levels being absent.