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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.

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source | link

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.