Why does carbon molecules and water have broad bandwidth of infrared absorption? I am kind of curious to know how a single element molecules can absorb many infrared frequencies not every single frequency in band but most of them.how is this possible because a molecule absorbs infrared photon if the photon frequency is in resonance with molecule frequency and all the transitions can be from v=1 TO V=v+1 and the frequency of each molecule should be unique because they don't have exactly same environment.please tell me how is this possible
 A: If you model the vibrational modes of molecules classically, in terms of masses and springs, you get the impression that molecules should vibrate at a few precisely defined frequencies.  However, according to QM, every spectral line is broadened by three phenomena: (1) inherent linewidth reciprocal to the lifetime of the initial state, in accord with the uncertainty principle, (2) collisions that shorten its effective lifetime before dephasing or deexcitation, and (3) doppler shifts from thermal motion.  
Even more importantly, there are QM selection rules that demand rotational fine structure, splitting each vibrational line into many resolvable lines.  There will also be purely rotational transitions that do not involve changes of vibrational state.  Dipole transitions require the angular momentum L to change by one unit, up or down.  It is easy to describe the fine structure for a diatomic or rigid linear molecule, but harder for water.  The rotational energy levels are ${{E}_{L}}=L(L+1)\ {{\hbar }^{2}}/2I$, where I denotes the moment of inertia.  If the basic vibrational frequency is $\nu $, traditionally stated as spatial frequency in units of cycles per cm, the observed lines will be at $\nu \pm \Delta {{E}_{L}}/hc$ where $\Delta E=(2L+1)\ {{\hbar }^{2}}/2I$.   The relative strength of the lines depends on degeneracy and populations of the initial states, hence $(2L+1)\exp (-{{E}_{L}}/kT)$ for an up-transition.  
Taking carbon dioxide as an example, its IR-active vibrational modes (those that make for an oscillating electric dipole moment) are its two degenerate bending modes at 667 and its asymmetric stretch mode at 2349 $c{{m}^{-1}}$.  The symmetric stretch mode at 1388 is IR-inactive for lack of an electric dipole moment.  It has no purely rotational dipole transitions because it is nonpolar.  Since no molecule is an ideal simple harmonic oscillator, you should not expect evenly spaced vibrational levels, and there will be sum, difference, and overtone lines as well.  Given its moment of inertia, the spacing of its rotational fine structure lines is about 0.9 $c{{m}^{-1}}$, but dozens of lines to either side of the vibrational frequency will be significant at atmospheric temperatures.  
A: $CO_2$ and $H_2O$ are light molecules so the have high vibration frequencies. They also have polar bonds or even a dipole so they couple well to the electromagnetic field. The other molecules in air, $O_2$ and $N_2$, lack polar bonds. Other molecules such as $NH_3% also have these properties but do not occur in air, unless something is drastically wrong. 
Each of these properties are not unique. It is the combination of these factors that make $CO_2$ and $H_2O$ prominent IR absorbers.
