Vibrational and rotational spectroscopy I don't understand why vibrational spectroscopy only has 1 intense absorption peak whereas the rotational spectroscopy has many separate peaks and the distance between the peaks is equal. $\Delta E\text{(vib)}$ is independent of quantum number so vibrational spectroscopy should instead have a graph of many separate peaks and the distance between which is the same. Is it due to the selection rule? But then both vibrational- and rotational spectroscopy share the same selection rule. $\Delta E\text{(rot)}$ depends on the quantum number $J$ which means that the rotational energy levels are not equally spaced in energy so its spectroscopy should not have equally spaced absorption peaks should it?
Summary of my question:


*

*Explanation for the the shape of vib- and rotational spectroscopy

*How does $\Delta E$ of vibration and rotation affect the number of peaks in the respective spectra, even though they share the same selection rule?

 A: There are several different issues conflated together here: selection rules, separation between energy levels, and energy level population (which you didn't mention).
For vibrational spectroscopy, in the approximation that a vibrational mode behaves like a quantum harmonic oscillator, the energy levels are equally spaced and the selection rule is $\Delta n=\pm 1$, where $n$ is the quantum number. The reason for this is explained here. So you expect to see (and do see) an absorption transition from $n=0$ to $n=1$. You might also expect to see a transition from $n=1$ to $n=2$ etc. This would occur at the same frequency since the gap between successive energy levels is the same. However, it relies on there being a thermal equilibrium population of molecules already in the $n=1$ state. For most molecules, at normal temperatures, the population of $n=1$ and higher levels (determined by the Boltzmann factor) is rather low. At elevated temperatures, you might see such transitions; also the frequency won't be exactly at the same frequency as the $n=0\rightarrow 1$ transition, because of anharmonicity effects.
For a linear rotor, the quantum levels are at $BJ(J+1)$ where $B$ is a constant and $J$ is the quantum number. These are not evenly spaced. The selection rule is $\Delta J=\pm 1$ (angular momentum conservation). So you expect to see (and do see) transitions between successive levels: $J=0\rightarrow 1$, $J=1\rightarrow 2$ etc. In this case, at normal temperatures, the spacing between rotational levels is typically small compared with the available thermal energy. So those higher states are populated, at least for $J$ not too high. The frequencies are not all the same, but the energy level spacings change linearly with $J$: 
$$\Delta E_{J\rightarrow J+1}=B(J+1)(J+2)-BJ(J+1)=2B(J+1).$$ So you see a spectrum with equally spaced lines for $J=0,1,2\ldots$ (in this rigid rotor approximation).

EDIT following OP comment.


*

*The first line, $J=0\rightarrow 1$ is at a frequency $\nu$ given by $h\nu=\Delta E=2B$ where $h$ is Planck's constant. 

*The second line, $J=1\rightarrow 2$ is at $h\nu=\Delta E=4B$.

*The third line, $J=2\rightarrow 3$ is at $h\nu=\Delta E=6B$.


And so on.
Hence the lines in the spectrum are equally spaced, $2B$ apart (in energy units) or $2B/h$ in frequency units. You can also see a diagram of this in the Linear Molecules section of the Rotational Spectroscopy Wikipedia page (reproduced below under the terms of the CC BY-SA 3.0 licence). The diagram shows the link between the energy levels and the lines in the spectrum (the only difference is that the transitions on the energy level diagram on that page are drawn for emission lines, $J\leftarrow J+1$, but exactly the same frequencies occur for the corresponding absorption lines $J\rightarrow J+1$).

