Electron Ionization and the Franck-Condon Diagram: vibrationally excited and vibrationally ground states I am currently studying Mass Spectrometry: A Textbook, third edition, by Jürgen H. Gross. On page 41, the author says the following:


  As explained by the Franck-Condon diagram, hardly any molecular ions will be generated in their vibrational ground state. Instead, the majority of the ions created by EI is vibrationally excited and many of them are well above the dissociation energy level, the source of this energy being the 70 eV electrons. 


On page 37, the author presented the following Franck-Condon diagram:


  


In this diagram, the energy is plotted on the ordinate, and the bond length is plotted on the abscissa.

My understanding is that the vertical transitions represent the electron ionization of the neutral $M$. 

Following this diagram, on page 38 the author said the following:


  The counterpart of the vertical ionization is a process where ionization of the neutral in its vibrational ground state would yield the radical ion also in its vibrational ground state, i.e., the $(0 \leftarrow 0)$ transition. This is termed adiabatic ionization and should be represented by a diagonal line in the diagram. The difference $IE_{\text{vert}} - IE_{\text{as}}$ can lead to errors in ionization energies in the order of 0.1 - 0.7 eV.


So, If i'm correctly interpreting the description below the Franck-Condon diagram, the only way for the molecular ions to be generated in their vibrationally ground state is if the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state? 

If my interpretation so far is correct, then two related questions immediately follow: 


Why is it unusual (as it seems to be implicitly implied) that the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state? 
How does a situation arise where the bond lengths between such molecular ions differ so significantly that some end up in a vibrationally excited state and others in a vibrationally ground state? (If I had to give an educated guess, I would say that this is just due to random or non-primary factors.)


I'm studying this from a non-chemistry background, so I'd appreciate it if people would please take the time to explain this in a more novice-friendly way.
Why is it unusual (as it seems to be implicitly implied) that the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state?
  


It's not so much about whether the bond length in $\ce{M^.+}$ is larger or smaller than that in $\ce{M}$; it's more about whether it's different. If you think pictorially about the vibration of a diatomic molecule or a spring, it makes sense that as you increase the energy (i.e. go to higher vibrational states), the amplitude of the vibration increases. The only way for a molecule to access bond lengths that are substantially different from its equilibrium bond length is therefore to be in a vibrational excited state.

Ionisation of $\ce{M}$ effectively always takes place from the vibrational ground state, so the initial bond length will always be very close to the equilibrium bond length of $\ce{M}$. If ionisation is almost instantaneous, such that the bond length cannot change during the ionisation, then $\ce{M^.+}$ is going to be generated with a bond length that is far away from its own equilibrium. This corresponds to an excited vibrational state. Whether it's larger or smaller than equilibrium doesn't matter; it just matters that it's different.

Of course, to say that this is the case presupposes that the equilibrium bond lengths of $\ce{M}$ and $\ce{M^.+}$ are substantially different. But this appears to be the assumption that the author makes. For more insight into this, @atbm's comments are spot on: the change in the equilibrium bond length will depend on the bonding character of the electron being removed. If a bonding electron is removed, then the bonds holding atoms together are weakened, and consequently $\ce{M^.+}$ will have a larger equilibrium bond length than $\ce{M}$; similar considerations hold for nonbonding and antibonding electrons.


  
  How does a situation arise where the bond lengths between such molecular ions differ so significantly that some end up in a vibrationally excited state and others in a vibrationally ground state? (If I had to give an educated guess, I would say that this is just due to random or non-primary factors.)
  


I suspect that the FC principle is no longer relevant to the adiabatic ionisation process. Adiabatic processes are associated with "slow" transitions: see e.g. https://en.wikipedia.org/wiki/Adiabatic_theorem, whereas the FC principle is to do with "instantaneous" processes where there is no possibility of the bond length changing during the ionisation.

This is not my area of expertise, but I will go out on a limb and say that the adiabatic ionisation process is largely hypothetical (at least in the present context). Wikipedia writes that 


  However, due to experimental limitations, the adiabatic ionization energy is often difficult to determine, whereas the vertical detachment energy is easily identifiable and measurable.


I am not sure what "error" the author is talking about, but my interpretation is that the adiabatic ionisation energy is the "true" or "actual" ionisation energy of the molecule; however, because physical ionisation does not take place in an adiabatic fashion, the measured ionisation energy differs from the "true" ionisation energy by ca. 0.1–0.7 eV.

So I would not interpret this as being two different physical processes that actually occur in the mass spectrometer; rather, it seems to be more of a thought experiment to me. Of course, I am open to being corrected on this.