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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:

enter image description here

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:

  1. 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?

  2. 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.

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    $\begingroup$ When the textbook says "hardly any molecular ions will be generated in their vibrational ground state", I think that implicitly assumes the electron ionised is not non-bonding (the details are too long to fit into a comment, but loss of a non-bonding electron does not significantly change the wavefunction, so the Franck-Condon principle means that the most probable transition is to the same vibrational state, as that has greatest overlap with the original wavefunction). I'd be interested as to others' thoughts. $\endgroup$ – atbm Dec 31 '19 at 16:30
  • $\begingroup$ @atbm Hmm, what does the electron ionised being non-bonding have to do with this? $\endgroup$ – The Pointer Dec 31 '19 at 16:31
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    $\begingroup$ The FC principle means that the probability of transition is dependent on the overlap between the initial and final vibrational wavefunctions. If the electron removed is non-bonding, then there is little change in the wavefunction so overlap between the initial and final vibrational ground states is almost 1 (and, because the eigenfunctions of the Hamiltonian are orthonormal, overlap with the other wavefunctions is almost 0). So the ion would be most likely to be generated in the ground state. $\endgroup$ – atbm Dec 31 '19 at 16:38
  • $\begingroup$ @atbm Ahh, interesting. There seems to be something implicit in this that I'm not sure I'm understanding correctly: Are you saying that the neutral molecule M (the original molecule) is in a vibrationally excited state? Because, based on what you just said, it seems that this is implied, since, as the author stated, most of the molecular ions generated will end up in their vibrationally excited state? $\endgroup$ – The Pointer Dec 31 '19 at 16:43
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    $\begingroup$ I'm assuming the neutral molecule is in the ground state - if it's in, say, the 1st excited state the molecule would be generated in the 1st excited state. This assumes the electron removed is non-bonding - which would mean the internuclear distance doesn't change, hence the link to your question. If a bonding electron is lost, the bond length will increase as in your example. $\endgroup$ – atbm Dec 31 '19 at 16:45
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  1. 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.

  1. 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.

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  • $\begingroup$ Thanks for the excellent answer! I’m going to make some comments as I read through your answer. $\endgroup$ – The Pointer Jan 1 at 7:46
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    $\begingroup$ The author then seems to emphasise the bond lengths becoming longer: "No matter where the electron has formally been taken from, ionization tends to cause weakening of the bonding within the ion as compared to the precursor neutral. Weaker bonding means longer bond lengths on the average and this goes with a higher tendency toward dissociation of a bond." Do you think that this is true most of the time? $\endgroup$ – The Pointer Jan 1 at 8:19
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    $\begingroup$ Regarding the seemingly contradictory quote, your reading of it is fairly accurate. The equilibrium bond length of M•+ is (typically) larger than the equilibrium bond length of M. However, because of the FC principle, during the physical process of ionisation, the bond length does not change: so immediately following ionisation, the bond length of M•+ is far away from its equilibrium, which corresponds to an excited vibrational state. If M•+ were to relax back to its ground state, then the bond would indeed lengthen. However, sometimes it does other stuff too, like fragment. $\endgroup$ – orthocresol Jan 1 at 12:05
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    $\begingroup$ Regarding your question: it is more complicated than that, I believe. Even bearing the FC principle in mind, ionisation will typically proceed to give a mixture of vibrational states. So, it doesn’t actually mean that every molecule of M•+ will be formed in an excited state; it only means that it’s most likely to be formed in an excited state. More formally, it depends on the Franck–Condon factors which atbm has mentioned. The probability of forming state x is related to the FC factor for state x. These in turn depend on the overlap between the vibrational wavefunctions of interest. $\endgroup$ – orthocresol Jan 1 at 12:11
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    $\begingroup$ So, with an instantaneous ionisation process, you get what is essentially a mixture of vibrational states of M•+. This mixture is, by virtue of the FC principle, biased towards excited states, but it does contain some ground state. On the other hand, an adiabatic process usually indicates one which is carefully controlled such that the wavefunction does not change during the course of the process. In this case, that refers to ground state (gs) to gs. And this will occur (almost) without fail, such that (nearly) all of the molecules go from gs to gs: no longer a mixture. $\endgroup$ – orthocresol Jan 1 at 12:16
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The reasons for the change in internuclear separation and the imporance of the Franck-Condon factors, as has been clearly pointed out in answers and comments. The FC factors determine the strengths of transitions from $M$ to $M^{+.}$ and to clarify this figure below shows a simple calculation based on harmonic oscillator wavefunctions of the effect of displacement on the spectrum produced.

FC factors

The FC factors with vibrational wavefunctions $\varphi$ are defined as $\displaystyle F=\left |\int \varphi_n^*(R-R_n)\varphi_m(R-R_m) \right|^2$ where $R$ is bond extension and $R_n,R_m$ equilibrium values. In the case of harmonic oscillator transitions from the ground state vibrational level on the lower (ground) electronic state to a higher electronic state is $\displaystyle F_{0,n}=\frac{x^n}{2^nn!}e^{-X/2}$ where the displacement parameter $X=\mu\omega(R_n-R_m)^2/\hbar$ with reduced mass $\mu$ and frequency $\omega$.

The log10 of the FC factors for different $X$ are shown in the figure. Small displacements lead to low quantum numbers in the excited state and large displacements to high quantum numbers.

Fc factor 2

The use of the word 'adiabatic' is strange here as the 0-0 transition will occur with a small or large probability, as determined by the FC factor; if small it might be 1000 times less intense than the most intense transition. Usually the term refers to situations when there is a interaction between potential energy surfaces that causes them to move apart (avoid one another or repel one another) in the adiabatic picture. This is the usual case that we draw for an energy profile leading to a transition state 'at the top of the hill'. The wavefunction then changes smoothly from that of reactant to product. This is not usually what happens in an optical transition as the interaction between states is minute in the absence of the photon. Thus to cause a adiabatic change, as shown in your picture, the two state have to be caused to mix at the position of the minimum of the $M^{+.}$ state which presumably means starting with energy high up in the ground state and using a photon to reach the minimum in the higher state but this is a lot like a normal transition. Alternatively a short pulse could be used, say 10 femtosec long, to couple the two states , and while they are in superposition ( about 10 fs before they de-phase) somehow to cause the transition to occur with a second pulse.

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