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The force constant of iodine in the first excited state is $k=41.78$ N/m, and in the ground state it is $170$ N/m.

I know that the formula for calculating $k$ is:

$$ k= (\frac{ΔE_{vib} *2π}{h})^2 µ$$

What causes this difference?

Is it that the energy difference for the ground state is greater then that of the excited state and so the force constant is also greater?

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    $\begingroup$ Likely due to the description of a harmonic oscillator not being as accurate as a Morse (or similar) potential, but you could be looking at something attributable to the Franck-Condon effect if the vibrational transition is accompanied by an electronic one. Where are your force constant data from? $\endgroup$ – Todd Minehardt Feb 20 at 19:04
  • $\begingroup$ @ToddMinehardt The excited state force constant was determined from experimental data from a lab we carried out ourselves, and the ground state force constant was given to use for comparison. $\endgroup$ – J.Se Feb 20 at 19:06
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    $\begingroup$ A bond in the excited state is typically weaker (and longer) than in the ground state, especially in diatomic molecules. $\endgroup$ – TAR86 Feb 20 at 19:12
  • $\begingroup$ @TAR86 If it is weaker, does that mean that the vibrational energy difference is smaller? $\endgroup$ – J.Se Feb 20 at 19:15
  • $\begingroup$ Rearranging the equation you provided should show this. $\endgroup$ – TAR86 Feb 20 at 19:18
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At the risk of providing a somewhat circular answer, in the harmonic approximation of an interatomic potential, the force constant is a measure of the force of attraction between the atoms (in a classical description they would obey Hooke's Law $|\vec{F}|=-k(x-x_\circ$)). The answer to the question is therefore that the attractive interatomic force in the electronically excited state of $\ce{I2}$ is weaker than in the ground state.

The following figure from Ref. 1 illustrates the potentials in the ground and excited states:

enter image description here

The potential in the excited (upper) state is broader. In the Morse functional approximation, the potential has the form $$V(r)=D_e\left(1-e^{-\beta(r-r_\circ)}\right)^2$$

where the width of the potential is proportional to $\beta$ and $$\beta = \left(\frac{k}{2D_e}\right)^{1/2}$$

A smaller force constant results in a smaller value of $\beta$ and a broader potential well. This fits nicely with the results noted in the OP and the above figure (the equations are also outlined in the accompanying text). The energy accompanying a vibrational transition is proportional to the square root of the force constant and therefore inversely proportional to the width of the potential well. This is analogous to the behavior observed for a QM particle in a box, wherein the spacing between energy levels narrows as the box size increases. Working backward from vibrational transition energies leads to the conclusion that smaller transition energies are associated with greater oscillatory amplitudes (broader potentials) and smaller harmonic force constant estimates.

References

  1. Bayram, S.B. and Freamat, M.V., A Spectral Analysis of Laser Induced Fluorescence of Iodine. arXiv eprint 1507.02600, 2015.
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