isotope - Are stronger bonds always shorter?

This question relates to a few earlier questions I have seen on the site (particularly this one about the bond strength of heavier isotopes). Does a bond necessarily have to be shorter to be stronger?

To my understanding, length and strength are definitely correlated, but I don't think it is necessarily a one to one correspondence. The examples I brought up in the comments of ron's answer in the linked question related to data tables that seemed to show no difference in bond length for deuterium substituted bonds despite significant changes in energy. (HCl NIST, DCl, data table)

To expand on my question, if length and strength are always directly related:

1. How do we explain experimental and theoretical cases where the length doesn't differ with bond energy?


2. If there is some direct relationship between length and strength, what is that relationship?

For the sake of not being too general, I'm looking for an answer focused on comparing bonds of the form $\ce{A-B}$ and $\ce{A-C}$ (similar to the examples I asked about in the linked question). So for a more specific case related to the linked question, does substituting a protium for a deuterium always make the bond shorter in addition to being stronger?

Answer

This is an interesting question and depending on how you define bond strength the answer is different. Let us for simplicity consider only diatomic molecules and let us assume that the electronic potential between the two atoms is well described by a Morse potential

\begin{align} V(r) &=D_e \left( 1 - \mathrm{e}^{-a(r-r_e)} \right)^2,& \text{with } a &=\sqrt{\frac{k_e}{2D_e}}. \end{align}

Here $D_e$ is the depth of the potential (at the minimum position $r=r_e$) and $k_e$ is the (harmonic) force constant. The potential depth is related to the dissociation energy $D_0$ by

\begin{align} D_0 &= D_e - \frac{1}{2}\omega_ehc,& \text{with } \omega_e &=\sqrt{\frac{k_e}{\mu}}, \end{align}

where $\omega_e$ is the harmonic wavenumber, $\mu$ is the reduced mass and $h$ and $c$ have their usual meaning.

We can define the strength of the bond by the magnitude of the dissociation energy $D_0$ or by the "spring constant" $k_e$ of the bond. In your example with deuterium you implicitly assumed the former definition. In the Born-Oppenheimer Approximation the potential (i.e. the Morse potential) does not depend on the mass of the atoms (they are assumed to be infinitely heavy) and the potential is the same for $\ce{H2}$ and $\ce{D2}$. However, because deuterium is heavier than hydrogen, the harmonic frequency is lower and therefore the dissociation energy is larger ($\ce{D2}$ has a smaller zero point energy). Using this definition, the bond length is not easily related to the bond strength (as the reduced mass of the system plays a role as well).

As stated earlier, we can also use $k_e$ as a measure for the bond strength. The larger the value of $k_e$, the steeper the harmonic part of the potential well and the more localized the nuclear wavefunctions will be. In other words, if we take $k_e$ as a measure of bond strength, then stronger bonds are indeed shorter.

As stated by @porphyrin, the separation between $k_e$ and $D_e$ is not very strict (see also the formula's above) and the explanation above implicitly assumed a constant $D_e$, just as we assumed constant $k_e$ for the different isotopologues.