# 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?

• Most of the time, shorter bonds mean more effective overlapping between the orbitals, which makes stronger bonds, although alot depends on the element and size of orbitals. Apr 27 '17 at 2:42
• Short answer is a definite no. I would have to dig around a bit to find where I read that, though... Apr 27 '17 at 3:03
• Is your question only meant to be for a given pair of elements? i.e. do you mean, "if the A-B bond in compound 1 is shorter than the A-B bond in compound 2, does that mean the first bond is stronger"? Or do you mean to include comparisons between different bonds A-B and C-D? Apr 27 '17 at 10:03
• @orthocresol I would say my question mainly relates to comparing A-B to A-C, where C could be the same as B or could be different. Comparing entirely different bonds (A-B to C-D) presumably adds in other factors that affect the overall length. Apr 27 '17 at 13:43
• Anytime I see "always" in a question about chemistry I get skeptical. It seems such propositions always lead to exceptions and/or "clarifications" of the question.
– MaxW
Aug 16 '17 at 6:44

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.

• It is appealing to try to separate force constant and dissociation energy but in the Morse potential they are linked as $\alpha = \omega_e\sqrt{2\pi^2c\mu/(Dh)}$ Apr 27 '17 at 8:41
• You are right of course. I've made an edit to make the assumption of a constant $D_e$ more explicit.
– Paul
Apr 27 '17 at 10:00
• @Paul I like this answer. The one thing I'm a little confused on is the bond length's relation with $k_e$. While the well could be more localized, if it is centered around the same point (i.e. reaches a minimum at the same $r_e$) wouldn't the bond length be the same? Apr 27 '17 at 21:19
• @Tyberius The minimum of the potential does not necessarily coincides with the expectation value for $r$ of the different vibrational states. In general the wavefunctions have more intensity toward the shallower part of the potential at larger values of $r$. The shift from $r_e$ for each vibrational state $|v\rangle$ is $\langle v|r-r_e|v\rangle$ and depends on $k_e$. However, this is not an easy expression. See this paper: inaesp.org/PublicJG/gallas_morse1980.pdf
– Paul
Apr 28 '17 at 8:08

I would like to add a few cases featuring A-A type bonds, where bond strength and bond length orders follow quite an anomaly.

• $\ce{N-N}$ and $\ce{P-P}$ bonds. The lone pairs present on nitrogen atoms repel each other and the large size of phosphorus atoms facilitates for charge distribution. This makes the $\ce{N-N}$ bond weaker than the $\ce{P-P}$ bond. This also explains why nitrogen portrays a lesser tendency for catenation. To sum up, both bond-length and bond-strength follow the same order, i.e. $\ce{N-N \lt P-P}$.
• Both strength and length follow the same order in these two cases as well:
• $\ce{O-O \lt S-S}$
• $\ce{F-F \lt Cl-Cl}$
• Lastly, I would like to highlight a third case, $\ce{C-C}$ and $\ce{Si-Si}$ bonds. On comparing the strengths, $\ce{C-C}$ bond takes the lead, and that's why we're all (atleast I am) a fan of organic chemistry! As for the length, $\ce{Si-Si}$ bond is longer.

I know this is kind of an old question, but I thought this would make for a valuable contribution as there is a canonical relationship between bond length and bond force constant, which is a measure of bond strength.

This relationship is something which very early chemists studied quite a bit because as more experimental data was gathered, they obviously wanted a way to make sense of it. The simplest rule which was found empirically, and is quite a useful qualitative tool, is known as Badger's rule [1], after Richard M. Badger who discovered the relationship in 1933.

The rule states:

$$k_e(r_e-d_{ij})^3=1.86\cdot10^5$$

with $k_e$ given in dynes/cm (old unit for force) and $r_e$ in angstrom.

There are two important points here. First, such an empirical relationship actually does exist and it is quite accurate. This can be plotted linearly with a simple rearrangement, and this is done in Badger's paper and in many other papers if you dig around.

The real important point here, however, is the constant $d_{ij}$. The $i$ and $j$ represent the row of the periodic table. Hydrogen is counted as it's own row (I guess helium would be included too). This is very powerful because it means that within each row, and within each combination of rows, the force constant and equilibrium bond length are characterized by the use of a single parameter. Badger even says that the whole reason for finding this relationship is so that the strength of bonds can be easily characterized using $k_e$. Note that this term $r_e-d_{ij}$ is described as an effective distance of approach for the nuclei.

You can also imagine how useful this would be when one couldn't simply calculate the bond lengths to within a tenth of an angstrom without thinking about it. If one had the vibrational frequency, you could estimate the force constant and hence determine the approximate equilibrium bond length.

Strictly speaking this equation was derived from diatomic data, but it can also be applied successfully to polyatomic molecules[2], though it is kind of inconsistent for transition metals from what I understand.

Some researchers revisited Badger's rule as recently as 2000 with the hopes of improving the rule as well as testing how it performs with charged species and other unusual diatomics[3]. Basically what they found is that without significantly complicating the form of the law, the rule cannot be significantly improved. The average error in determining $k_e$ from their set of over 100 diatomics was about $17\%$. Which isn't that bad considering they had $\ce{He_2^{2+}}$ and things like this in the set.

Basically, this rule relating bond length to bond strength, as measured by force constant, is very robust for diatomics (it probably never goes the other way), and is a very good rule of thumb in polyatomic systems. The whole question of isotopes actually could be described by a law like this if corrections were made either for anharmonicity or to use the vibrationally-averaged coordinate rather than $r_e$.

[1] Badger, R. M. (1934). A relation between internuclear distances and bond force constants. The Journal of Chemical Physics, 2(3), 128-131.

[2] Badger, R. M. (1935). The relation between the internuclear distances and force constants of molecules and its application to polyatomic molecules. The Journal of Chemical Physics, 3(11), 710-714.

[3] Cioslowski, J., Liu, G., & Castro, R. A. M. (2000). Badger's rule revisited. Chemical Physics Letters, 331(5), 497-501.