# For a given element do heavier isotopes form stronger bonds than lighter isotopes?

I know that $\ce{C-D}$ and $\ce{C-T}$ bonds are stronger than $\ce{C-H}$ bonds, but is this generally true for isotopes of all elements? (By isotopes I mean stable ones with large half lives.)

Isovalent isotopes will have the same force constant. However the different masses of the isotope will affect the position of the vibrational state in its potential well. You can rationalise the difference in well depth briefly using the vibrational frequencies of a classical oscillator as the harmonic approximation to the asymmetric well for low lying states. If the stretching frequency is given by; $$\nu \propto \frac{1}{\sqrt{\mu}}, \ \ \ \ \ \ \ \ where \ \ \ \ \ \ \ \mu =\frac{m_1\cdot m_2}{m_1+m_2}$$ Then as the reduced mass increases the frequency decreases and the vibrational energy decreases $E=\hbar\omega$ with $\omega\cdot 2\pi=\nu$. We can see that if we were to double the reduced mass then the energy would decrease by root 2.

So we need to examine the reduced masses of some simple systems, lets say the diatomics you suggest. If we let the masses of $H=1$, $D=2$, and $T=3$ then we calculate the reduced masses to be $1/2$, $2/3$ and $3/4$. The reduced mass is increasing as we go along the series and hence we expect the successive molecules to have stationary states lower in the well.

In vibrational spectroscopy this is best seen by the decrease in zero point energies for HCl and DCl. It is often easiest to take the ratios of the frequencies $\omega _{HCl}/\omega _{DCl}$, where using $m_{Cl}=35$ and H and D as above we obtain 1.395. indicating that the deuterated molecule is deeper in its well.

Edit

I have been advised to explain zero point energy. In short it is a irremovable quantity of energy from a quantum oscillator. The classical oscillator has no parallel to this. It is a manifestation of the uncertainty principle in position and momentum. If you want the truth of where it comes from you will need to consult the physics SE, but I will do my best :)

The quantum oscillator has a Hamiltonian of the classical oscillator except just replace the classical position and momentum variables with their quantum operator analogues $p\mapsto \hat p$ and $x\mapsto \hat x$.

The classical Hamiltonian (or total energy) is learnt at school to be kinetic + potential, $$H=\frac{p^2}{2\mu}+\frac 12 kx^2$$ For the reduced two body system. It is a postulate of quantum mechanics that in the position representation (just a way of formulating things) that wherever you see a $p$ and an $x$ you replace them by $\hat p$ and $\hat x$ where, $$\hat p =-i\hbar \frac{\partial }{\partial x}, \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \hat x =\cdot x$$ Therefore with these changes in the above classical expression we generate the quantum Hamiltonian of the quantum harmonic oscillator, $$\hat H=\frac{\hbar^2}{2\mu}\frac{\partial ^2}{\partial x^2}+\frac 12 kx^2$$ Where $i=\sqrt -1$ so $i^2=-1$ and $--1=1$.

It is another principle of quantum mechanics that the Schrodinger equation is the equation to solve for quantum systems, just like in classical mechanics you use Newton's second law, $F=ma$, This equation has a bit of a different shape but in essence (decoupled from time) takes the following form. $$\hat H\psi =E\psi$$ Substituting the above expression for $\hat H$ in and doing a bit of maths (wayyyy to advanced for us here) we can solve for the energy $E$ in the Schrodinger equation, when we do this we get,

$$E_n=(n+\frac 12)\hbar \omega$$ What this tells us is that the vibrational levels are not continuous like the classical oscillator (above) but can only take discrete values. We see in the maths of the solution for $E$ the need for discrete integers $n$ which we call quantum numbers. These may take in this case values of $0,1,2,...$ etc etc. Each corresponding to the next adjacent energy solution.

Something interesting happens if we choose $n=0$ however. You would think that the lowest energy would be zero right? well no... quantum mechanics says we can never remove a little bit of energy from the system. (try substituting $n=0$ in the above expression and we see that the answer is non zero (1/2 $\hbar \omega$).

This is a zero point energy.

As promised here is the home-made graph of the molecular potential energy. Im sorry its not very good but honestly the best I could do with the available tools! Apologies.

The main points to follow are:

• At the base of the well the graph looks like a parabola, i.e. the harmonic oscillator approximation to the molecule works best at low lying states since the anharmonicity term grows with excitation (well increasing quantum number).
• Where the harmonic oscillator is valid you can see that the energy levels of are evenly spaced, each one corresponds to the quantum numbers $n$ as given above. You can calculate $\Delta E=E_{n+1}-E_{n}$ to get the difference in energy as $\frac 12\hbar \omega$. In this approximation the equilibrium bond length does not change with excitation. However we can see it does in the anharmonic case.
• To answer your question, the energy levels corresponding to the heavier isotope are the blue ones since they sit deeper in the well. So you could imagine the blue ones as C-D while the red ones would be C-H. (normally you would just be looking at one species per graph, so only one set of coloured lines.

• Maybe it'd be useful if you could expand slightly on zero point energy for a harmonic oscillator (add the formula?), and perhaps add one of those graphs with potential energy versus internuclear separation, with the vibrational levels shown for bonds involving different isotopes. – Nicolau Saker Neto May 11 '15 at 15:35
• Sure thing, are we allowed to use images off the web in answers? :) – AngusTheMan May 11 '15 at 15:41
• Why wouldn't we? If the image isn't generic, you can always cite the source! – Nicolau Saker Neto May 11 '15 at 15:50
• I strongly suggest not using images off the web due to copyright concerns; the major exceptions are images on Wikimedia Commons which are freely licenced. "Generic" images may well still be copyrightable. – J. LS May 11 '15 at 15:53
• @HritikNarayan I have updated the answer with the requested graph, I feel it goes slightly off topic but it is just background information to help you, the main part of the answer is above the edit section. Sorry the graph is so shoddy but I'm terrible at art! hope that helps :) – AngusTheMan May 11 '15 at 18:27

When comparing isotopes, the different nuclear masses control the bond length and bond strength.

The radius of the $n^\text{th}$ Bohr orbit is given by

$$r_{n} = {n^2\hbar^2\over Zk_\text{c} e^2 m_\text{e}}$$

where $Z$ is the atom’s atomic number, $k_\text{c}$ is Coulomb’s constant, $e$ is the electron charge, and $m_\text{e}$ is the mass of the electron. In this equation it is assumed that the electron orbits the nucleus and the nucleus remains stationary. Given the mass difference between the electron and nucleus, this is generally a reasonable assumption. However, in reality the nucleus does move too. It is relatively straightforward to remove this assumption and make the equation more accurate by replacing $m_\text{e}$ with the electron’s reduced mass, $\mu_\text{e}$

$$\mu_\text{e} = \frac{m_\text{e} \cdot m_\text{nucleus}}{m_\text{e} + m_\text{nucleus}}$$

Now the equation for the Bohr radius becomes

$$r_n = {n^2\hbar^2\over Zk_\text{c} e^2 \mu_\text{e}}$$

Since the reduced mass of an electron orbiting a heavy nucleus is always larger than the reduced mass of an electron orbiting a lighter nucleus

$$r_\text{heavy} < r_\text{light}$$

and consequently an electron will orbit closer to the nucleus of a heavier isotope than it will orbit the nucleus of a lighter isotope. In turn, this means that the heavy isotope will have shorter, stronger bonds.

• Does the heavy isotope necessarily have a shorter bond length? Taking HCl [webbook.nist.gov/cgi/… and DCl[webbook.nist.gov/cgi/inchi?ID=C7698057&Mask=1000#Diatomic] as the common examples, NIST has their bond lengths as nearly equal despite the large difference in reduced mass. In fact, DCl actually has a slightly longer equilibrium bond length. – Tyberius Apr 25 '17 at 20:16
• @Tyberius Both answers indicate that the compound with the heavy isotope will be lower in the potential well and hence have a shorter bond length. The NIST data is pretty old. – ron Apr 25 '17 at 20:28
• But lower in the potential well the equilibrium bond length shouldn't change significantly. Its effectively a harmonic potential when you are deep in the well, so shifting it slightly down shouldn't change anything. I also found another link with more recent values that also includes $\ce{H^37Cl}$ and $\ce{D^37Cl}$ and it too shows larger bond lengths for the cases with larger isotopes. – Tyberius Apr 25 '17 at 21:24
• @Tyberius Look at the primary isotope effect with carbon, $\ce{K_H > K_D}$, the $\ce{C-D}$ bond is shorter and stronger. This second reference is pushing 30+ years from when the data was actually collected. Also, the effect is just harder to see (smaller) in $\ce{HCl}$ because the effect of the reduced mass is much less than with lighter nuclei such as carbon. – ron Apr 25 '17 at 21:33
• Even 30 years ago, this level of quantum mechanics would have been known and if I were them I wouldn't bother publishing results that not only don't reflect what you say should be the observed trend, but actually show the exact opposite. I'm fine with the idea $\ce{C-D}$ is shorter, but does the heavier isotope have to be shorter? Alternatively, I know it should usually be stronger, but does stronger = shorter always? It seems like there are very few cases where the observed $\ce{X-D}$ bond length is measurably shorter. – Tyberius Apr 26 '17 at 2:21