What is the bond energy of Carbon-Deuterium bond?
Answers
Fair warning - I wanted to explain this without having to invoke quantum mechanics, but I ultimately couldn't. I'll post an additional response afterwards using an analogy which I have that might work better.
The amount of energy required to break a chemical bond is a function of the zero-point energy of the system, in this case, a C-H or C-D bond. In their ground state, any quantum mechanical system retains some finite amount of energy - its translational kinetic energy may be zero, but it will still have a certain amount of vibrational kinetic energy. This is true, even at absolute zero.
In quantum mechanical systems, energy is quantized and one can use the Schrodinger equation to calculate what those are for various potentials. In this case, if you model the C-X bond as a quantum oscillator, you can calculate the allowable energy levels and you get that
This means, that there are n energy levels and they all come in half integer multiples of Planck's constant and the frequency. Some more math leads you to this relationship for the frequency
where k is the spring constant (we modeled this as a simple harmonic oscillator, remember) and μ is the reduced mass of the system. I calculated the reduced mass of both systems and for C-H it was 0.923 u and for C-D it was 1.71 u. So, you can see from this that, an increase in the reduced mass of the oscillator leads to a decrease in the frequency of oscillation. And, since the energy level of the oscillator is proportional to the frequency, we can see that ground state energy level for C-D is lower than it is for C-H.
Since the ground state of the oscillator is lower, more energy has to be put in to separate it, which results in a larger bond energy for C-D than for C-H.