# Bond frequency vs Bond Strength in Kinetic Isotope Effect

I am confused by the following: in Ansyln's Modern Physical Organic Chemistry, a higher vibrational frequency defines a stronger bond due to the deeper/sharper potential well (page 76). However, the kinetic isotope effect is explained as being due to the lower frequency of C-D bonds in comparison to C-H, resulting in a lower zero point energy and subsequently a higher bond dissociation energy (page 423). These two concepts seem at odds with one another, and I was wondering what factor I am not considering.

• The first statement is false. Compare for instance the dissociation energies and vibrational frequencies of $\ce{H_2}$ and $\ce{N_2}$. The vibrational frequency is determined both by the steepness of the potential and the reduced mass of the atoms. Note that the potential for different isotopologues is the same within the Born-Oppenheimer approximation.
– Paul
Jan 3, 2022 at 21:55
• Since a higher vibrational frequency leads to a higher ZPE, how can I rationalize the fact that a stronger bond (C-C vs C=C for instance) has a higher ZPE, yet the ZPE of C-D is lower than C-H (and the bond is still stronger)? Jan 3, 2022 at 23:05
• Increasing the the force constant for a given pair of masses increases the frequency. [That's what the first statement refers to.] Increasing the mass for a given force constant decreases the frequency. [That's the kinetic isotope effect you describe—or, more precisely, there you are significantly increasing one of the masses while leaving the force constant about the same.] See equation here (pictured at right) relating masses, force constant, and frequency: www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/… Jan 3, 2022 at 23:42
• Sure, but since the energy of the vibrational level is proportional to the vibrational frequency, why can one state that the lower zero point energy of C-D vs C-H leads to greater energy needed to break that bond while at the same time a stronger bond (C=C vs C-C) has a higher zero point energy yet still needs more energy to break its bond? Jan 4, 2022 at 0:03
• The experimentally relevant dissociation energy is $D_0$, that is $D_0=D_e-\tfrac{1}{2}\omega_0\hbar$. A lower ZPE thus gives a larger dissociation energy.
– Paul
Jan 4, 2022 at 4:40