# Comparing the zero point vibrational energies of DCl and HCl

(a) Compare the zero point energy (i.e. $$E_0)$$ in joules for $$\ce{H^{35}Cl}$$ and $$\ce{D^{35}Cl}$$ (assuming both are anharmonic oscillators).

(b) Explain the effect this will this have on their chemistry (you may wish to use a diagram to help here).

I have calculated a value for the vibrational frequency and have a value of the anharmonicity constant for $$\ce{H^{35}Cl}$$ (but not the anharmonicity constant for $$\ce{D^{35}Cl}).$$ I am unsure on how to proceed with these as all I know is an equation linking vibrational energy to the vibrational frequency and anharmonicity constant. I would be so grateful for any insight into this.

• Hint: look at the values of H$_2$ and D$_2$ in the X state and see if you can find the mass scaling of $\omega_e$ and $\omega_e x_e$.
– Paul
Apr 1, 2021 at 18:55
• Marina, I have edited the answer a little bit, the previous answer was slightly wrong btw. Apr 2, 2021 at 9:39

I don't know which equations you are using, but this is the one I was taught: $$E(v)=hc\tilde{\nu}\left( v+\frac{1}{2}\right) -hc\tilde{\nu}\chi\left(v+\frac{1}{2}\right)^2.\tag{1}$$ Here, $$\tilde{\nu}$$ is the wavenumber of the vibration; $$v$$ is the quantum number that represents the vibrational levels $$(v= 0, 1, 2,\ldots);$$ $$\chi$$ is the anharmonicity constant: $$\chi=\frac{\tilde{\nu}}{4\,D_\mathrm{e}},\tag{2}$$ where $$D_\mathrm{e}$$ is the depth of the lowest point of the potential well (not the depth of the $$v = 0$$ vibrational level!).
The zero-point energy is the energy when $$v=0$$ i.e. the molecule is in the lowest vibrational state. So you have to substitute that in the energy equation.
Now, we can assume that the well depth $$(D_\mathrm{e})$$ of the $$\ce{H-Cl}$$ and $$\ce{D-Cl}$$ bonds are same.
The only difference between $$\ce{H-Cl}$$ and $$\ce{D-Cl}$$ would be in the value of the vibrational wavenumber $$(\tilde{\nu}),$$ which you can get from the vibrational frequencies that you already have, as you wrote in the question. Once you substitute that in the equation, you will get your answer.