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(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.

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    $\begingroup$ 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 $\omega_e x_e$. $\endgroup$
    – Paul
    Apr 1, 2021 at 18:55
  • $\begingroup$ Marina, I have edited the answer a little bit, the previous answer was slightly wrong btw. $\endgroup$
    – S R Maiti
    Apr 2, 2021 at 9:39

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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.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – andselisk
    Apr 2, 2021 at 9:37

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