# Calculating the zero-point energy for the C2N2 molecule

I shall calculate the zero-point energy for the $$\ce{C2N2}$$ molecule in the $$^{1}\Sigma_{g}^{+}$$ electron ground state with the following vibrational wavenumbers (symmetries in parentheses) which were found in IR and Raman spectra:

• $$v_1 = 2330~\mathrm{cm}^{-1}$$ ($$\sigma_g$$)
• $$v_2 = 854~\mathrm{cm}^{-1}$$ ($$\sigma_g$$)
• $$v_3 = 2158~\mathrm{cm}^{-1}$$ ($$\sigma_u$$)
• $$v_4 = 507~\mathrm{cm}^{-1}$$ ($$\pi_g$$)
• $$v_5 = 233~\mathrm{cm}^{-1}$$ ($$\pi_u$$)

I have no clue about how to do it - maybe somebody can help me.

• Do you know what a zero-point energy is? – orthocresol Jan 29 at 16:35

The adiabatic electronic energy only contains the energy of the electrons. You are missing the energy that the molecule has in form of the vibrational groundstate of the nuclei. This energy can be calculated trivially in the harmonic approximation. Within the harmonic approximation we can define normal modes and each mode corresponds to a 1D harmonic oscillator with its respective normal mode frequency. The eigenenergy of a harmonic oscillator is given by $$E_{n} = \hbar \omega(n+1/2), \ n\geq0$$ Each normal mode of your molecule contributes thus such an energy when you include vibrational energy. The zero point energy refers to the system where each mode is in the groundstate i.e. $$n_i=0$$, where $$i$$ is an index for the normalmode. The zero-point energy correction is then simply $$E_{ZP} = \sum_i^{3N-5} \hbar\omega_i 1/2$$ where $$N$$ is the number of atoms and $$3N-5$$ is the number of normal modes in a linear molecule. In a nonlinear molecule you would have $$3N-6$$ normal modes. $$n$$ is the quantum number of the oscillator state.
In your case, you have some degenerate eigenvalues due to symmetry. The $$\pi_u$$ and $$\pi_g$$ modes are actually sets of two degenerate modes, which means that their frequency would appear twice in the sum that i denoted. This also works out when we consider your example, we have the following number of modes $$3\cdot 4 -5=7$$. Since we have to sets of doubly degenerate modes we see only 5 distinct frequencies although there are 7 normal modes.
$$E_{ZP} = 1/2(\hbar \omega_1+\hbar \omega_2+\hbar\omega_3 + 2\hbar \omega_4 + 2\hbar \omega_5)$$ with $$\omega_i = 2\pi \nu_i.$$
It's sometimes a bit vague what people mean with zero point energy. Some people use the term for the correction that i defined above, but technically it should include also the electronic adiabatic energy. Which means that you have to add the electronic state energy to the $$E_{ZP}$$ value.