The vibrational partition function for a molecule which can be described as a simple harmonic oscillator with fundamental frequency $$\nu $$ is given by
A. $$\exp \left( { - \frac{{h\nu }}{{{K_B}T}}} \right)$$
B. $${\left[ {1 - \exp \left( { - \frac{{h\nu }}{{{K_B}T}}} \right)} \right]^{ - 1}}$$
C. $$\exp \left( { - \frac{{h\nu }}{{{K_B}T}}} \right){\left[ {1 - \exp \left( { - \frac{{h\nu }}{{{K_B}T}}} \right)} \right]^{ - 1}}$$
D. $$\exp \left( { - \frac{{h\nu }}{{2{K_B}T}}} \right){\left[ {1 - \exp \left( { - \frac{{h\nu }}{{{K_B}T}}} \right)} \right]^{ - 1}}$$
Answer: Option B
Solution (By Examveda Team)
The vibrational partition function describes how molecules are distributed among different vibrational energy levels at thermal equilibrium.For a molecule that behaves as a quantum mechanical simple harmonic oscillator, the vibrational energy levels are given by Ev = (v + 1/2)hν, where v = 0, 1, 2, ....
When the zero-point energy is treated separately, the vibrational partition function becomes:
qvib = 1 / [1 − exp(−hν/kBT)]
This expression is equivalent to:
qvib = [1 − exp(−hν/kBT)]−1
Therefore, Option B is the correct expression for the vibrational partition function.
Why the other options are incorrect:
Option A: It contains only the exponential term and is not the complete partition function.
Option C: It includes an unnecessary extra exponential factor, so it does not represent the standard vibrational partition function.
Option D: It contains the factor exp(−hν/2kBT), which accounts for the zero-point energy. This form is obtained only when the zero-point energy is explicitly included in the partition function and is not the standard expression generally used in engineering chemistry and statistical thermodynamics.

Please the solutions
Please the solutions
Option b kase aya
Solution
Option b ka solution dn