Final answer:
The ratio of the fundamental frequencies for ethylene and deuterated ethylene depends on the square root of the inverse ratio of their reduced masses. Since deuterium is twice as heavy as hydrogen, deuterated ethylene has a higher reduced mass and thus a lower vibrational frequency compared to normal ethylene.
Step-by-step explanation:
To calculate the ratio of the fundamental frequencies for ethylene and deuterated ethylene, we use the formula for the frequency of vibration of a diatomic molecule, which can be approximated as a harmonic oscillator:
ν = (1/2π) √(k/μ)
where ν is the frequency, k is the force constant for the C-C bond, and μ is the reduced mass of the molecule. The reduced mass (μ) is calculated using the masses of the atoms in the molecule:
μ = (m1*m2)/(m1+m2)
For ethylene (C2H4), the reduced mass when considering the C-C stretch is calculated using the masses of carbon (C) atoms. However, for deuterated ethylene (C2D4), the deuterium atoms (D) are twice as massive as hydrogen (H) atoms. Assuming the force constant k is the same for both molecules because the C-C bond strength isn't altered by isotopic substitution, the ratio of the two frequencies relies solely on the difference in reduced mass.
Since the mass of deuterium is double that of hydrogen, the reduced mass of deuterated ethylene would be greater than that of ethylene. This results in a lower frequency for deuterated ethylene compared to regular ethylene. The exact ratio can be represented as the square root of the inverse ratio of the reduced masses:
Ratio = √(μ_ethylene / μ_deuterated_ethylene)
The result is that the fundamental frequency of vibration for deuterated ethylene is lower than that for normal ethylene.