Question

The translational degrees of freedom (\(f_t\)) and rotational degrees of freedom (\(f_r\)) of \( \text{CH}_4 \) molecule are:

• A. \( f_t = 2 \, \text{and} \, f_r = 2 \)
• B. \( f_t = 3 \, \text{and} \, f_r = 3 \)
• C. \( f_t = 3 \, \text{and} \, f_r = 2 \)
• D. \( f_t = 2 \, \text{and} \, f_r = 3 \)

Answer 1

The Correct Option is B

For the polyatomic, non-linear molecule CH₄, the degrees of freedom are as follows:

Total Degrees of Freedom = Translational Degrees of Freedom + Rotational Degrees of Freedom = 3 + 3 = 6

Methane (CH₄) is classified as a polyatomic molecule with a non-linear geometry.

For non-linear polyatomic molecules:

The translational degrees of freedom (\(f_{t}\)) are 3, accounting for movement along the x, y, and z axes.

The rotational degrees of freedom (\(f_{r}\)) are also 3, enabling rotation around three independent, perpendicular axes.

Therefore, for CH₄:

\[ f_{t} = 3 \, \text{and} \, f_{r} = 3. \]