The translational degrees of freedom (\(f_t\)) and rotational degrees of freedom (\(f_r\)) of \( \text{CH}_4 \) molecule are:
- \( f_t = 2 \, \text{and} \, f_r = 2 \)
- \( f_t = 3 \, \text{and} \, f_r = 3 \)
- \( f_t = 3 \, \text{and} \, f_r = 2 \)
- \( f_t = 2 \, \text{and} \, f_r = 3 \)
The Correct Option is B
Solution and Explanation
Since CH4 is polyatomic Non-Linear D.O.F of CH4:
T. DOF = 3 R DOF = 3
The molecule CH4 (methane) is a polyatomic molecule with a non-linear structure.
For non-linear polyatomic molecules:
The translational degrees of freedom (ft) are 3, corresponding to motion along the x, y, and z axes.
The rotational degrees of freedom (fr) are also 3, as the molecule can rotate about three mutually perpendicular axes.
Thus, for CH4, we have:
\[ f_{t} = 3 \, \text{and} \, f_{r} = 3. \]
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