STANDARD Physics Degrees of Freedom Assertion Reason Question

By law of equipartition of energy, the energy for each degree of freedom in thermal equilibrium is $\frac{1}{2} k_B T$. Each quadratic term form in the total energy expression of a molecule is to be counted as a degree of freedom. Thus each vibrational mode gives 2 degrees of freedom i.e., kinetic and potential energy modes, corresponding to the energy $2 \left( \frac{1}{2} k_B T \right) = k_B T$.