Why does vibrational motion not have a significant impact on heat capacity? The bonds of a molecule are rigid - they can stretch and bend, which leads to additional degrees of freedom. However, according to (Young, Freedman, & Ford, 2011),

Resulting vibrations lead
  to additional degrees of freedom and additional energies. For most diatomic gases,
  however, vibrational motion does not contribute appreciably to heat capacity. The
  reason for this is a little subtle and involves some concepts of quantum mechanics.

Source: Young, H., Freedman, R., & Ford, L. (2011). University physics with modern physics (1st ed.). Boston, Mass.: Addison-Wesley.
 A: The explanation that says something like

because thermal energy is to low to excite transitions between vibrational states then it can not be stored in vibrational degrees of freedom and so it does not contribute to heat capacity

does not really explain anything and I think it is kind of wrong.
If the energy gap between the first two vibrational states($n=0$ and $n=1$) of a molecule is $w$ then we only need the total energy of the system to be greater than $w$ for the mode $n=1$ to have a non zero probability. This is because each non forbidden state for the whole system should be occupied equiprobably.
So even for low temperature the vibrational modes do store thermal energy. But they contribute very little to the heat capacity.
Proof:
The heat capacity per degree of freedom can derived using Maxwell–Boltzmann statistics: the probabilty $ p(n) $ for a small subsystem (or a molecule) to have the energy $E(n)$ is proportional to,
$$\begin{align}
p(n) &\propto g(n)\ e^{-E(n)/k_B T}
\end{align}$$
where $ k_B $ is Boltzmann’s constant and $ g(n) $ is the degeneracy ,i.e., the number of state for this subsystem with energy $E(n)$. It can be zero.
We can apply this formula to translational, rotational or vibrational modes for one or several molecules. Let's use it for a vibrational mode (=one degree of freedom) and we can take this mode to be an harmonic oscillator with energy  $ E(n) = nw $. Then the probability to be in state $ n $ is $ p(n)=N Exp(-nw/k_B T) $ With $ N $ the normalization constant $ N=1-Exp(-w/k_B T)$.
The energy stored in this mode is $ Q(w,T) = \sum_{n=0}^\infty p(n) n w = w/(Exp(w/k_BT)-1) $
$ Q(0,T)=kT $ but $ Q(w,T) $ decreases almost exponentially and approaches zero for large $ w $.  There's no  transition , it's all smooth
If instead we vary T then as expected $Q(w,T)$ becomes almost linear as $T$ increases.
What is interesting is to look at $ Q(w,T)/k_B T $ which is a function of $ k_B T/w $ with a transition between $ k_B T=w/10 $ and $ k_B T \sim w 2 $ as seen on this image

