I hear there might be a countable ill-founded model of ZFC, i.e. one in which there is an infinite decreasing sequence $\alpha_0 \ni \alpha_1 \ni \alpha_2 \cdots$. This is possible when there is no function in the model which corresponds to said sequence.
Is it possible that all such $\alpha_i$ are ordinals of the model?
On
The notion of the von Neumann rank is internal. That is, $\sf ZF$ proves that there is a definable function $r$ such that $r(x)=\bigcup\{r(y)\mid y\in x\}$ for all $x$. We can show that the range of this function is exactly the class of ordinals.
Now, if we have an ill-founded model $M$, with a decreasing sequence of elements $(x_n)_{n<\omega}$ (the sequence is not in the model, of course), we can apply this $r$ function pointwise to the sequence and obtain $(r(x_n))_{n<\omega}$ which is a sequence of ordinals of $M$, and by induction it is easy to see that it is indeed a decreasing sequence of ordinals.
In fact, every ill-founded model $M$ of $\mathsf{ZFC}$ contains an infinite descending sequence of ordinals. If $s_0\ni^M s_1\ni^M s_2\ni^M ...$ is a descending sequence in $M$, consider the sequence of ranks of the $s_i$s: letting $\alpha_i=rank(s_i)$, we have $\alpha_0\ni^M \alpha_1\ni^M \alpha_2\ni^M ...$
Put another way, we can tell if a model of $\mathsf{ZFC}$ is ill-founded just by looking at its ordinals.