Could chromatic number be easy to calculate when colouring is hard for some graph class? Similar question was asked before, but there was an error in it so it was left unanswered Graph class with easy chromatic number, but NP-hard coloring

Is there any infinite set of graphs $C$ such as:


There is a polynomial algorithm recognizing for every graph $G$ whether $G$ belongs to $C$
There is a polynomial time algorithm computing chromatic number $\chi(g)$ for every $g \in C$
Humankind does not know a polynomial time algorithm for computing proper colouring for $C$, or (which is better) there is a proof that such an algorithm (under reasonable assumptions) does not exist.
Yes. Since 3-COLORING is FNP-complete, for any problem in FNP we can construct a graph G that is 3-colorable if and only if the problem has a solution. So you can pick your favorite problem from TFNP\FP (if it's not empty!), like finding a factor of a composite number, or a second Hamiltonian-cycle in a 3-regular graph, and convert it into a graph.

In more detail with the example of finding a factor.
We will fix an algorithm that converts a composite number $N$ into a graph $G(N)$ such that from $G(N)$ you can decode the value of $N$.
Compositeness of $N$ is a necessary for $G(N)$ to be in our class.
Moreover, any proper $3$-coloring of $G(N)$ can be converted into a proper divisor $d$ of $N$.
This can be done, e.g., by writing up a CNF that describes whether a given $d$ encoded in binary as $x_1\ldots x_{n}$ divides a fixed $N$ (that also has $n$ digits in binary, but these are not variables).
From satisfiability of a CNF there is a standard reduction to 3-COLORABILITY such that you can easily convert a proper $3$-coloring back to an $x_i$ satisfying assignment.
Thus solving the coloring problem, you would find a divisor $d$.
Moreover, by adding some simple extra gadgets, you can also decide whether a graph is of this form or not.
Our class is formed by graphs created this way, which are thus all $3$-colorable.