We will prove the general case: In a $ n \times n$ board filled with integers $1$ to $n^2$, then there exist 2 neighboring squares whose difference is at least $n$.
Explanation of main idea: Find $n$ distinct pairs of neighbors where one term is $\leq K$ and the other neighboring term is $\geq K+1$. Then, the smallest of this terms is at most $K-n+1$, and it's neighbour is at least $K+1$, so their difference is at least $n$.
For each row, let $r_i$ be the largest number in that row.
For each column, let $c_i$ be the largest number in that column.
Let $N$ be the smallest of these $2n$ values. WLOG $N=r_I$, and cell $(I,J)$ is equal to $N$. Then, every other integer in this row is $\leq N-1$.
For every column not equal to $J$, we can find adjacent cells where one is $\leq N-1$ (starting with row $I$) and one is $\geq N+1$ (otherwise, this contradicts the minimality of $N$).
In column $J$, either $N$ is adjacent to a number $<N$, or to a number $>N$.
Case 1:$N$ is adjacent to a number $< N$
We now have $n$ distinct pairs of neighbors (since they are in different columns) where one term is $\leq N-1$ and the other is $\geq N$. Let the smallest of the terms be $S$, then $S \leq N-n$. Hence, the difference between $S$ and its neighbor is at least $n$.
Case 2:$N$ is adjacent to a number $> N$.
We now have $n$ distinct pairs of neighbors (since they are in different columns) where one term is $\leq N$ and the other is $\geq N+1$. Let the smallest of the terms be $S$, then $S \leq N-n+1$. Hence, the difference between $S$ and its neighbor is at least $n$.
Within a $2\times 2$ square, the king can move from any directly to eevery other field, hence we need at least four colours. If we assign colour $(x\bmod 2, y\bmod 2)$ to field $(x,y)$, ew see that $4$ colours also suffice.
For the rook, all fields in one column are mutually adajacent and hence must have different colours, thus requiring at least $8$ colours. But eight colours also suffice if you assign $x+y\bmod 8$ to $(x,y)$.
Similarly, the bishop has adjacent fields along a diagonal, thus also requiring eight colours. But eight colours also suffice by assigning colour $x$ to $(x,y)$.
For the knight, the original checssboard colouring witnesses that the chromatic numbre is 42$.
Best Answer
I'll just give some hints that will allow you to easily deduce the formula that user14111 gave in a comment. Call any pair of adjacent squares a domino, which can be horizontal or vertical. Call two dominos neighbours if their union is a $2\times2$ square. Call a domino monochromatic for a colouring if its squares have the same colour.