Let $F$ be a finite field with many elements, say more than 7 for example, and $X$ be the corresponding projective plane $\mathbb{P}^2(F)$.

For a set of points in $X$, if any three of them are noncollinear (so they are automatically different points), we say this set of points are noncollinear.

Define the following configuration complex:
$C_n$ is the free abelian group generated by all such noncollinear $n$-tuples $(x_0,x_1,\dots,x_n)\in X^{n+1}$ (i.e. points $x_0,x_1,\dots,x_n$ are noncolinear); the differentials are defined as
$$ d(x_0,\dots,x_n)=\sum_{i=0}^n(-1)^i(x_0,\dots,\hat{x}_i,\dots,x_n). $$
So we get a chain complex 
$$ C_*:\dots\mathop{\to}\limits^{d} C_1\mathop{\to}\limits^{d} C_0\mathop{\to}\limits^{\epsilon}\mathbb{Z}\to0, $$
where $\epsilon$ is the augmentation map.

My question is whether this chain complex is almost acyclic, i.e. its homology is always zero except for one term (the last nonzero term).

Or if $|F|$ is sufficiently large can we deduce that $H_n(C_*)=0$ for $n\ll|F|$?

Here is something I know about the last nonzero term: if $|F|=q$ is odd, then the longest arc of $X$ is a conic going through $q+1$ points. So now $C_q\neq0, C_{q+1}=0$. If $|F|=q$ is even, then the longest arc of $X$ is a $(q+2)$-arc. So now $C_{q+1}\neq0, C_{q+2}=0$.