Without any prior exposure to the cohomology of groups, one might naively proceed by replacing a group by a free resolution.

For instance, let's take $G = \mathbb{Z}^2$, and resolve:    
$$ 0 \to \mathbb{Z} \overset{1 \mapsto xyx^{-1}y^{-1}}\longrightarrow \mathrm{F}_2 \to \mathbb{Z}^2 \to 0. $$
This mirrors exactly the topology of the torus $\mathrm{T} = \mathbf{B} \mathbb{Z}^2$, as obtained by attaching two loops, and then a disc along the commutator.     
The same works for $G = \mathbb{Z}^3$, as there is a relation between the relations, given by the Jacobi–Witt–Hall identity (this is not strictly true, as I need to allow some twisting to conjugate the relations to get a relation between the relations). The cohomology of the three torus, and its cell structure, is mirrored by a sequence looking somewhat like   
$$ 0 \to \mathbb{Z} \to \mathrm{F}_3 \to \mathrm{F}_3 \to 0. $$

A different example would be to take $G = \mathbb{Z}/2\mathbb{Z}$:
$$ 0 \to \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0.$$
This is not so good... following this on the nose would get us to construct $\mathbb{RP}^2$ instead of $\mathbf{B} \mathbb{Z}/2\mathbb{Z} = \mathbb{RP}^\infty$. The problem is that the relation $x^2 = 1$ is not imposed once, but infinitely many times, once for each pair $(x^{2n}=1, x^{2n+1}=1)$. To account for this, one has to extend the sequence by another map $ \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z}$, and a similar argument gives us that we should in fact be looking at    

$$ \cdots \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \cdots \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0.$$
This of course mirrors the cell structure of $\mathbf{B} \mathbb{Z}/2\mathbb{Z} = \mathbb{RP}^\infty$.    

With this in mind, suppose we now try to define group cohomology while staying in the category of groups (rather than using $\mathbb{Z}[G]$-modules for instance). We have a bar construction    

$$ \cdots F^3 G \underset{\underset{\longrightarrow}{\longleftarrow}}{\overset{\overset{\longrightarrow}{\longleftarrow}}{\longrightarrow}} F^2 G \underset{\longrightarrow}{\overset{\longrightarrow}{\longleftarrow}} F G \longrightarrow G \to 0,$$

where $F$ is the functor (comonad) sending a group to the free group on the underlying set. This corresponds to looking at all possible elements of $G$ as generators, then imposing every equation $g_1 \ldots g_j = k$ as a relation, etc. Mimicking the previous, we only remember the outer evaluation maps and get a sequence

$$ \cdots \to F^3 G \to F^2 G \to FG \to G \to 0. $$

In the case of $G = \mathbb{Z}/2\mathbb{Z}$, after cleaning up by removing extraneous terms corresponding to the identity (for instance replacing $FG = \langle e_0, e_1 \rangle$ by $\langle e_1 \rangle$), we get precisely the sequence
$$ \cdots \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \cdots \overset{\times 2}{\longrightarrow} \mathbb{Z} \overset{\times 2}{\longrightarrow} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0.$$

**Question:** Is the sequence    
$$ \cdots \to F^3 G \to F^2 G \to FG \to 0 $$
always the "correct" replacement for $G$? Can it be used to compute the group cohomology of $G$?    
(Perhaps it would be wiser to apply a functor valued in some abelian category to the simplicial group obtained by the bar construction, and then use Dold–Kan to turn it into a genuine chain complex... but the sequence above for $\mathbb{RP}^\infty$ really does feel like the correct object to consider.)

More generally, I would like to understand why we have run into the problems we have (e.g. those with $ G = \mathbb{Z}/2\mathbb{Z}$), whereas doing the analogous thing for the comonad associated to the adjunction between $\mathbb{Z}[G]$-modules and abelian groups does compute the group cohomology with no hiccups.    
I understand the need to distinguish between groups and $\infty$-groups (where imposing the same relation multiple times is different from imposing it once), but somehow that problem seems to have vanished when considering $\mathbb{Z}[G]$-modules...?   
(Aside: do these objects under consideration bear any relation to crossed complexes?)