Why do iron(II) ions and chromium have different electronic configurations?

What I know:

1. Empty 3d orbitals are higher in energy than empty 4s orbitals
2. Aufbau Principle (electrons always go into an empty orbital with the lowest energy)
3. Partially/half/fully filled 3d orbital are lower in energy than 4s orbitals
4. $\ce{Cr}$ is an exception: $\ce{[Ar] 3d^5 4s^1}$. It breaks the "rule" because the extra stability from half-filled 3d orbitals is favoured.
5. From (1) and (2) => After the $\ce{[Ar]}$ core, electrons will fill 4s before occupying 3d
6. When forming ions, electrons of the highest energy are removed first.
7. Iron(II) ions exist as $\ce{[Ar] 3d^6}$ in reality.

What I understand:

I understand why Fe is $\ce{[Ar] 3d^6 4s^2}$. [from (5)]
I understand why Cr is $\ce{[Ar] 3d^5 4s^1}$.

From (4), (5) and (3), I believe for $\ce{Cr}$ the 4s is filled before 3d too, i.e. it is $\ce{[Ar] 3d^3 4s^2}$ before the last electron is added. After the last electron is added to form $\ce{[Ar] 3d^4 4s^2}$, a 4s electron immediately moves to the 3d orbital which is of lower energy.

From (6) and (3), for transition metals, 4s electrons are removed first. i.e., I understand why iron(II) ions may be $\ce{[Ar] 3d^6}$

What I don't understand:

I know $\ce{[Ar] 3d^6}$ is correct for iron(II), but I don't understand why?

Why doesn't iron ions adjust itself as Cr does to achieve the extra stability? Is it because, in this case, it needs to excite an electron to an orbital of higher energy, whereas in Cr, the 4s electron just moves down to an orbital of lower energy?

P.S.

• I've read What is the electronic configuration of Fe(II) ion? and I am not satisfied with the answers in that thread.

• I've read Do ions also form special electron configurations? (just as Copper and Chromium do), and it only tells me "it is possible for ions to have special electronic configurations too!"