At the end of electronic configuration, we were taught that, electron orbitals are most stable when they are either fully filled or half filled. E.g., the final valence configuration of chromium is $\ce{(4s)^1 (3d)^5}$ and not $\ce{(4s)^2 (3d)^4}$. But the final electronic configuration of chlorine is $\ce{(3s)^2 (3p)^5}$ and not $\ce{(3s)^1 (3p)^6}$.
Why doesn't chlorine follow the half filled or full filled theory?
Having a half-filled or filled subshell is stabilising, as you say. But for Cl, the difference in energy between 3s and 3p is greater than the additional stabilisation from filling the 3p subshell. 3d and 4s are closer in energy, so for Cr it is favourable to promote a 4s electron and have a 4s1 configuration.
In the case of 3$d$ and 4$s$ energies, there is an "anomaly" in the ordering, being the 3$d$ orbital energy higher than that of 4$s$. However, you are comparing the configuration of chromium (with 3$d$ and 4$s$ orbitals) with chlorine. The valence shell of the latter contains 3$s$ and 3$p$ orbitals, and the energy of 3$s$ orbitals are lower than that of 3$p$.
The reason we see these Aufbau's principle exceptions in transition metals is because the $4s$ and $3d$ orbitals are very similar in energy.
In chromium, having a $4s^2$ $3d^4$ configuration results in electron-electron repulsion due to the two electrons in the $4s$ orbital. For this reason, chromium adopts a $4s^1$ $3d^5$ configuration, in which each electron occupies its own orbital. Recall that Hund's rule essentially states that you fill each orbital once before going back with the second electron in regards to orbitals of the same energy (in the same subshell). In the case of chromium, we are dealing with orbitals that are almost in the energy ($3d$ and $4s$) so you can essentially view this as a special case of Hund's rule that extends to orbitals of nearly the same energy.
For what it's worth, the other notable exception is copper, which adopts a $4s^1$ $3d^9$ configuration over $4s^2$ $3d^8$. For reasons that go beyond the scope of this answer, before you hit the transition metal elements ($Z \le 21$), $3d$ is higher in energy than $4s$, so we fill $4s$ before $3d$. But beyond this point, $3d$ becomes lower in energy than $4s$, so you lose $4s$ electrons first when we're talking about ionization. Knowing this, it is energetically more favorable to have $4s^1$ $3d^9$ because the higher energy orbital only has one unpaired electrons and the lower energy orbitals have paired ones.
The important thing to note from these two examples is that there is no special added stability form merely having a half full electron shell (i.e., something is not magically more stable because of having a half filled subshell). Adopting this form of configuration has consequences that are perfectly explainable to students of intro chemistry—do not brush it off as mere "magic."
