An $n$-type semiconductor is made up of impurity atoms that could donate electrons to the lattice, which causes conduction. Hence they are called donor atoms. Since they donate electrons, these donor atoms become positively ionized. On the other hand, in a $p$- type semiconductor, the impurity atoms are called acceptor atoms, as they create a deficiency in the sea of electrons and hence appears a s a vacant site left by a electron. These entities are called holes and are assumed to be positive. Hence the acceptor atoms are negatively ionized as these vacant sites have a high probability of capturing electrons nearby.
First of all, don't think that a $pn$ junction is not formed by taking individually a $p$-type and an $n$ type and combining them. They are developed on a single semiconductor crystal. Now, the answer to your question.
Can someone let me know why wouldn't the electron just beside the $(+)$ move towards it and neutralize everything......
The concentration of electrons is higher than that of holes in the $n$-type semiconductor and reverse for the holes. In other words, electrons are majority carriers in an $n$-type semiconductor and holes are majority carriers in a $p$-type semiconductor. So, at the junction or the interface between the two types, there is a concentration gradient of charge carriers and hence they would diffuse across the junction and recombine, thereby depleting the electrons in the $n$- region and hols in the $p$-region. Hence the name depletion region.
Then why all charges just don't recombine?
Because the flow of carriers across the junction is stopped sometime. The ions are immobile and you can see positive ions in the $n$-side and negative ions in the $p$-side as discussed above.
Suppose our crystal is free from all defects. Then the electrons from the $n$ side has to encounter with the periodic negative potential of the ions in the $p$-side. It's like a potential well problem. Only those electrons having energy sufficient to overcome that potential barrier will diffuse across the junction and recombine. So all charge carriers will not diffuse across the junction and neutralize the polarity. To overcome this potential barrier we need to supply energy to the electrons, which is what we do by forward biasing the $pn$ junction. You need to overcome the electric field in the junction, by an external field.
So, the next question would be, then why the electrons could neutralize the ions in the $n$-side. Of course, the electrons are mobile.
Consider an intrinsic semiconductor. At a given temperature, electron-hole pairs are formed in it and it is called thermal excitation. When you dope it and construct an $n$-type material, then the so called $5^{th}$ valence electron is loosely bound to the parent atom and can be activated at very low temperatures than that required for thermal activation of intrinsic carriers. Hence at ordinary room temperatures, obviously, the $5^{th}$ electron will be detached from the atom and the atom becomes ionized and the electron contributes to conduction.
So, for the ion to re-capture these electrons, and neutralize, it is no possible at ordinary temperatures. At ordinary temperatures, the electron energy level is very higher than the bound energy level of the valence electron. So such a possibility is no there. But that doesn't mean the electrons will not interact with the ions. The electrons have a probability of scattering by the positive potentials. A similar argument can be made at the $p$-side also, which is left as a homework.
