Diffusion current
When a p-n junction is formed, a diffusion phenomena causes electrons from the n-doped region to diffuse to the p-doped region. At the same time (even if it's an abstraction) holes diffuse from the p-type region to the n-type one. The atoms that lose a carrier (electron or hole) become ions, which means that instead of being neutral, they have a positive or negative net charge. This happens because the ideal equilibrium would have the same concentration of mobile carriers equal all over the region.
Ohmic current
However, this diffusion causes the growth of a region, populated by ions, called depletion region, because all atoms have lost their carrier. These ions, as we said, are electrically charged, and cause an electric field directed from the n-region to the p-region, pushing carriers in the opposite way than diffusion. Therefore an equilibrium is reached in which the current (movement of carriers) caused by diffusion is perfectly balanced by the current caused by the electric field (ohmic current).
Effect of biasing
Applying a potential to the junction causes a perturbation on this equilibrium, making one of the currents dominant on the other. Reverse biasing the junction causes the ohmic current to prevail, while forward biasing increases the diffusion current.
Now, the diffusion current is a much stronger phenomena, from which derives the exponential growth of the forward bias current with the bias voltage. Ohmic current, on the other side, is much weaker, and saturates quite soon (neglecting avalanche effect) because the width of the depletion region (which determines the resistivity) is proportional to the reverse bias voltage.
Why is the positive charge not distributed in the entire N type region (as there are free electrons for conduction in the N type region)?
It actually is. The diagrams are shown that way because we're primarily interested in the electric field produced by that charge, which is strongest near the junction, since that's where it's nearest the corresponding negative charge.
Why are the charges fixed at junction despite the presence of holes and free electrons?
Once the diffusion has stabilized, there are no free carriers in the central region, which becomes insulating. This insulating barrier becomes the dielectric between the two conductive regions, effectively forming a capacitor.
Best Answer
A hole is not the result of a charge-neutral atom losing an electron. A hole is created when an "acceptor" atom is located in a silicon crystal but that atom does not have as many electrons available for bonding as do the silicon atoms. Silicon atoms bond by sharing a pair of electrons, each atom contributes one electron to the bond. The acceptor atom leaves one bond unfilled, and it's that unfilled bond that constitutes the hole. Note that although this hole exists the atoms are completely charge-neutral. It's easy for a wandering electron to get stuck in the hole, and when that happens the acceptor atom actually has one more electron than it normally would...thus it has become a negative ion. The captured electron came from somewhere...some atom that was also previously charge-neutral...so that atom has become a positive ion. Since we have an immobile negative ion and an immobile positive ion, an electric field exists between them.
As holes are filled with wayward electrons the e-field increases in strength until it prevents any more movement of electrons. At this point the depletion region has been created. This region is depleted of free (mobile) charge carriers but the impurity (non-silicon) atoms are ionized.
I've mentioned silicon but the same thing can be done with some other materials, such as germanium and gallium-arsenide.