Depletion layer thinner in p-side I found this diagram on the Internet. In all such diagrams in class, we have drawn the depletion layer to be equally wide in both p- and n-side of the pn junction diodes. This is counterintuitive because even if the depletion layer is not equally wide on both sides, it should be wider in the p-side because anions are generally bigger than the cations. Why then is the diagram drawn like this?

 A: 
In all such diagrams in class, we have drawn the depletion layer to be equally wide in both p- and n-side of the pn junction diodes.

This isn't correct. The width of the depletion regions will be related to the dopant densities on the two sides. If the p-side is more heavily doped, it will have a narrower depletion region. If the p-side is more lightly doped, it will have a wider depletion region.
In the example you found online, they have evidently assumed the p-side is more heavily doped than the n-side, whether they've stated this assumption or not. You can see this is consistent in their plot of Q vs x.

it should be wider in the p-side because anions are generally bigger than the cations.

The space charge in the depletion region is determined by the number of ionized dopant sites per unit volume. Since we measure dopant density on a volume basis (i.e. in units of cm-3), the charge density in the depletion region will follow directly. If we doped with 1016cm-3 acceptors, the charge density in in the depletion region will be -q x 1016cm-3. If we doped with 1017cm-3 donors, the charge density in in the depletion region will be +q x 1017cm-3.
