Final answer:
The ionic radius of a positive ion (cation) is smaller than the element's atomic radius because removing electrons decreases electron-electron repulsions and increases effective nuclear charge, both of which draw the electron cloud closer to the nucleus.
Step-by-step explanation:
The question pertains to the change in ionic radius as compared to the atomic radius when an atom becomes a positive ion (cation). A cation is smaller than its parent neutral atom because when one or more electrons are removed, the electron-electron repulsions in the valence shell decrease, and the effective nuclear charge (Zeff) felt by the remaining electrons increases. This increased Zeff causes the electron cloud to be drawn closer to the nucleus, thereby decreasing the ionic radius.
As an example, for the element vanadium (V), when it loses one electron to become V²+, its ionic radius is 79 pm, and it becomes even smaller, 64 pm, as V³+ when it loses another electron. Down the groups of the periodic table, cations with the same charge have increasing radii due to the principal quantum number (n) increasing, which results in a larger distance between the nucleus and the valence shell of the ion.