Why do neutral, unbonded atoms shrink in size as they approach having 8 electrons in their valence shells? A good example is elements 3 through 10 in this table, that is, lithium (1 valence electron) through neon (8 valence electrons).
For posterity, below is my earlier incorrect version of my question. It's incorrect because it based on a table that used the atomic radii of bonded atoms to describe their sizes. That is a very different measurement from the sizes of neutral atoms, since for bonded atoms you are tossing in additional electrons via the bonds. Very small atomic radii for halogens are easy to explain in that case, since the incomplete shells must fight very hard over the shared electrons to complete their own octets, making their bonding radii anomalously short. This question had an interesting back-and-forth, and I thank the two contributors both who both caught my chart interpretation error and answered the real and interesting question that lurked beneath my incorrect one.
Why are nearly-complete atomic shells, such as the seven valence electrons of fluorine, so compact in comparison to the bloated-but-very-stable situation created by adding one more electron to make neon?
I thought I could answer that in terms of Pauli exclusion and wound up completely baffled. Can anyone help on this one? Why does "almost complete" equal "very small" in a charge-neutral atom?