Question #65200

1 Answer
May 13, 2016


Most likely, #F^-# will have a slightly smaller ionic radius than #N^(3-)#.


For ions derived from elements in different groups, a size comparison is meaningful only if the ions are isoelectronic.

From the periodic Table of the elements we see that N is two columns to the left of F, so their ground-state configuration would mean F is slightly larger.

Adding 3 electrons to N would increase its radius to slightly more than F, and probably just a bit more than the extra electron will add to the ionic radius of F- because they will not be bound as tightly to the nucleus.