Question #7eca9

We know that the atoms generally decrease in size as they move left-to-right in a row of the Periodic Table of the Elements, and increase as they move down in a column, or Group. There is a sharp reversal going from the last element in a row to the first one of the next row.

Once the initial atomic sized are determined, we can observe that adding electrons (- charges) will increase the effective size, and removing them (+ charges) will decrease the effective size.

Given:
#N^(3-) Mg^(2+) Na^+, F^-, O^(2-) Ne#.

"Normal" sequence (decreasing):
First Row: #Ne, N, O, F#
Second Row: #Na, Mg#

Ionic changes in each row:
#Ne .("no change"), N -> N^(3-) "larger", O -> O^(2-) "larger", F -> F^- ,"larger"#
#Na -> Na^+ "smaller",Mg -> Mg^(2+) "smaller", #.

So, relatively, #N^(3-)# should be larger than #O^(2-)#, which will still be larger than #F^-# , #Na^(+1)# remains larger than #Mg^(2+)# and #Ne# is unchanged as the smallest.

Considering the effect of charge balance, more electrons relative to protons (-) expand the radius, and fewer electrons (+) decrease the effective radius.

Thus, "isoelectronic" atoms can be compared:
#N -> N^(3-)# "expands" just a bit more than #O^(2-) #
#O -> O^(2-)# "expands" just a bit more than #F^- , #
#F -> F^- ,# slight expansion.

#Na -> Na^+ #"contracts" just a bit
#Mg -> Mg^(2+)# "contracts" a bit more than #Na^+ #

Thus, the final order of the relative size of the ions listed is:
#N^(3-) > O^(2-) > F^-, > Na^+ > Mg^(2+) > Ne#.

Actual values (picometer):
#N^(3-) = 150#
#O^(2-) = 140#
#F^-, = 133#
#Na^+ = 102#
#Mg^(2+) = 70#
#Ne = 62#
http://abulafia.mt.ic.ac.uk/shannon/ptable.php

https://www.webelements.com/periodicity/ionic_radius/

http://chemguide.co.uk/atoms/properties/atradius.html