Question

Question #7eca9

Answer 1

You may work it out from the "size" of each atom's nucleus and the electrons around it, or look it up in a table.
`N^(3-) > O^(2-) > F^-, > Na^+, > Mg^(2+) > Ne`.

Explanation:

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