# Question 7eca9

Nov 22, 2017

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.
${N}^{3 -} > {O}^{2 -} > {F}^{-} , > N {a}^{+} , > M {g}^{2 +} > N e$.

#### 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:
${N}^{3 -} M {g}^{2 +} N {a}^{+} , {F}^{-} , {O}^{2 -} N e$.

"Normal" sequence (decreasing):
First Row: $N e , N , O , F$
Second Row: $N a , M g$

Ionic changes in each row:
Ne .("no change"), N -> N^(3-) "larger", O -> O^(2-) "larger", F -> F^- ,"larger"#
$N a \to N {a}^{+} \text{smaller",Mg -> Mg^(2+) "smaller} ,$.

So, relatively, ${N}^{3 -}$ should be larger than ${O}^{2 -}$, which will still be larger than ${F}^{-}$ , $N {a}^{+ 1}$ remains larger than $M {g}^{2 +}$ and $N e$ 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:
$N \to {N}^{3 -}$ "expands" just a bit more than ${O}^{2 -}$
$O \to {O}^{2 -}$ "expands" just a bit more than ${F}^{-} ,$
$F \to {F}^{-} ,$ slight expansion.

$N a \to N {a}^{+}$"contracts" just a bit
$M g \to M {g}^{2 +}$ "contracts" a bit more than $N {a}^{+}$

Thus, the final order of the relative size of the ions listed is:
${N}^{3 -} > {O}^{2 -} > {F}^{-} , > N {a}^{+} > M {g}^{2 +} > N e$.

Actual values (picometer):
${N}^{3 -} = 150$
${O}^{2 -} = 140$
${F}^{-} , = 133$
$N {a}^{+} = 102$
$M {g}^{2 +} = 70$
$N e = 62$
http://abulafia.mt.ic.ac.uk/shannon/ptable.php