Do polyatomic ions act as individuals during an ionic reaction.?

1 Answer
Dec 4, 2015

I assume you mean to ask whether polyatomic ions necessarily react or don't react with each other in an ionic reaction. And well, that would conditionally go either way.

If you just mean whether they split into individual monatomic ions... no, that never happens to my knowledge. Polyatomic ions exist as polyatomic ions. Their bonds don't randomly break apart because that would waste too much energy.

Here's an example where there is no reaction happening:

#"LiC"_2"H"_3"O"_2(aq) + "NaF"(aq) -> "NaC"_2"H"_3"O"_2(aq) + "LiF"(aq)#

In this reaction, nothing truly reacts and every ion in solution is known as a spectator ion (meaning that it does not react at all and merely dissociates in solution). So then it works out like this for the net ionic reaction:

#cancel("Li"^(+)(aq)) + cancel("C"_2"H"_3"O"_2^(-)(aq)) + cancel("Na"^(+)(aq)) + cancel("F"^(-)(aq)) ->#
#cancel("Na"^(+)(aq)) + cancel("C"_2"H"_3"O"_2^(-)(aq)) + cancel("Li"^(+)(aq)) + cancel("F"^(-)(aq))#

But when you form a precipitate according to some solubility "rules" you may have learned in General Chemistry, it is going to make two ions associate with each other strongly enough to form a solid in solution and not exist as individual ions.

Here's an example of that:

#"Ba"("C"_2"H"_3"O"_2)_2(aq) + "Na"_2"SO"_4(aq) -> "BaSO"_4(s) + 2"NaC"_2"H"_3"O"_2(aq)#

In this case, we see that barium easily forms a precipitate with sulfate at room temperature in water. So, the net ionic reaction is, after canceling out spectator ions:

#"Ba"^(2+)(aq) + cancel(2"C"_2"H"_3"O"_2^(-)(aq)) + cancel(2"Na"^(+)(aq)) + "SO"_4^(2-)(aq) ->#
#"BaSO"_4(s) + cancel(2"Na"^(+)(aq)) + cancel(2"C"_2"H"_3"O"_2^(-)(aq))#

#color(blue)("Ba"^(2+)(aq) + "SO"_4^(2-)(aq) -> "BaSO"_4(s))#

And in this case, one polyatomic ion (acetate) doesn't participate in the reaction, while the other (sulfate) does.