# Question #1ffde

Ok, lots to talk about here

First, let's look at the reactants for this equation.

$C a {\left(N {O}_{3}\right)}_{2} + K F \to$ ???

You need the subscript 2 after the nitrate ($N {O}_{3}$) because the charge of the calcium ion is +2 and a single nitrate ion is -1, so you need two nitrate ions to balance the positive charge of the calcium ion to produce a neutral compound.

You might like to review

video from: Noel Pauller

$C a {\left(N {O}_{3}\right)}_{2} + K F \to$ no chemical reaction

You might predict the following...
$C a {\left(N {O}_{3}\right)}_{2} + K F \to C a {F}_{2} + K N {O}_{3}$

Review reaction prediction here:

video from: Noel Pauller

In this case, both predicted products ($C a {F}_{2} \mathmr{and} K N {O}_{3}$) are soluble in water, so no reaction will take place.

No net ionic equation because no reaction takes place with these reactants.

Review net ionic equations here:

video from: Noel Pauller

Hope this helps!

Mar 16, 2015

Well, in your case, the nitrate ion receives a subscript of 2 because that is the charge of calcium cation.

Because calcium is in group 2 of the periodic table, it tends to lose two electrons from its outermost shell and form cations that have a (+2) charge.

I won't go into details about how polyatomic ions get their charges, but as I'm sure you know, the nitrate ion has a (-1) charge.

This means that you need two nitrate anions to balance the positive charge of the calcium cation and form the neutral calcium nitrate, or $C a {\left(N {O}_{3}\right)}_{2}$.

Not all polyatomic anions have a (-1) charge; for example, the sulfate ion has a (-2) charge, which means that when it combines with the calcium cation, you'll only need one sulfate to balance the positive charge and form calcium sulfate, or $C a S {O}_{4}$.

As a conclusion, you need to become familiar with the charges on the more common cations and anions, and with the charges on the more common polyatomic ions.

Once you develop a feel for what ion can have what charge, ionic compounds will be a walk in the park.