Question

Question #47c50

Answer 1

Here's how you can do that.

Explanation:

I'm assuming that you're not really familiar with how coordination compounds work...

You're dealing with a coordination compound, so right from the start, you should look for a complex ion.

Now, notice that you have

`color(red)([color(black)("Cr"("H"_2"O")_6)]color(black)("Cl"_color(purple)(3)))`

If the chemical formula for your coordination compound starts with a bracket, then you're dealing with a positively charged complex ion and a negatively charged ion or anion.

So, you know that the red brackets separate the complex ion from the anion. Remove the brackets to separate the two

`"Cr"("H"_2"O")_6^(color(blue)(?+)" "` and `" " color(purple)(3)"Cl"^color(darkorange)(?-)`

As you know, chlorine is located in group 17 of the Periodic Table, which implies that it forms `1-` anions. This means that you have

`color(darkorange)(?-) = 1-`

The overall negative charge coming from the `color(purple)(3)` chloride anions will be

`color(purple)(3) * (1-) = 3-`

Since coordination compounds are neutral, the overall positive charge coming from the cation must be balanced by the overall negative charge of the anion.

This means that you have

`color(blue)(?+) = 3+`

And so

`["Cr"("H"_2"O")_6]"Cl"_3 -> "Cr"("H"_2"O")_3^(3+) + 3"Cl"^(-)`

Therefore, you can say that every hexaaquachromium(III) chloride coordination compound contains

• one hexaaquachromium(III) complex ion, `1 xx "Cr"("H"_2"O")_6^(3+)`
• three chloride anions, `3 xx "Cl"^(-)`