# Question 47c50

May 28, 2017

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

$\text{Cr"("H"_2"O")_6^(color(blue)(?+)" }$ 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 $\textcolor{p u r p \le}{3}$ chloride anions will be

$\textcolor{p u r p \le}{3} \cdot \left(1 -\right) = 3 -$

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, $3 \times {\text{Cl}}^{-}$