Question #47c50

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"^(-)#