Question #47c50
1 Answer
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
#color(darkorange)(?-) = 1-#
The overall negative charge coming from the
#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"^(-)#