Question

A coordination compound \( \mathrm{CrCl_3 \cdot 6H_2O} \) is mixed with excess of \( \mathrm{AgNO_3} \) solution, three moles of AgCl are precipitated per mole of the compound. Write the structural formula of the coordination compound.

Hint:

Three moles of AgCl mean all three chloride ions are free. So all six water molecules are bonded to Cr inside the bracket.

Answer 1

Step 1: Understand what the precipitate tells us:
Silver nitrate reacts only with free chloride ions, which are outside the coordination sphere.
Chloride ions that are bonded to the metal inside the square brackets do not give a precipitate.
So the moles of AgCl formed equal the moles of ionisable chloride.

Step 2: Count the ionisable chloride:
One mole of the compound gives 3 moles of AgCl.
This means all three chloride ions are free and lie outside the coordination sphere.
No chloride is bonded to chromium.

Step 3: Fill the coordination sphere:
Chromium(III) shows a coordination number of 6 in this complex.
The six water molecules are neutral ligands, so all six sit around Cr.
The sphere is \( \mathrm{[Cr(H_2O)_6]^{3+}} \), which carries the 3+ charge of Cr(III).
Three \( \mathrm{Cl^-} \) ions balance this charge outside the bracket.
\[ \mathrm{CrCl_3 \cdot 6H_2O \;\rightarrow\; [Cr(H_2O)_6]Cl_3} \]
\[ \mathrm{[Cr(H_2O)_6]Cl_3 + 3AgNO_3 \rightarrow [Cr(H_2O)_6](NO_3)_3 + 3AgCl\downarrow} \]

Final Answer:
The structural formula is hexaaquachromium(III) chloride.
\[ \boxed{\mathrm{[Cr(H_2O)_6]Cl_3}} \]