Question

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

Hint:

One AgCl per mole means one free chloride, so two Cl and four H2O sit inside the bracket with two water molecules outside.

Answer 1

Step 1: Use the precipitation clue:
Only free chloride ions outside the coordination sphere react with silver nitrate. Chloride bound to the metal does not react.
One mole of AgCl per mole of compound means only one \( \mathrm{Cl^-} \) is free.

Step 2: Place the other chlorides inside:
The compound has three Cl in all. One is free, so two Cl are bound to chromium as ligands.
\[ \mathrm{Ag^+ + Cl^- \rightarrow AgCl \downarrow} \]

Step 3: Fix the coordination number:
Chromium(III) shows a coordination number of 6. Two Cl are already inside, so four more ligands are needed.
Water supplies them. Four \( \mathrm{H_2O} \) go inside the bracket.
The two remaining water molecules are water of crystallisation and stay outside.

Step 4: Write the formula:
Inside the bracket: Cr, four \( \mathrm{H_2O} \) and two Cl. Outside: one \( \mathrm{Cl^-} \) and two \( \mathrm{H_2O} \).
The complex ion has charge +3 - 2 = +1, which one outside chloride balances.

Final Answer:
The structural formula is \( \mathrm{[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O} \).
\[ \boxed{\mathrm{[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O}} \]