A coordination compound \(CrCl_3 \cdot 6H_2O\) is mixed with excess of \(AgNO_3\) solution, two moles of \(AgCl\) are precipitated per mole of the compound. Write the structural formula of the coordination compound.
Show Hint
For hydrated chromium chloride complexes:
\[
[Cr(H_2O)_6]Cl_3
\]
gives \(3\) moles of \(AgCl\).
\[
[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O
\]
gives \(2\) moles of \(AgCl\).
\[
[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O
\]
gives \(1\) mole of \(AgCl\).
The number of moles of \(AgCl\) formed equals the number of chloride ions present outside the coordination sphere.
Step 1: Count ionisable $Cl^-$ ions from the AgCl test. 2 moles of $AgCl$ precipitate means 2 $Cl^-$ ions per formula unit are free in solution (outside the coordination sphere). The remaining $3 - 2 = 1$ $Cl^-$ is coordinated to Cr inside the sphere. Step 2: Determine the composition of the coordination sphere. Cr(III) has coordination number 6. With 1 $Cl^-$ inside, the remaining 5 positions are filled by $H_2O$ molecules: complex ion is $[Cr(H_2O)_5Cl]^{2+}$. One more $H_2O$ stays outside as crystal water (total $H_2O = 5 + 1 = 6$). Step 3: Write and verify the structural formula. Total $Cl = 1$ (inside) $+ 2$ (outside) $= 3$; Total $H_2O = 6$: consistent with $CrCl_3 \cdot 6H_2O$. \[ \boxed{[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O} \]