A coordination compound \( CrCl_3.4H_2O \) precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. What is the formula of the compound?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \([Cr(H_2O)_4Cl_2].Cl\)
In a coordination compound, only the chloride ions outside the coordination sphere (counter ions) are free in solution. They alone are precipitated as AgCl by silver nitrate and they count as separate ions in conductance. Chlorides bonded to the metal stay in the complex ion.
Chromium(III) shows a coordination number of 6. The molar conductance shows two ions in total: one complex cation and one free \( \ce{Cl-} \). So of the three chlorides, two are ligands inside the sphere and one is outside, and all four water molecules complete the six positions:
\( \displaystyle \ce{[Cr(H2O)4Cl2]Cl -> [Cr(H2O)4Cl2]^+ + Cl^-} \quad (2\ \text{ions}) \)
\( \displaystyle \ce{Cl^- + AgNO3 -> AgCl v + NO3^-} \)
So the compound is \( \ce{[Cr(H2O)4Cl2]Cl} \).
Why the other options are incorrect
Option (1) \([Cr(H_2O)_4]Cl_3\) — Incorrect. All three chlorides would be free: \( \ce{[Cr(H2O)4]^3+ + 3Cl-} \), four ions in total.
Option (2) \([CrCl_3].4H_2O\) — Incorrect. No chloride would be free, so there would be no AgCl precipitate and almost no conductance.
Option (3) \([Cr(H_2O)_3Cl_3].H_2O\) — Incorrect. This is also a neutral complex with all three chlorides bonded to Cr; it would give no ions and no precipitate.