When 1 mol \( CrCl_3.6H_2O \) is treated with excess of \( AgNO_3 \), 3 mol of \( AgCl \) are obtained. The conductivity of the solution will correspond to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1:3 electrolyte
In a complex, only the ions outside the coordination sphere (counter ions) are free in solution. Silver nitrate precipitates only these free chloride ions as AgCl; chloride bound to the metal is not precipitated. So the moles of AgCl formed give the number of ionisable chlorides.
Here 1 mol of the compound gives 3 mol of AgCl, so all three chlorides are outside the sphere and the six water molecules are the ligands: the complex is \( \ce{[Cr(H2O)6]Cl3} \).
\( \displaystyle \ce{[Cr(H2O)6]Cl3 + 3AgNO3 -> [Cr(H2O)6](NO3)3 + 3AgCl v} \)
In water it dissociates as
\( \displaystyle \ce{[Cr(H2O)6]Cl3 -> [Cr(H2O)6]^3+ + 3Cl-} \)
one cation and three anions, 4 ions in all. Its conductivity therefore corresponds to a 1:3 electrolyte.