A coordination compound CrCl_3.4H_2O precipitates silver chloride whe…

The correct answer is Option (4) → $[Cr(H_2O)_4Cl_2].Cl$

Complete statement:

The coordination compound is $[Cr(H_2O)_4Cl_2]Cl$.

Reasoning:

1. Information from molar conductance

It corresponds to a total of two ions in solution.

So the compound must dissociate as:

$\text{[complex]}^+ + \text{Cl}^{-}$

This means only one chloride ion is outside the coordination sphere (as counter-ion).

2. Information from $AgNO_3$ test

It precipitates $AgCl$, so at least one $Cl^-$ must be free (ionic) in solution.

Hence, exactly one chloride is outside the complex.

3. Total composition is $CrCl_3.4H_2O$

So total:

$Cr = 1$

$Cl = 3$

$H_2O = 4$

If $1\, Cl^-$ is outside, then $2\, Cl^-$ must be inside the coordination sphere.

Thus the complex part is:

$[Cr(H_2O),Cl_2]^+$

and the full compound is:

$[Cr(H_2O)_4Cl_2]Cl$

Why other options are incorrect:

• $[Cr(H_2O)_4]Cl_3$ → gives 4 ions (1 complex + 3 $Cl^-$)
• $[CrCl_3].4H_2O$ → no free $Cl^-$ ions for $AgCl$ precipitation
• $[Cr(H_2O)_3Cl_3].H_2O$ → no ionic chloride

So the correct answer is:

$[Cr(H_2O)_4Cl_2]Cl$