Read the passage carefully and answer the Questions.

Werner in 1898, propounded his theory of coordination compounds. In coordination compounds metals show two types of linkages (valences)-primary and secondary. The primary valences are normally ionisable and are satisfied by negative ions. The secondary valences are non-ionisable. These are satisfied by neutral molecules or negative ions. The secondary valence is equal to the coordination number and is fixed for a metal. The ions or group bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to the different coordination numbers. Both double salts as well as complexes are formed by the combination of two or more stable compounds in stoichiometric ratio. However, they differ in the fact that double salts such as carnallite, Mohr's salt, potash alum etc. dissociate into simple ions completely when dissolved in water. But complex ions do not dissociate completely into simple ions.

When $1\, mol\, CrCl_3.6H_2O$ is treated with excess of $AgNO_3$, 3 mol of AgCl are obtained. The formula of the complex is:

$[Cr(H_2O)_6]Cl_3$

The correct answer is Option (4) → $[Cr(H_2O)_6]Cl_3$

The experiment involves treating $1 \text{ mol}$ of the coordination compound $\text{CrCl}_3\cdot 6\text{H}_2\text{O}$ with excess silver nitrate ($\text{AgNO}_3$) and obtaining $3 \text{ mol}$ of silver chloride ($\text{AgCl}$) precipitate.

1. Ionizable Chlorides

The amount of $\text{AgCl}$ precipitate formed is directly proportional to the number of ionizable chloride ions ($\text{Cl}^-$) present in the solution. These are the chloride ions that are outside the coordination sphere (i.e., acting as counter-ions, or primary valencies).

$\text{Moles of } \text{AgCl} = \text{Moles of Complex} \times \text{Number of Ionizable } \text{Cl}^-$

$3 \text{ mol AgCl} = 1 \text{ mol Complex} \times X$

$X = 3$

This means that all three chloride ions ($\text{Cl}_3$) must be outside the coordination sphere as counter-ions.

2. Coordination Sphere

The total number of ligands required to satisfy the secondary valence (coordination number, which is typically 6 for $\text{Cr}^{3+}$) must be 6.

• Total $\text{H}_2\text{O}$ molecules: 6
• Total $\text{Cl}$ ions: 3

Since the 3 chloride ions are outside the sphere, the coordination sphere must be filled by the available water molecules.

$\text{Coordination Sphere} = 6 \text{ water molecules}$

$\text{Ionization Sphere} = 3 \text{ chloride ions}$

3. Final Formula

The formula where the central metal ($\text{Cr}$) is coordinated by all six water molecules, and all three chloride ions are ionizable, is:

$\mathbf{[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3}$