Three moles of AgCl get precipitated when one mole of an octahedral co-ordination compound with empirical formula CrCl$$_3$$.3NH$$_3$$.3H$$_2$$O reacts with excess of silver nitrate. The number of chloride ions satisfying the secondary valency of the metal ion is ___.

We have an octahedral co-ordination compound whose empirical formula is $$\text{CrCl}_3\cdot3\text{NH}_3\cdot3\text{H}_2\text{O}$$ and whose reaction with excess silver nitrate gives three moles of $$\text{AgCl}$$ per mole of the compound.

First, recall the basic ideas of Werner’s theory:

$$$\text{Primary valency} = \text{ionisable (charge) chloride ions outside the square brackets}$$$ $$$\text{Secondary valency} = \text{coordination number} = 6 \text{ for an octahedral Cr(III) complex}$$$

Whenever chloride ions are present outside the coordination sphere, they behave as free anions. When such a salt is treated with $$\text{AgNO}_3$$, each ionisable $$\text{Cl}^-$$ gives one mole of $$\text{AgCl}$$ precipitate according to the reaction

$$\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}\downarrow$$

In the present case, three moles of $$\text{AgCl}$$ are obtained from one mole of the complex. Therefore, by direct stoichiometry,

$$$\text{Number of ionisable chloride ions} = 3$$$

These three chloride ions must originate from outside the coordination sphere because only such chloride ions can react with silver ion. Consequently, all the chloride ions appearing in the empirical formula are primary-valency ions.

The empirical formula already contains exactly three chloride ions ($$\text{CrCl}_3\cdot3\text{NH}_3\cdot3\text{H}_2\text{O}$$). Since we have just established that all three are outside the brackets, none of them remains to act as a ligand inside the sphere.

Next, we must ensure that the chromium still achieves its required coordination number of six. Inside the coordination sphere we therefore place the six neutral ligands that are left, namely $$3\text{NH}_3$$ and $$3\text{H}_2\text{O}$$. Symbolically this is written as

$$\big[\text{Cr}(\text{NH}_3)_3(\text{H}_2\text{O})_3\big]\text{Cl}_3$$

Thus,

$$$\underbrace{\text{Number of chloride ions outside the brackets}}_{=3} + \underbrace{\text{Number of chloride ions inside the brackets}}_{=?} = 3$$$

The outside ions are already 3; hence

$$$\text{Number of chloride ions satisfying secondary valency} = 0$$$

Therefore, no chloride ion is bonded directly to the metal through the secondary valency.

Hence, the correct answer is Option 0.