On the basis of following observations arrange the following compound…

The correct answer is Option (2) → (B), (C), (A), (D)

Only ionisable (outer sphere) $\text{Cl}^-$ ions react with $\text{AgNO}_3$ to form $\text{AgCl}$.

Electrolyte type reveals number of free chloride ions.

Step 1: Key Concept

$\text{AgNO}_3$ forms $\text{AgCl}$ only with free (ionisable) $\text{Cl}^-$ ions

$\text{Cl}^-$ inside coordination sphere $\rightarrow$ does NOT precipitate

$\text{Cl}^-$ outside coordination sphere $\rightarrow$ gives $\text{AgCl}$

So we must find:

How many $\text{Cl}^-$ are outside the coordination sphere?

Electrolyte type helps us determine this.

Step 2: Interpret Electrolyte Type

Electrolyte type	Meaning after dissociation
$1:1$	gives 1 anion outside
$1:2$	gives 2 anions outside
$1:3$	gives 3 anions outside
$2:1$	gives least free $\text{Cl}^-$ (since $\text{Cl}^-$ mostly coordinated)

Number of anions formed $=$ number of free $\text{Cl}^-$ ions

Step 3: Analyse Each Compound

(B) $\text{NiCl}_2 \cdot 2\text{KH}_2\text{O} \rightarrow \text{2:1 electrolyte}$

Dissociation produces more cations than anions

This implies very few $\text{Cl}^-$ are free

So $\rightarrow$ minimum $\text{AgCl}$ formed

(C) $\text{CoCl}_3 \cdot 4\text{NH}_3 \rightarrow \text{1:1 electrolyte}$

Dissociates into:

$1 \text{ complex ion} + 1 \text{ Cl}^-$

So:

$1 \text{ free Cl}^- \rightarrow \text{forms 1 mol AgCl}$

(A) $\text{PdCl}_2 \cdot 4\text{NH}_3 \rightarrow 1:2$ electrolyte

Dissociates into:

$1 \text{ complex ion} + 2 \text{ Cl}^-$

So:

$2 \text{ free Cl}^- \rightarrow \text{forms 2 mol AgCl}$

(D) $\text{CrCl}_3 \cdot 6\text{H}_2\text{O} \rightarrow 1:3$ electrolyte

Dissociates into:

$1 \text{ complex ion} + 3 \text{ Cl}^-$

So:

$3 \text{ free Cl}^- \rightarrow \text{forms 3 mol AgCl}$

$\rightarrow \text{Maximum AgCl formed}$

Step 4: Arrange in Increasing Order

$\text{AgCl}$ formed:

$(B) < (C) < (A) < (D)$