1. Principle involved

Aqueous $AgNO_3$ gives $AgCl$ precipitate only with ionisable/free chloride ions present outside the coordination sphere.

• Chloride ions inside the coordination sphere are not immediately precipitated by aqueous $AgNO_3$.
• Therefore, the complex that has the maximum number of counter-ion $Cl^-$ will consume the maximum equivalents of $AgNO_3$.

Since, $Ag^+ + Cl^- \rightarrow AgCl \downarrow$ 1 mole of free $Cl^-$ consumes 1 mole (1 equivalent) of $AgNO_3$.

2. Check each complex

Option A: $Na_3[CrCl_6]$

Here all 6 chloride ligands are inside the coordination sphere: $[CrCl_6]^{3-}$ The ions outside are $3Na^+$, not $Cl^-$.

So, free chloride ions = $0$.

Thus, $AgNO_3$ consumed = $0$ equivalents.

Option B: $[Cr(H_2O)_5Cl]Cl_2$

Here one $Cl^-$ is coordinated inside the sphere, and 2 chloride ions are outside as counter ions.

So, free chloride ions = $2$.

Thus, $AgNO_3$ consumed = $2$ equivalents.

Option C: $[Cr(H_2O)_6]Cl_3$

All six ligands are neutral $H_2O$ molecules inside the sphere, and 3 chloride ions are outside as counter ions.

So, free chloride ions = $3$.

Thus, $AgNO_3$ consumed = $3$ equivalents.

Option D: $Na_2[CrCl_5(H_2O)]$

Here 5 chloride ligands are coordinated inside the sphere: $[CrCl_5(H_2O)]^{2-}$ Outside ions are $2Na^+$, not $Cl^-$.

So, free chloride ions = $0$.

Thus, $AgNO_3$ consumed = $0$ equivalents.

3. Comparison

• A: $0$
• B: $2$
• C: $3$
• D: $0$

Maximum consumption is for: $[Cr(H_2O)_6]Cl_3$

So the correct option is C.

4. Verification with stored answer

Stored correct answer: C

My derived answer: C

They match.