We need the total moles of $\mathrm{AgCl}$ formed when 1 mole each of the given complexes is treated with excess $\mathrm{AgNO_3}$.

Key idea

$\mathrm{AgNO_3}$ precipitates only those chloride ions which are present as counter ions (ionisable chloride) outside the coordination sphere.

Chloride ligands inside the coordination sphere do not immediately precipitate with $\mathrm{Ag^+}$ in this context.

1. $\mathrm{[Co(NH_3)_4Cl_2]Cl}$

Here, the complex is:

• Coordination sphere: $\mathrm{[Co(NH_3)_4Cl_2]^+}$
• Outside chloride: $1$

So, from $1$ mole of this complex: $1 \text{ mole of } \mathrm{Cl^-} \Rightarrow 1 \text{ mole of } \mathrm{AgCl}$

2. $\mathrm{[Ni(H_2O)_6]Cl_2}$

Here, the complex is:

• Coordination sphere: $\mathrm{[Ni(H_2O)_6]^{2+}}$
• Outside chloride: $2$

So, from $1$ mole of this complex: $2 \text{ moles of } \mathrm{Cl^-} \Rightarrow 2 \text{ moles of } \mathrm{AgCl}$

3. $\mathrm{[Pt(NH_3)_2Cl_2]}$

This is a neutral complex. Both chloride ions are inside the coordination sphere as ligands. There is no ionisable chloride outside.

So, from $1$ mole of this complex: $0 \text{ mole of } \mathrm{AgCl}$

4. $\mathrm{[Pd(NH_3)_4]Cl_2}$

Here, the complex is:

• Coordination sphere: $\mathrm{[Pd(NH_3)_4]^{2+}}$
• Outside chloride: $2$

So, from $1$ mole of this complex: $2 \text{ moles of } \mathrm{Cl^-} \Rightarrow 2 \text{ moles of } \mathrm{AgCl}$

5. Total moles of $\mathrm{AgCl}$

Adding all contributions: $1 + 2 + 0 + 2 = 5$

Final Answer

$\boxed{5}$

Comparison with stored correct answer

Stored correct answer = $5$

Our derived answer = $5$

So, the answer agrees with the stored correct answer.