Total number of moles of AgCl precipitated on addition of excess of AgNO$$_3$$ to one mole each of the following complexes [Co(NH$$_3$$)$$_4$$Cl$$_2$$]Cl, [Ni(H$$_2$$O)$$_6$$]Cl$$_2$$, [Pt(NH$$_3$$)$$_2$$Cl$$_2$$] and [Pd(NH$$_3$$)$$_4$$]Cl$$_2$$ is

We need to find the total number of moles of AgCl precipitated on addition of excess AgNO$$_3$$ to one mole each of the given complexes.

Key Principle: Only ionic (free) chloride ions outside the coordination sphere react with AgNO$$_3$$ to form AgCl precipitate. Chloride ions inside the coordination sphere (bonded to the metal) do not precipitate.

Complex 1: [Co(NH$$_3$$)$$_4$$Cl$$_2$$]Cl

Coordination sphere: [Co(NH$$_3$$)$$_4$$Cl$$_2$$]$$^+$$ (2 Cl inside).

Free chloride: 1 Cl$$^-$$ outside.

Moles of AgCl = 1

Complex 2: [Ni(H$$_2$$O)$$_6$$]Cl$$_2$$

Coordination sphere: [Ni(H$$_2$$O)$$_6$$]$$^{2+}$$ (no Cl inside).

Free chloride: 2 Cl$$^-$$ outside.

Moles of AgCl = 2

Complex 3: [Pt(NH$$_3$$)$$_2$$Cl$$_2$$]

This is a neutral complex with all chlorides inside the coordination sphere.

Free chloride: 0.

Moles of AgCl = 0

Complex 4: [Pd(NH$$_3$$)$$_4$$]Cl$$_2$$

Coordination sphere: [Pd(NH$$_3$$)$$_4$$]$$^{2+}$$ (no Cl inside).

Free chloride: 2 Cl$$^-$$ outside.

Moles of AgCl = 2

Total moles of AgCl = 1 + 2 + 0 + 2 = $$\boxed{5}$$