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(D) The number of $AgCl$ moles precipitated depends on the number of ionizable chloride ions $(Cl^-)$ outside the coordination sphere.$1$. $[Co(NH_3)_

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$5$

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(D) The number of $AgCl$ moles precipitated depends on the number of ionizable chloride ions $(Cl^-)$ outside the coordination sphere.$1$. $[Co(NH_3)_4Cl_2]Cl$ dissociates to give $1$ $Cl^-$ ion,so it produces $1$ mole of $AgCl$.$2$. $[Ni(H_2O)_6]Cl_2$ dissociates to give $2$ $Cl^-$ ions,so it produces $2$ moles of $AgCl$.$3$. $[Pt(NH_3)_2Cl_2]$ has no ionizable $Cl^-$ ions outside the coordination sphere,so it produces $0$ moles of $AgCl$.$4$. $[Pd(NH_3)_4]Cl_2$ dissociates to give $2$ $Cl^-$ ions,so it produces $2$ moles of $AgCl$.Total moles of $AgCl = 1 + 2 + 0 + 2 = 5$ moles.

Total number of moles of $AgCl$ precipitated on addition of excess of $AgNO_3$ to one mole each of the following complexes $[Co(NH_3)_4Cl_2]Cl$,$[Ni(H_2O)_6]Cl_2$,$[Pt(NH_3)_2Cl_2]$ and $[Pd(NH_3)_4]Cl_2$ is $...............$ (in $.00$)

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