Question

Total number of moles of AgCl precipitated on addition of excess of AgNO₃​ to one mole each of the following complexes [Co(NH₃​)4​Cl₂​]Cl, [Ni(H₂​O)_6​]Cl_2​,[Pt(NH₃​)2​Cl₂​] and [Pd(NH₃​)₄​]Cl₂​ is

Answer 1

Correct Answer: 5

The correct answer is 5.
$\left[Co{\left(N{H}_3\right)}_{4}C{l}_2\right]Cl\Rightarrow$ Gives 1 mole $AgCl$
$\left[Ni{\left(H_{2}O\right)}_6\right]C{l}_2\Rightarrow$ Gives 2 moles $AgCl$
$\left[Pt{\left(N{H}_3\right)}_{2}C{l}_2\right]\Rightarrow$ Gives No $AgCl$
$\left[Pd{\left(N{H}_3\right)}_4\right]C{l}_2\Rightarrow$ Gives 2 moles $AgCl$
Total number of moles of $AgCl=5$ mole.

Answer 2

We need to analyze each complex and determine how many chloride ions are outside the coordination sphere (i.e., are ionizable). Only these chloride ions will react with AgNO$_3$ to form AgCl precipitate. \begin{enumerate} 

[Co(NH$_3$)$_4$Cl$_2$]Cl: This complex has one chloride ion outside the coordination sphere. When AgNO$_3$ is added, it will react with this chloride ion: [Co(NH$_3$)$_4$Cl$_2$]Cl + AgNO$_3$ $\rightarrow$ [Co(NH$_3$)$_4$Cl$_2$]$^+$ + AgCl + NO$_3^-$ Thus, 1 mole of AgCl will be precipitated. 

[Ni(H$_2$O)$_6$]Cl$_2$: This complex has two chloride ions outside the coordination sphere. When AgNO$_3$ is added, it will react with both chloride ions: [Ni(H$_2$O)$_6$]Cl$_2$ + 2AgNO$_3$ $\rightarrow$ [Ni(H$_2$O)$_6$]$^{2+}$ + 2AgCl + 2NO$_3^-$ Thus, 2 moles of AgCl will be precipitated. 

[Pt(NH$_3$)$_2$Cl$_2$]: This complex has no chloride ions outside the coordination sphere. Therefore, it will not react with AgNO$_3$ and no AgCl will be precipitated. 

[Pd(NH$_3$)$_4$]Cl$_2$: This complex has two chloride ions outside the coordination sphere. When AgNO$_3$ is added, it will react with both chloride ions: [Pd(NH$_3$)$_4$]Cl$_2$ + 2AgNO$_3$ $\rightarrow$ [Pd(NH$_3$)$_4$]$^{2+}$ + 2AgCl + 2NO$_3^-$ Thus, 2 moles of AgCl will be precipitated. \end{enumerate} Now, let's sum up the moles of AgCl precipitated: 1 mole (from [Co(NH$_3$)$_4$Cl$_2$]Cl) + 2 moles (from [Ni(H$_2$O)$_6$]Cl$_2$) + 0 moles (from [Pt(NH$_3$)$_2$Cl$_2$]) + 2 moles (from [Pd(NH$_3$)$_4$]Cl$_2$) = 5 moles Therefore, the total number of moles of AgCl precipitated is 5. 

Final Answer: 5.00