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
Given below are two statements regarding conformations of n-butane. Choose the correct option. 
Consider a weak base \(B\) of \(pK_b = 5.699\). \(x\) mL of \(0.02\) M HCl and \(y\) mL of \(0.02\) M weak base \(B\) are mixed to make \(100\) mL of a buffer of pH \(=9\) at \(25^\circ\text{C}\). The values of \(x\) and \(y\) respectively are
Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to

A coordination compound holds a central metal atom or ion surrounded by various oppositely charged ions or neutral molecules. These molecules or ions are re-bonded to the metal atom or ion by a coordinate bond.
A coordination entity composes of a central metal atom or ion bonded to a fixed number of ions or molecules.
A molecule, ion, or group which is bonded to the metal atom or ion in a complex or coordination compound by a coordinate bond is commonly called a ligand. It may be either neutral, positively, or negatively charged.