Question

Palladium $(II)$ tends to form complexes with a coordination number of $4$ . One such compound was originally formulated as $PdC{l}_2\cdot 3N{H}_3$. Suppose an aqueous solution of the compound is treated with excess $AgN{O}_{3\text{ (aq)}}$, how many moles of $AgC{l}_{(s)}$ are formed per mole of $PdC{l}_2\cdot 3N{H}_3$ ?

Answer 1

Answer: 1

Because of their charge, the $C{l}^{-}$ions can be either in the coordination sphere, where they are bonded directly to the metal, or outside the coordination sphere, where they are bonded ionically to the complex. Because the $N{H}_3$ ligands are neutral, they must be in the coordination sphere.
So, the correct formula is $\left[Pd{\left(N{H}_3\right)}_{3}Cl\right]Cl$.
The chloride ion that serves as a ligand will not be precipitated as $AgC{l}_{(s)}$. Thus, only the single $C{l}^{-}$, which is outside the coordination sphere can react with $AgN{O}_3$.
Therefore, $1mol$ of $AgC{l}_{(s)}$ is formed per mole of the complex.