Q119CP - Chemistry: Molecular Nature Of Matter And Change

Question

The metal ion in platinum(IV) complexes, like that in cobalt(III) complexes, has a coordination number of 6 and often has ${\mathbf{Cl}}^{\mathbf{-}}$ ions and ${\mathbf{NH}}_{\mathbf{3}}$ molecules as ligands. For the traditional (before the work of Werner) formulas ${\mathbf{PtCl}}_{\mathbf{4}}\mathbf{·}\mathbf{6}{\mathbf{NH}}_{\mathbf{3}}$  and ${\mathbf{PtCl}}_{\mathbf{4}}\mathbf{·}\mathbf{4}{\mathbf{NH}}_{\mathbf{3}}$, (a) give the modern formula and charge of the complex ion, and (b) predict the moles of ions per mole of compound dissolved and moles of AgCl formed with excess ${\mathbf{AgNO}}_{\mathbf{3}}\mathbf{.}$

Step-by-Step Solution

Answer

(a) The modern formula and charge of the complex ion is $\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_4$and charge: 4+ and $\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_2$ and charge 2+

(b) 5 moles of ion produced, 4 moles of AgCl formed; 3 moles of ion produced, and 3 moles of AgCl formed.

Step 1: Given data

The given compounds are ${\mathrm{PtCl}}_4·6{\mathrm{NH}}_3$ and  ${\mathrm{PtCl}}_4·4{\mathrm{NH}}_3$

Step 2: Concept Introduction

A coordination complex consists of a central atom or ion, typically metallic and termed the coordination center, and a surrounding array of bound molecules or ions, which are successively called ligands or complexing agents.

Step 3: Find the modern formula

Since it is known that Pt(IV) has a coordination number of 6, it is expected that there are 6 ligands attached to Pt(IV). The ${\mathrm{NH}}_3$ outside can now be placed inside the complex ion bracket, and displace the amount of Cl atoms as needed.

${\mathrm{PtCl}}_4·6{\mathrm{NH}}_3: 6{\mathrm{NH}}_3$ atoms will attach to the metal ion and the 4Cl atoms will counter the ions of the compound.

$\mathrm{Pt}{{\left({\mathrm{NH}}_3\right)}_6}^{4+}+4{\mathrm{Cl}}^{-}\to {\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_6\right]}_6{\mathrm{Cl}}_4$ Charge: 4+

${\mathrm{PtCl}}_4·4{\mathrm{NH}}_3$ : $4{\mathrm{NH}}_3$ atoms will attach to the metal ion and the 2Cl atoms will counter the ions of the compound.

$\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_4{{\mathrm{Cl}}_2}^{2+}+2{\mathrm{Cl}}^{-}\to \left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_2$ Charge: 2+

Step 4: Prediction of moles

For every mole of a compound, 2 ions are formed, and their respective ratios are based on the equations below. Each mole of ${\mathrm{Cl}}^{-}$released by the complex ion forms 1 mole of AgCl when reacted with ${\mathrm{AgNO}}_3$.

${\mathrm{Cl}}^{-}+{\mathrm{Ag}}^{+}\to \mathrm{AgCl}$

For every 1 mole of $\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_4$, 5 moles of ions are formed 4 moles of ${\mathrm{Cl}}^{-}$ and 1 mole $\mathrm{Pt}{{\left({\mathrm{NH}}_3\right)}_6}^{4+}$

For 4 moles of ${\mathrm{Cl}}^{-}$ there will be 4 moles of AgCl such that,

$\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_4\to {\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_6\right]}^{4-}+4{\mathrm{Cl}}^{-}$

For every 1 mole of $\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_2$ , 3 moles of ions are formed 2 moles of ${\mathrm{Cl}}^{-}$ and 1 mole of $\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_6{{\mathrm{Cl}}_2}^{2+}$.

For the 2 moles of ${\mathrm{Cl}}^{-}$ procude 2 mole of AgCl

$\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_2\to \mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_4{{\mathrm{Cl}}_2}^{2+}+2{\mathrm{Cl}}^{-}$