Chapter 19, Problem 1DQ

Interpretation Introduction

Interpretation:

The formula for the compound showing the complex ion present needs to be determined and it should be named.

Concept Introduction :

A coordinate complex is formed when an electron efficient ion or molecule share electron/s with and electron deficient metal ion.

Answer to Problem 1DQ

• Hexachloroplatinateion
• Potassium hexachloroplatinate (IV)

Explanation of Solution

  ${\left[{\text{PtCl}}_{\text{6}}\right]}^{\text{2-}}$ is the formula of the compound that displays the complex ion.

When ${\text{AgNO}}_{\text{3}}$ is added into the compound’s aqueous solution, no chloride ion is given into the solution by ${\text{PtCl}}_{\text{4}}\text{.2KCl}$ . In fact, within the solution, no chloride ion is produced since all of the Cl-ion areremaining within the coordination sphere.

  $\text{Pt}{\text{. Cl}}_{\text{4}}\text{.2KCl }\left(\text{s}\right){\text{ + H}}_{\text{2}}\text{O }\left(\text{l}\right)\to {\left[{\text{PtCl}}_{\text{6}}\right]}^{\text{2-}}\left(\text{aq}\right){\text{ + 2K}}^{\text{+}}\left(\text{aq}\right)$

In the compound two potassium and six chloride ions are present. Therefore, to provide a neutral compound platinum must carry +4 charge. Hence, platinum has +4 oxidation state and the name of the compound is platinum (IV), and the ligand’s number is 6 chloride ions. Cl- is selected as chloro, hexa prefix displays six Cl- ions are present.

Therefore, the name of the complex ion is thus Hexachloroplatinateion and the name of the compound is potassium hexachloroplatinate (IV).