Question

Which of the following complex produces ppt with ${\text{AgN}}{{\text{O}}_{\text{3}}}$ and exist in two geometrical isomeric forms?
(A) ${\text{PtC}}{{\text{l}}_{\text{2}}}{\text{.2N}}{{\text{H}}_{\text{3}}}$
(B) ${\text{PtC}}{{\text{l}}_{\text{2}}}{\text{.3N}}{{\text{H}}_{\text{3}}}$
(C) ${\text{PtC}}{{\text{l}}_4}{\text{.4N}}{{\text{H}}_{\text{3}}}$
(D) ${\text{PtC}}{{\text{l}}_3}{\text{.N}}{{\text{H}}_{\text{3}}}$

Answer 1

Any kind of reagent reacts with the anions that are required to satisfy the primary valences of a complex. Hence silver nitrate will also react with the primary chloride anions.

Complete step by step answer:
The platinum ion can have two valences, $+ {\text{II}}$ and $+ {\text{IV}}$ . In the first complex, the valence of platinum is $+ 2$ which is satisfied by the two chloride anions. Hence there are no free chloride anions satisfying the secondary valance that could silver nitrate.
In the second complex too there are two chloride anions both satisfying the primary valance of the platinum cation. Hence this will also not form any precipitate with silver nitrate.
In the third complex, there are four chloride anions out of which two satisfy the primary valance of the platinum cation, while two are the secondary valance. So this will give curdy white precipitate of silver chloride with silver nitrate. This isomer also exists as two geometrical isomers with the two chloride anions satisfying the primary valance being is the planer (cis) position and (trans) or the axial position.
In the third complex, there being three chloride ions, two of them satisfy the primary valance while the third chloride ion is replaceable. But this compound cannot show geometrical or cis-trans isomerism.
Hence, the correct answer is option C.

Note:
Geometrical isomerism is defined as the arrangement of different ligands in a heterocyclic complex. When two identical ligands are in adjacent positions then it is called the cis isomer while those in the axial position are called axial isomers.