JEE Advanced 2021 Paper 1 · Q19 · Nomenclature & Isomerism

Correct answer:6

Solution

The compound [Pt(NH$_3$)$_4$Cl$_2$]Br$_2$ can show both geometrical and ionisation isomerism.

Ionisation isomers (which anions sit inside the coordination sphere vs outside): 1. [Pt(NH$_3$)$_4$Cl$_2$]Br$_2$ — Cl coordinated, Br outside. 2. [Pt(NH$_3$)$_4$Br$_2$]Cl$_2$ — Br coordinated, Cl outside. 3. [Pt(NH$_3$)$_4$ClBr]ClBr — one of each coordinated, one of each outside.

Each of these complex cations is octahedral with the formula [Pt(NH$_3$)$_4$X$_2$]$^{2+}$ (or [Pt(NH$_3$)$_4$XY]$^{2+}$), which exhibits cis/trans geometrical isomerism: - 1: cis-[Pt(NH$_3$)$_4$Cl$_2$]Br$_2$ and trans-[Pt(NH$_3$)$_4$Cl$_2$]Br$_2$ (2 isomers). - 2: cis-[Pt(NH$_3$)$_4$Br$_2$]Cl$_2$ and trans-[Pt(NH$_3$)$_4$Br$_2$]Cl$_2$ (2 isomers). - 3: cis-[Pt(NH$_3$)$_4$ClBr]ClBr and trans-[Pt(NH$_3$)$_4$ClBr]ClBr (2 isomers).

All six are achiral (an octahedral [Pt(NH$_3$)$_4$X$_2$] has a plane of symmetry in both cis and trans forms when the four NH$_3$ are equivalent). Total = 2 + 2 + 2 = 6 isomers.

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