Question

Total no. of possible isomers for complex compounds $[Cu{{(N{{H}_{3}})}_{4}}]$, $[PtC{{l}_{4}}]$ are:
A) 3
B) 2
C) 4
D) None of these

Answer 1

Hint: The answer here is based upon the concept of inorganic chemistry that includes the general definition of isomers that has types that is based on coordination isomerism. This fact helps you to approach the correct answer.

Complete answer:
We are familiar with concepts that with the chapters of types of isomerism and how can that be found.
Let us now refresh those facts and approach to the required answer.
- Isomerism is the term used for the compounds that have the same molecular formula but different arrangement of atoms in the space.
- Isomers have several types and one among those is the coordination isomerism which is defined as the isomerism in which complexes with same ligands can be exchanged mutually based on the charge of the ins and thus can form isomers of those complexes.
Here, in the above two complexes $[PtC{{l}_{4}}]$ and $[Cu{{(N{{H}_{3}})}_{4}}]$ that are coordinated together in the two spheres.
The possibilities of isomerism is as shown below,
(i) $[Cu{{(N{{H}_{3}})}_{3}}Cl]$ and $[PtN{{H}_{3}}C{{l}_{3}}]$ where charge of the chlorine are -1 and -3 respectively
(ii) $[CuN{{H}_{3}}{{(Cl)}_{3}}]$ and $[Pt{{(NH)}_{3}}Cl]$ where charges on chlorine are -3 and -1 respectively
(iii)$[CuC{{l}_{4}}]$ and $[Pt{{(N{{H}_{3}})}_{4}}]$ where charges on chlorine atoms are -4 and 0 respectively
But, the isomer $[CuN{{H}_{2}}{{(Cl)}_{2}}]$ and $[Pt{{(NH)}_{2}}C{{l}_{2}}]$ is not as charges on chlorine gets equal of -2 charge each and therefore complex becomes neutral.
Therefore, the correct answer is the possibility of isomers are 4 (including isomers of question).

The correct option therefore is option C) 4.

Note: Note that when questions are given in the form of two complexes and asked for the number of isomers formed as a whole then coordination isomerism applies. If question includes term ‘each’ then other type of isomers (cis or trans) applies.