Question

How many more co-ordination isomers are possible of the compoun $\left[Cu\right(N{H}_3{)}_4\left]\right[PtC{l}_4]$?

• A. 3
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is C 6

Coordination Isomerism: This type of isomerism is shown by compounds in which both cation and the anion are complexes. The isomers differ as a result of different groups being coordinated about a particular coordination center.

Example: $\left[Co\right(N{H}_3{)}_6\left]\right[Cr\left(CN{)}_6\right]$ = hexaaminecobalt (III) hexacyanochromate (III)

$\left[Cr\right(N{H}_3{)}_6\left]\right[Co\left(CN{)}_6\right]$ = hexaaminechromium (III) hexacyanocobalt (III)

All possible co-ordination isomers:

$\left[Cu\right(N{H}_3{)}_4\left]\right[PtC{l}_4]$

$\left[Cu\right(N{H}_3{)}_{3}Cl\left]\right[Pt\left(N{H}_3\right)C{l}_3]$

$\left[Cu\right(N{H}_3{)}_{2}C{l}_2\left]\right[Pt\left(N{H}_3{)}_{2}C{l}_2\right]$ 2 isomers possible here.

$\left[Cu\right(N{H}_3\left)C{l}_3\right]\left[Pt\right(N{H}_3{)}_{3}Cl]$

$\left[CuC{l}_4\right]\left[Pt\right(N{H}_3{)}_4]$