Coordination Compounds - Result Question 26
####28. The total number of possible isomers for the complex compound $[Cu^{II}(NH_3)_4][Pt^{II} Cl_4]$
(a) 3
(b) 6
(c) 5
(d) 4
[1998]
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Solution:
- (d) The total number of isomers for the complex compound
$[Cu{ }^{II}(NH_3)_4][Pt^{II} Cl_4]$ is four.
These four isomers are
$[Cu(NH_3)_3 Cl][Pt(NH_3) Cl_3]$,
$[Pt(NH_3)_3 Cl][Cu(NH_3) Cl_3]$, $[Pt(NH_3)_4][CuCl_4]$
and $[Cu(NH_3)_4][PtCl_4]$.
The isomer $[Cu(NH_3)_2 Cl_2][Pt(NH_3)_2 Cl_2]$ does not exist as both the parts are neutral.
For writing all combination of structural isomers, we have to keep the same value of primary valence and secondary valence of central atom/ion for all the isomers.
