A) 3
B) 6
C) 5
D) 4
Correct Answer: D
Solution :
[d]The total number of isomers for the complex compound \[\left[ C{{u}^{II}}{{\left( N{{H}_{3}} \right)}_{4}} \right]\left[ P{{t}^{II}}C{{l}_{4}} \right]\]is four. |
These four isomers are |
\[\left[ Cu{{\left( N{{H}_{3}} \right)}_{3}}Cl \right]\left[ Pt\left( N{{H}_{3}} \right)C{{l}_{3}} \right]\], |
\[\left[ Cu\left( N{{H}_{3}} \right)C{{l}_{3}} \right]\left[ Pt{{\left( N{{H}_{3}} \right)}_{3}}Cl \right],\] |
\[\left[ CuC{{l}_{4}} \right]\left[ Pt{{\left( N{{H}_{3}} \right)}_{4}} \right]\] and \[\left[ Cu{{\left( N{{H}_{3}} \right)}_{4}} \right]\left[ PtC{{l}_{4}} \right].\] |
The isomer \[\left[ Cu{{\left( N{{H}_{3}} \right)}_{2}}C{{l}_{2}} \right]\left[ Pt{{\left( N{{H}_{3}} \right)}_{2}}C{{l}_{2}} \right]\]does not exist due to both parts being neutral. |
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