Coordination CompoundHard
Question
The total number of possible isomers of the compound [CuII(NH3)4] [PtIICl4] are:
Options
A.3
B.5
C.4
D.6
Solution
(C) There are four possible coordination isomers as given below :
(i)[CuII(NH3)4] [PtII Cl4]
(ii)[PtII (NH3)4] [CuIICl4]
(iii)[CuII (NH3)3 Cl] [PtII (NH3)Cl3]
(vi) [PtII (NH3)3 Cl] [CuII (NH3)Cl3]
(i)[CuII(NH3)4] [PtII Cl4]
(ii)[PtII (NH3)4] [CuIICl4]
(iii)[CuII (NH3)3 Cl] [PtII (NH3)Cl3]
(vi) [PtII (NH3)3 Cl] [CuII (NH3)Cl3]
Create a free account to view solution
View Solution FreeMore Coordination Compound Questions
How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with a Ca2+ ion?...The magnetic moment of [NiX4]2– ion is found to be zero. Then the metal of the complex ion is (X = monodentate anionic l...Among the following metal carbonyls, the C - O bond order is lowest in...For the reaction, Ni2+ + 4NH3 ⇋ [Ni(NH3)4]2+ at equilibrium, if the solution contains 1.6 ×10-4% of nickel in...Which of the following is not onsidered as an organometallic compound ?...