SATHEE: Co-Ordination Compounds Question 221

Co-Ordination Compounds Question 221

Question: The total number of possible isomers of the complex compound $[C u^{I I} {(N H_{3})}_{4}] [P t^{I I} C l_{4}]$ is

Options:

A) 3

B) 6

C) 5

D) 4

Show Answer

Answer:

Correct Answer: D

Solution:

For $[C u^{I I} {(N H_{3})}_{4} [P t^{I I} C l_{4}]$ four isomers are possible which are $[C u {(N H_{3})}_{4} [P t C l_{4}]$ , $[C u C l_{4}], [P t {(N H_{3})}_{4}] [P t C l_{3} (N H_{3})] [C u {(N H_{3})}_{3} C l]$ and $[P t {(N H_{3})}_{3} C l] [C u (N H_{3}) C l_{3}]$

Co-Ordination Compounds Question 221

Question: The total number of possible isomers of the complex compound $[C u^{I I} {(N H_{3})}_{4}] [P t^{I I} C l_{4}]$ is

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Co-Ordination Compounds Question 221

Question: The total number of possible isomers of the complex compound [CuII(NH3)4][PtIICl4] is

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: The total number of possible isomers of the complex compound $[C u^{I I} {(N H_{3})}_{4}] [P t^{I I} C l_{4}]$ is