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Geometrical Isomerism in Square Planar Geometry

Total number ...

Question

Total number of geometrical isomers for the complex $[R h C l (C O) (P P h_{3}) (N H_{3})]$ is

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Solution

[Rh Cl(CO) $(P P h_{3} (N H_{3})$ ] is a square planar complex with four different and hence it will have three geometrical isomers

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Similar questions

Q. Total number of geometrical isomers for the complex

[R h C l (C O) (P P h_{3}) (N H_{3})]

Q. The possible number of geometrical isomers for the complex

[P t (N H_{3}) (N H_{2} O H) (N O_{2}) (p y)] N O_{2}

Q. The number of geometrical isomers of the complex

[C o (N O_{2})_{3} (N H_{3})_{3}]

Q. For the given complex

{[C o {C l}_{2} (e n) {({N H}_{3})}_{2}]}^{+}

, the number of geometrical isomers, the number of optical isomers and total number of isomers of all type possible respectively are:

Q. The number of geometrical isomers for octahedral

[C o C l_{4} (N H_{3})_{2}]^{-}

complex is _____

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