The number of geometric isomers that can exist for square planar $[P t (C l) (p y) (N H_{3}) (N H_{2} O H)]^{+}$ is:

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Solution

The correct option is B 3
The complex is of the type [Mabcd] as all are monodentate ligands.

There are a total of 3 geometrical isomers.

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Similar questions

The number of geometric isomers that can exist for square planar $[P t (C l) (p y) (N H_{3}) (N H_{2} O H)]^{⨁}$ is (py = pyridine) :

[P t (C l) (p y) (N H_{3}) (N H_{2} O H)]^{+}

The number of geometric isomers that can exist for square planar $[P t (C l) (p y) (N H_{3}) (N H_{2} O H)]^{⨁}$ is (py = pyridine) :

[P t (C l) (p y) (N H_{3}) (N H_{2} O H)]^{+}

=

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