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) :

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Solution

The correct option is B
3

No. of geometrical isomers of $[P t (C l) (P y) (N H_{3}) (N H_{2} O H)]^{⨁} = 3$

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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)]^{+}

=

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

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

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