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

4

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6

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2

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3

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Solution

The correct option is C $3$
As shown above there are 3 possible geometrical isomers possible for the given compound.Hence the correct option is 3.

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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)]^{+}

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

[A u C l_{2} B r_{2}]

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