The number of geometrical isomers for octahedral $[C o C l_{4} (N H_{3})_{2}]^{-}$ complex is

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Solution

The correct option is A 2
In octahedral $[C o C l_{4} (N H_{3})_{2}]^{-}$ , all the ligand are monodentate. The two $N H_{3}$ ligands can be arranged either adjacent to or opposite to each other to form cis and trans isomers respectively, making it a total of two possible isomers.
Hence, the answer is 2.

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