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Question

The total number of geometrical isomers possible for [M(HO(CH2)2NH2)2Br2] is:


A
2
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B
4
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C
5
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D
6
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Solution

The correct option is A 2

(HO(CH2)2NH2) is a bidentate ligand. The given complex has square bipyramidal shape. Its isomers are as shown.

1171762_1101266_ans_ad0c018b46a241fc87622c00fd40435a.JPG

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