The ratio of cis and trans-isomers of the complex $[M a_{2} b c d e]^{n \pm}$ is

5 : 3

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2 : 1

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7 : 3

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3 : 1

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Solution

The correct option is A $2 : 1$
The complex $[M a_{2} b d c e]^{n \pm}$ contains six cis isomers and three trans-isomers.

Thus the ratio of the cis and trans isomers of the complex is $2 : 1.$

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