The number of geometrical isomers of

$C H_{3} - C H = C H - C H = C H - C H = C H C L$ is

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Solution

The correct option is D
8

As the ends of the compounds are different and three double bonds exhibit geometrical isomerism, are $2^{3} = 8$

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C H_{3} - C H = C H - C H = C H - C H = C H C l

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