The correct option is D 3 and 4
1. Compound (i) is a symmetrical compound. It has same methyl terminal groups.
The formula used to determinne the number of geometrical isomers in compound having same terminal groups with even number of double is, 2(n−1)+2[(n/2)−1]
where, n is the number of double bond present in the molecule.
In the given molecule we have two double bond so,
No. of geometrical isomers = 2(n−1)+2[(n/2)−1]
= 2(2−1)+2[(2/2)−1]
=21+20
=2+1
=3
2. Compound (ii) is a unsymmetrical compound. It has different terminal groups (methyl and chlorine).
The formula used to determinne the number of geometrical isomers in compound having different terminal groups is, 2n, where n is the number of dounle bond present in the molecule.
In the given compound we have 2 double bond so n = 2
Therefore,
No. of geometrical isomers = 2n
= 22
= 4.
Therefore, the correct option is (d).