(i) `CH_(2)=CH-CH_(2)-CH=CH_(2)`
(ii) `CH_(2)=CH-CH=CH-CH_(3)`
(iii) `CH_(3)-CH=CH-CH=CH-CH_(3)`
The numbers of possible geometrical isomers for the above compounds are

Options

A. 0, 2, 4
B. 2, 2 4
C. 0, 3, 3
D. 0,23

Answer

To determine the number of possible geometrical isomers for the given compounds, we need to analyze each compound based on the conditions for geometrical isomerism. The key conditions are:

1. The compound must be planar (which typically involves sp² hybridized carbon atoms).
2. Each sp² hybridized carbon atom must have different groups attached to it.

Let's analyze each compound step by step:

### Compound (i): `CH₂=CH-CH₂-CH=CH₂`

1. Identify the double bonds: There are two double bonds in this compound.
2. Analyze the first double bond (C1=C2):
- C1 has two hydrogen atoms (H) attached (C1: H₂C).
- C2 has one hydrogen (H) and one carbon (C3) attached (C2: H-CH₂).
- Since C1 has two identical groups (H), it cannot exhibit geometrical isomerism.
3. Analyze the second double bond (C4=C5):
- C4 has one hydrogen (H) and one carbon (C3) attached (C4: H-CH).
- C5 has two hydrogen atoms (H) attached (C5: H₂C).
- Again, C5 has two identical groups (H), so this double bond cannot exhibit geometrical isomerism.
4. Conclusion: There are **0 geometrical isomers** for this compound.

### Compound (ii): `CH₂=CH-CH=CH-CH₃`

1. Identify the double bonds: There are two double bonds in this compound.
2. Analyze the first double bond (C1=C2):
- C1 has two hydrogen atoms (H) attached (C1: H₂C).
- C2 has one hydrogen (H) and one carbon (C3) attached (C2: H-CH).
- C1 has two identical groups (H), so it cannot exhibit geometrical isomerism.
3. Analyze the second double bond (C3=C4):
- C3 has one hydrogen (H) and one carbon (C2) attached (C3: H-CH).
- C4 has one hydrogen (H) and one methyl group (C5: CH₃).
- Since both C3 and C4 have different groups, this double bond can exhibit geometrical isomerism (cis and trans).
4. Conclusion: There are **2 geometrical isomers** (cis and trans) for this compound.

### Compound (iii): `CH₃-CH=CH-CH=CH-CH₃`

1. Identify the double bonds: There are two double bonds in this compound.
2. Analyze the first double bond (C2=C3):
- C2 has one hydrogen (H) and one carbon (C1) attached (C2: H-CH).
- C3 has one hydrogen (H) and one carbon (C4) attached (C3: H-CH).
- Both C2 and C3 have different groups, so this double bond can exhibit geometrical isomerism (cis and trans).
3. Analyze the second double bond (C4=C5):
- C4 has one hydrogen (H) and one carbon (C3) attached (C4: H-CH).
- C5 has one hydrogen (H) and one carbon (C6) attached (C5: H-CH).
- Again, both C4 and C5 have different groups, so this double bond can exhibit geometrical isomerism (cis and trans).
4. Conclusion: There are **4 geometrical isomers** (cis and trans for each double bond).

### Summary of Results:
- Compound (i): 0 geometrical isomers
- Compound (ii): 2 geometrical isomers
- Compound (iii): 4 geometrical isomers

### Final Answer:
The total number of possible geometrical isomers for the three compounds is:
- **0 + 2 + 4 = 6 geometrical isomers.**