The structures to two compound `(I)` and `(II)` are given below. Which statement is correct ?
`F_(2)C=C=CF_(2)(I)` and `F_(2)C=C=C=CF_(2)(II)`

Options

A. Compound `(I)` is resolvable but `(II)` is non`-` resolvable.
B. Compound `(II)` is resolvable but `(I)` is non`-` resolvable.
C. Both are resolvable.
D. Neither of them is resolvable.

Explanation

To determine the correct statement regarding the compounds `(I)` and `(II)`, we will analyze their structures and stereochemistry.

### Step-by-Step Solution:

1. **Identify the Structures**:
- Compound `(I)` is `F₂C=C=CF₂`, which has a structure of a carbon-carbon double bond followed by another carbon-carbon double bond.
- Compound `(II)` is `F₂C=C=C=CF₂`, which has a structure of a carbon-carbon double bond followed by another carbon-carbon double bond, but with an additional carbon in between.

2. **Determine Hybridization**:
- In compound `(I)`, the first two carbons are sp² hybridized (due to the double bond), and the last carbon is sp hybridized (due to the double bond with another carbon).
- In compound `(II)`, the first two carbons are also sp² hybridized, while the last two carbons are sp hybridized.

3. **Analyze the Geometry**:
- For compound `(I)`, the geometry around the sp² hybridized carbons will be trigonal planar, while the sp hybridized carbon will have a linear geometry.
- For compound `(II)`, the geometry around the sp² hybridized carbons will also be trigonal planar, but the sp hybridized carbons will have a linear geometry, creating a different spatial arrangement.

4. **Check for Symmetry**:
- Compound `(I)` has a plane of symmetry because the fluorine atoms are arranged symmetrically around the central carbon-carbon double bond.
- Compound `(II)` also has a plane of symmetry due to the arrangement of fluorine atoms and the linear arrangement of the carbon atoms.

5. **Optical Activity**:
- A compound is optically active if it lacks a plane of symmetry and can rotate plane-polarized light. Since both compounds `(I)` and `(II)` have a plane of symmetry, they are both optically inactive.

6. **Conclusion**:
- Since both compounds are optically inactive, they are non-resolvable. Therefore, the correct statement is that neither compound `(I)` nor compound `(II)` is resolvable.

### Final Answer:
Both compounds `(I)` and `(II)` are optically inactive and non-resolvable.