Polar covalent molecules exhibit dipole moment. Dipole moment is equal to the product of charge separation , q and the bond length d for the bond. Unit of dipole moment is debye. One debye is equal to `10^(-18)` esu cm.
Dipole moments is a vector quantity. It has both magnitude and direction. Hence, dipole moment of a molecule depends upon the relative orientation of the bond dipoles, but not on the polarity of bonds alone. A symmetrical structure shows zero dipole moment. Thus, dipole moment helps to predict the geometry of a molecules. Dipole moment values can be distinguish between cis- and trans- isomers, ortho, meta and pare-forms of a substance, etc.
Q. Which of the following statement is correct regarding `C_(2)H_(2)F_(2)` molecule?

Options

A. One isomer is polar, one is non-polar
B. Two isomers are polar, one is non-polar
C. Two isomers are planar, one is non-planar
D. Two isomers are only possible and planar

Explanation

To determine the correct statement regarding the molecule \( C_2H_2F_2 \), we need to analyze its structure and the resulting dipole moments. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Possible Isomers
The molecule \( C_2H_2F_2 \) can exist in different structural forms (isomers). The two main isomers are:
1. **Cis-isomer**: Both fluorine atoms are on the same side of the carbon-carbon bond.
2. **Trans-isomer**: The fluorine atoms are on opposite sides of the carbon-carbon bond.

### Step 2: Draw the Structures
1. **Cis-isomer**:
```
F F
\ /
C = C
/ \
H H
```
2. **Trans-isomer**:
```
F H
\ /
C = C
/ \
H F
```

### Step 3: Analyze Polarity
- **Cis-isomer**: The dipole moments of the C-F bonds do not cancel out due to their orientation, resulting in a net dipole moment. Therefore, this is a polar molecule.
- **Trans-isomer**: The dipole moments of the C-F bonds cancel out because they are oriented in opposite directions, resulting in a net dipole moment of zero. Therefore, this is a non-polar molecule.

### Step 4: Conclusion on Dipole Moments
From the analysis:
- The cis-isomer of \( C_2H_2F_2 \) is polar (has a dipole moment).
- The trans-isomer of \( C_2H_2F_2 \) is non-polar (has no dipole moment).

### Step 5: Identify the Correct Statement
Based on the above analysis, the correct statement regarding \( C_2H_2F_2 \) is that it has both polar and non-polar isomers.

### Final Answer
The correct statement regarding \( C_2H_2F_2 \) is that it exhibits both polar (cis) and non-polar (trans) isomers.

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