`O_(2)F_(2)` is an unstable yellow change solid and `H_(2)O_(2)` is a colourless liquid, both have `O-O` bond and `O-O` bond length in `H_(2)O_(2)` and `O_(2)F_(2)` respectively is `:`

Options

A. 1.22 Å, 1.48 Å
B. 1.48Å , 122 Å
C. 1.22 Å , 1.22Å
D. 1.48 Å , 1.48 Å

Answer

To solve the question regarding the bond lengths of the O-O bond in \( H_2O_2 \) and \( O_2F_2 \), we can follow these steps:

### Step 1: Understand the Compounds
- \( O_2F_2 \) (dioxygen difluoride) is an unstable yellow solid.
- \( H_2O_2 \) (hydrogen peroxide) is a colorless liquid.
- Both compounds contain an O-O bond.

### Step 2: Analyze the O-O Bond
- The O-O bond is present in both compounds, and we need to compare their bond lengths.
- Bond length can be influenced by the hybridization of the atoms involved and the surrounding electronegative atoms.

### Step 3: Apply Bent's Rule
- Bent's rule states that the more electronegative atoms will pull electron density towards themselves, affecting the hybridization of the central atom.
- In \( H_2O_2 \), the O-O bond has a higher p-character due to the presence of the OH groups, which increases the s-character in the O-O bond.

### Step 4: Compare the O-O Bonds
- In \( O_2F_2 \), the presence of fluorine (a highly electronegative atom) pulls electron density away from the O-O bond, resulting in a bond with more s-character and less p-character.
- Therefore, the O-O bond in \( H_2O_2 \) has more p-character compared to the O-O bond in \( O_2F_2 \).

### Step 5: Conclusion on Bond Lengths
- Since the O-O bond in \( H_2O_2 \) has more p-character, it results in a longer bond length compared to the O-O bond in \( O_2F_2 \).
- Thus, we conclude that the O-O bond length in \( H_2O_2 \) is greater than that in \( O_2F_2 \).

### Final Answer
The O-O bond length in \( H_2O_2 \) is greater than that in \( O_2F_2 \).

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