Compare the $\mathrm{O-O}$ bond lengths in $\mathrm{H_2O_2}$ and $\mathrm{F_2O_2}$

In peroxide-type molecules, the $\mathrm{O-O}$ bond is a single bond.

• In $\mathrm{H_2O_2}$, the substituent on oxygen is $\mathrm{H}$.
• In $\mathrm{F_2O_2}$, the substituent on oxygen is highly electronegative $\mathrm{F}$.

Fluorine shows a strong $-I$ effect and withdraws electron density from oxygen. This reduces lone-pair repulsion around the two oxygen atoms, allowing the $\mathrm{O-O}$ bond to become shorter in $\mathrm{F_2O_2}$ than in $\mathrm{H_2O_2}$.

Therefore, for the comparison \text{$\mathrm{O-O}$ bond length in } \mathrm{H_2O_2} \quad \text{is longer than that in } \mathrm{F_2O_2}.}

So, $X = \text{longer}.$

Compare the $\mathrm{O-H}$ bond length in $\mathrm{H_2O_2}$ with the $\mathrm{O-F}$ bond length in $\mathrm{F_2O_2}$

Bond length depends strongly on the sizes of the bonded atoms.

• The $\mathrm{O-H}$ bond involves hydrogen, which is very small.
• The $\mathrm{O-F}$ bond involves fluorine, which is larger than hydrogen.

Typical bond lengths:

• $\mathrm{O-H} \approx 0.96\,\text{Å}$
• $\mathrm{O-F} \approx 1.4\,\text{Å}$

Hence, $\mathrm{O-H} \text{ bond length } < \mathrm{O-F} \text{ bond length.}$

So, $Y = \text{shorter}.$

Choose the correct option

We found:

• $X = \text{longer}$
• $Y = \text{shorter}$

This matches: $\boxed{\text{Option B}}$

Verification with stored answer

Stored correct answer: $\mathrm{B}$

Our derived answer is also $\mathrm{B}$, so they agree.