Question

Consider the hydrolysis of xenon tetrafluoride and choose the correct statement(s).

• A. Xenon is only oxidised during hydrolysis
• B. Xenon is only reduced during hydrolysis
• C. It is an example of redox hydrolysis
• D. There are three $p_\pi - d_\pi$ bond present in final xenon compound of hydrolysis

Answer 1

Answer: (C), (D)

C, D

Answer 2

The hydrolysis of xenon tetrafluoride ($\text{XeF}_4$) is a complex reaction that involves disproportionation of xenon.

1. Balanced Hydrolysis Reaction: The complete hydrolysis of $\text{XeF}_4$ can be represented by the following balanced chemical equation: $3\text{XeF}_4 + 6\text{H}_2\text{O} \rightarrow 2\text{XeO}_3 + \text{Xe} + 12\text{HF}$

2. Oxidation States of Xenon:

   • In $\text{XeF}_4$, the oxidation state of $\text{Xe}$ is +4 (since $\text{F}$ is -1).
   • In $\text{XeO}_3$, the oxidation state of $\text{Xe}$ is +6 (since $\text{O}$ is -2). This represents oxidation of $\text{Xe}$ from +4 to +6.
   • In elemental $\text{Xe}$, the oxidation state of $\text{Xe}$ is 0. This represents reduction of $\text{Xe}$ from +4 to 0.

3. Evaluation of Statements A, B, C:

   • A. Xenon is only oxidised during hydrolysis: False. Xenon is also reduced to elemental $\text{Xe}$.
   • B. Xenon is only reduced during hydrolysis: False. Xenon is also oxidised to $\text{XeO}_3$.
   • C. It is an example of redox hydrolysis: True. Since xenon undergoes both oxidation (from +4 to +6 in $\text{XeO}_3$) and reduction (from +4 to 0 in $\text{Xe}$), it is a redox reaction. As water is involved, it is a redox hydrolysis. Specifically, it's a disproportionation reaction of xenon.

4. Evaluation of Statement D:

   D. There are three $\text{p}_\pi - \text{d}_\pi$ bond present in final xenon compound of hydrolysis: The "final xenon compound" refers to $\text{XeO}_3$ (xenon trioxide), as $\text{Xe}$ is an element and $\text{HF}$ does not contain xenon. Let's determine the structure of $\text{XeO}_3$:

   • Xenon (Group 18) has 8 valence electrons.
   • In $\text{XeO}_3$, $\text{Xe}$ forms bonds with three oxygen atoms.
   • To satisfy the octet of oxygen, each oxygen forms a double bond with xenon.
   • The total number of valence electrons is $8 (\text{Xe}) + 3 \times 6 (\text{O}) = 26$.
   • 3 double bonds account for $3 \times 4 = 12$ electrons.
   • Remaining electrons for lone pairs on oxygen: $3 \times 2 (\text{lone pairs}) \times 2 (\text{electrons/pair}) = 12$ electrons.
   • Total electrons used: $12 + 12 = 24$.
   • Remaining electrons: $26 - 24 = 2$. These form one lone pair on the central xenon atom.
   • Thus, $\text{Xe}$ in $\text{XeO}_3$ has 3 bond pairs (double bonds) and 1 lone pair.
   • Steric number = 4, so the hybridization of $\text{Xe}$ is $\text{sp}^3$.
   • The geometry is trigonal pyramidal.
   • Each $\text{Xe}=\text{O}$ double bond consists of one $\sigma$ bond and one $\pi$ bond.
   • Xenon is a Period 5 element and has vacant $\text{d}$ orbitals. Oxygen is a Period 2 element and has $\text{p}$ orbitals. The $\pi$ bonds in $\text{XeO}_3$ are formed by the lateral overlap of the filled $\text{p}$ orbitals of oxygen and the empty $\text{d}$ orbitals of xenon. These are $\text{p}_\pi - \text{d}_\pi$ bonds.
   • Since there are three $\text{Xe}=\text{O}$ double bonds, there are three $\text{p}_\pi - \text{d}_\pi$ bonds in $\text{XeO}_3$.
   • Therefore, statement D is True.

Both statements C and D are correct.