Q.Why does the following reaction occur ? XeO6 4–(aq) + 2F–(aq) + 6H+(aq) → XeO3(g) + F2(g) + 3H2O(l) What conclusion about the compound Na4XeO6 (of which XeO6 4– is a part) can be drawn from the reaction.

The reaction occurs because of the relative oxidation states and stability of xenon compounds in acidic medium.

Key reasoning:

1. In $\ce{XeO6^{4-}}$, xenon is in the +8 oxidation state — the highest possible for xenon and extremely unstable. This species is a powerful oxidising agent.

2. In acidic solution, $\ce{XeO6^{4-}}$ oxidises fluoride ions ($\ce{F^-}$, oxidation state $-1$) to elemental fluorine ($\ce{F2}$, oxidation state $0$), while itself being reduced to $\ce{XeO3}$ where xenon is in the more stable +6 state.

3. The reaction is thermodynamically favorable because it converts the highly unstable Xe(VIII) to the relatively more stable Xe(VI), releasing the excess oxidising power by generating $\ce{F2}$ gas. …

Why the reaction proceeds: oxidation state and thermodynamic driving force

Xenon in $\ce{XeO6^{4-}}$ sits in the +8 oxidation state, the highest known for xenon. This is an extraordinarily electron-deficient state for a noble gas. Even though xenon can form stable compounds (unlike lighter noble gases), the +8 state pushes the limits of stability. The species is desperate to gain electrons and drop to a lower, more comfortable oxidation state.

Fluoride ion, meanwhile, is normally very reluctant to be oxidised—fluorine is the most electronegative element and $\ce{F^-}$ holds its electrons tightly. Yet the oxidising power of $\ce{XeO6^{4-}}$ is so extreme that it can rip electrons even from $\ce{F^-}$, converting it to $\ce{F2}$ gas. This tells us immediately that we're dealing with one of the most powerful oxidising agents in chemistry.

Step-by-step analysis of the reaction

1. Identify the oxidation state changes

In $\ce{XeO6^{4-}}$, xenon has oxidation state +8 (oxygen is $-2$, so $x + 6(-2) = -4 \implies x = +8$).

In $\ce{XeO3}$, xenon has oxidation state +6 (since $x + 3(-2) = 0 \implies x = +6$).

Xenon is reduced from +8 to +6, gaining 2 electrons per xenon atom.

Fluoride $\ce{F^-}$ (oxidation state $-1$) is oxidised to $\ce{F2}$ (oxidation state $0$), losing 1 electron per fluoride ion.

2. Write the half-reactions

Reduction half-reaction:

Oxidation half-reaction:

The electrons balance perfectly: 2 electrons released by two fluoride ions are consumed by one $\ce{XeO6^{4-}}$ ion.

3. Thermodynamic feasibility

Fluoride is oxidised only by reagents that out-oxidise the $\ce{F2/F^-}$ couple ($E^\ominus = +2.87$ V) — the strongest common oxidant there is. That this reaction runs at all tells us perxenate in acid is one of the very few species that clears that bar. The reaction is driven by:

• The instability of the +8 oxidation state of xenon
• The acidic medium, which stabilises the products
• The escape of gaseous products ($\ce{XeO3}$ and $\ce{F2}$), shifting equilibrium forward

4. Role of acid

The $\ce{H^+}$ ions are essential. They protonate the oxide ligands, allowing water molecules to leave and stabilising the lower oxide $\ce{XeO3}$. Without acid, the reaction would not proceed efficiently.

Conclusions about $\ce{Na4XeO6}$

From this reaction, we can draw several important conclusions about sodium perxenate: …