Why does the following reaction occur?

$$XeO^{ 4- }_{ 6 \; (aq) } \; + \; 2 \; F^{ – }_{ (aq) } \; + \; 6 \; H^{ + }_{ (aq) } \; \rightarrow \; XeO_{ 3 \; (g) } \; + \; F_{ 2 \; (g) } \; + \; 3 \; H_{ 2 }O_{ (l) }$$

What conclusion about the compound $$Na_{ 4 }XeO_{ 6 }$$​ (of which $$XeO_{ 6 }^{ 4- }$$​ is a part) can be drawn from the reaction?

Asked by Pragya Singh | 1 year ago |  66

##### Solution :-

$$XeO^{ 4- }_{ 6 \; (aq) } \; + \; 2 \; F^{ – }_{ (aq) } \; + \; 6 \; H^{ + }_{ (aq) } \;$$

$$\rightarrow \; XeO_{ 3 \; (g) } \; + \; F_{ 2 \; (g) } \; + \; 3 \; H_{ 2 }O_{ (l) }$$

The oxidation no. of Xe reduces from +8 in $$XeO_{ 6 }^{ 4- }$$​ to +6 in $$XeO_{ 3 }$$

The oxidation no. of F increases from -1 in $$F^{ – }$$ to 0 in $$F_{ 2 }$$

Hence, $$XeO_{ 6 }^{ 4- }$$ is reduced on the other hand $$F^{ – }$$ is oxidized.

As $$Na_{ 2 }XeO_{ 6 }^{ 4- }−​ (or XeO_{ 6 }^{ 4- })$$is a stronger oxidizing agent compared to $$F_{ 2 }$$​, this reaction occurs.

Answered by Abhisek | 1 year ago

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