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 |  51

1 Answer

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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