Question

Why does the following reaction occur? $bold XeO subscript bold 6 superscript bold 4 bold minus end superscript bold left parenthesis bold aq bold right parenthesis bold space bold plus bold 2 bold F to the power of bold minus bold left parenthesis bold aq bold right parenthesis bold space bold plus bold 6 bold H to the power of bold plus bold left parenthesis bold aq bold right parenthesis bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold rightwards arrow bold XeO subscript bold 3 bold left parenthesis bold g bold right parenthesis bold space bold plus bold F subscript bold 2 bold left parenthesis bold g bold right parenthesis bold plus bold 3 bold H subscript bold 2 bold O bold left parenthesis bold l bold right parenthesis$

conclusion about the compound Na₄XeO₆ (of which XeO₆^4– is a part) can be drawn from the reaction ?

Solution

Writing the oxidation number of all the atoms above their respective symbols, $bold Xe with bold plus bold 8 on top bold O subscript bold 6 superscript bold 4 bold minus end superscript bold left parenthesis bold aq bold right parenthesis bold space bold plus bold 2 stack bold F to the power of bold minus with bold minus bold 1 on top bold left parenthesis bold aq bold right parenthesis bold plus bold 6 bold H to the power of bold plus bold left parenthesis bold aq bold right parenthesis end exponent bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold rightwards arrow bold Xe with bold plus bold 6 on top bold O subscript bold 3 bold left parenthesis bold g bold right parenthesis bold plus stack bold space bold F subscript bold 2 with bold 0 on top bold left parenthesis bold g bold right parenthesis bold plus bold 3 bold H subscript bold 2 bold O bold left parenthesis bold l bold right parenthesis$

Since the O.N.of F increases from -1 in F^- to zero in F₂ and therefore it is oxidised and hence acts as a reductant. The O.N. of Xe decreases from +8 in XeO₆^4- to 6+ in XeO₃and therefore it is reduced and acts as an oxidant. This reaction occurs since Na4XeO₆^4-(or XeO₆^4-isa stronger oxidant than F₂).

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Redox Reactions

Some More Questions From Redox Reactions Chapter

What are indirect redox reactions?

Write formulas for the following compounds:
(a) Mercury (II) chloride
(b) Nickel (II) sulphate
(c) Tin (IV) oxide
(d) Thallium (I) sulphate
(e) Iron (III) sulphate
(f) Chromium (III) oxide

What are electrolytes?

What are cations and anions?

What is electrolysis?

Mention two uses of electrolysis.

Define oxidation number of an element.

Differentiate between an ion and an atom.

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