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Why does the following reaction occur? X e O_{6(a q)}^{4-}+2 F_{(a q)}^{-}+6 H_{(a q)}^{+} \rightarrow X e O_{3}(g)+F_{2(g)}+3 \mathrm{H}_{2} \mathrm{O}_{(l)} What conclusion about the compound \mathrm{Na}_{4} \mathrm{XeO}_{6} (of which \mathrm{XeO}_{6}^{4-} is a part) can be drawn from the reaction?
CBSE | Chemistry | Class 11 | Exercise 1 | NCERT | Redox Reactions
Why does the following reaction occur? X e O_{6(a q)}^{4-}+2 F_{(a q)}^{-}+6 H_{(a q)}^{+} \rightarrow X e O_{3}(g)+F_{2(g)}+3 \mathrm{H}_{2} \mathrm{O}_{(l)} What conclusion about the compound \mathrm{Na}_{4} \mathrm{XeO}_{6} (of which \mathrm{XeO}_{6}^{4-} is a part) can be drawn from the reaction?

Solution:

X e O_{6(a q)}^{4-}+2 F_{(a q)}^{-}+6 H_{(a q)}^{+} \rightarrow X e O_{3(g)}+F_{2(g)}+3 H_{2} O_{(l)}

The oxidation no. of Xe decreases from +8 in \mathrm{XeO}_{6}^{4-} to +6 in \mathrm{XeO}_{3}.

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

Henceforth, \mathrm{XeO}_{6}^{4-} is diminished then again F^{-}is oxidized. As \mathrm{Na}_{2} \mathrm{XeO}_{6}^{4-} (or \left.\mathrm{XeO}_{6}^{4-}\right) is a more grounded

oxidizing specialist contrasted with F_{2}, this response happens.

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