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Solved: In halogen displacement reactions a halogen
Chapter 8, Problem 116P(choose chapter or problem)
In halogen displacement reactions a halogen element can be generated by oxidizing its anions with a halogen element that lies above it in the periodic table. This means that there is no way to prepare elemental fluorine, because it is the first member of Group 7A. Indeed, for years the only way to prepare elemental fluorine was to oxidize \(\mathrm{F^-}\) ions by electrolytic means. Then, in 1986, a chemist reported that by reacting potassium hexafluoromanganate(IV) \((\mathrm{K_2MnF_6})\) with antimony pentafluoride \((\mathrm{SbF_5})\) at \(\mathrm{150^\circ C}\), he had generated elemental fluorine. Balance the following equation representing the reaction:
\(\mathrm{K}_{2} \mathrm{MnF}_{6}+\mathrm{SbF}_{5} \longrightarrow \mathrm{KSbF}_{6}+\mathrm{MnF}_{3}+\mathrm{F}_{2}\)
Questions & Answers
QUESTION:
In halogen displacement reactions a halogen element can be generated by oxidizing its anions with a halogen element that lies above it in the periodic table. This means that there is no way to prepare elemental fluorine, because it is the first member of Group 7A. Indeed, for years the only way to prepare elemental fluorine was to oxidize \(\mathrm{F^-}\) ions by electrolytic means. Then, in 1986, a chemist reported that by reacting potassium hexafluoromanganate(IV) \((\mathrm{K_2MnF_6})\) with antimony pentafluoride \((\mathrm{SbF_5})\) at \(\mathrm{150^\circ C}\), he had generated elemental fluorine. Balance the following equation representing the reaction:
\(\mathrm{K}_{2} \mathrm{MnF}_{6}+\mathrm{SbF}_{5} \longrightarrow \mathrm{KSbF}_{6}+\mathrm{MnF}_{3}+\mathrm{F}_{2}\)
ANSWER:
Step 1 of 5
We need to balance the given equation
