A) \[Sb{{F}_{5}}\]
B) \[Mn{{F}_{3}}\]
C) \[KSb{{F}_{6}}\]
D) \[Mn{{F}_{4}}\]
Correct Answer: A
Solution :
\[{{K}_{2}}Mn{{F}_{6}}\] with \[Sb{{F}_{5}}\] forms \[{{F}_{2}}\]. In this reaction stronger Lewis acid \[Sb{{F}_{5}}\] displaces the weaker one \[(Mn{{F}_{4}})\] from its salt. \[Mn{{F}_{4}}\] is unstable and readily decompose to give \[Mn{{F}_{3}}\] and fluorine \[{{K}_{2}}Mn{{F}_{6}}+2Sb{{F}_{5}}\xrightarrow{{}}2KSb{{F}_{6}}+\underset{(unstable)}{\mathop{Mn{{F}_{4}}}}\,\] \[Mn{{F}_{4}}\xrightarrow{{}}Mn{{F}_{3}}+\frac{1}{2}{{F}_{2}}\]
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