MCQ
$Xe{F_4}$ on partial hydrolysis produces
- A$Xe{F_2}$
- ✓$XeO{F_2}$
- C$XeO{F_4}$
- D$Xe{O_3}$
Complete hydrolysis;
$2Xe{F_4} + 3{H_2}O\, \to $$Xe + Xe{O_3} + {F_2} + 6HF$
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$CH_3-CH = CH -CH_2 -CH_2 -CH(CH_3)_2$
$(A)$ $(B)\,\,\,(C)$ $(D)$ $(E)\,\,\,(F)$
arrange them in decreasing order of reactivity towards free radical substitution
$ {I_2} + 2{e^ - } \to \,2{I^ - }\,;\,\,{E^o}\, = \,\,0.54\,\,V $
$ MnO_4^ - \, + \,8{H^ + }\, + \,5{e^ - } \to \,M{n^{2 + }}\, + \,4{H_2}O\,;\,{E^{o\,}} = 1.52\,\,V $
$ F{e^{3 + }} + {e^ - } \to \,\,F{e^{2 + }}\,;\,\,{E^o}\, = \,\,0.77\,\,V $
$ S{n^{4 + }} + 2{e^ - } \to \,\,S{n^{2 + }}\,;\,\,{E^o}\, = \,\,0.1\,\,V $
The strongest reducant and oxidant respectively are
$A$ is