MCQ
$P{H_4}I + NaOH$ forms
- ✓$P{H_3}$
- B$N{H_3}$
- C${P_4}{O_6}$
- D${P_4}{O_{10}}$
$PH _4 I + NaOH \rightarrow PH _3+ H _2 O + NaI$
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$(i)\,CH_3CH_2OH$
$(ii)\ CH_3COCH_3$
$(iii)\ \begin{array}{*{20}{c}}
{C{H_3} - CHOH} \\
{\,\,|} \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(iv) \,CH_3OH$
Which of the above compound(s), on being warmed with iodine solution and $NaOH,$ will give iodoform ?
$ {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