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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$A$ :
Statement $I :$ Aniline is less basic than acetamide.
Statement $II :$ In aniline, the lone pair of electrons on nitrogen atom is delocalised over benzene ring due to resonance and hence less available to a proton.
Choose the most appropriate option:
$Pb \left( NO _{3}\right)_{2} \stackrel{673 K }{\longrightarrow} A + PbO + O _{2}$
$A \stackrel{\text { Dimerise }}{\longrightarrow} B$
Which forms di-iodide on reaction with $HI $ (excess) ?