MCQ
When aluminium phosphide is treated with dil. sulphuric acid
- A$S{O_2}$ is liberated
- ✓$P{H_3}$ is evolved
- C${H_2}S$ is evolved
- D${H_2}$ is evolved
$2 AlP +\text { dil. } 3 H _2 SO _4 \rightarrow Al _2\left( SO _4\right)_3+2 PH _3 \text {. }$
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$\begin{array}{*{20}{c}}
{{C_6}{H_5} - C - H} \\
{\,\,\,\,\,\,\,\,\,\,||} \\
{\,\,\,\,\,\,\,\,\,\,\,\,O}
\end{array}\xrightarrow{{N{H_2}OH/{H^ \oplus }}}[X]$
$[X] $ will be :