- ✓Acetone
- BAcetaldehyde
- CPropionaldehyde
- DFormaldehyde
$\begin{array}{*{20}{c}}
{C{H_3}} \\
{C{H_3}}
\end{array} > C = O\, + \,$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{{H_2} - CH - C - C{H_3}}
\end{array}$ $\xrightarrow[{ - \,{H_2}O}]{{HCl}}$ $\mathop {{{(C{H_3})}_2}C = CH.COC{H_3}}\limits_{Mesityl\,\,oxide} $
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$(A)$ $\left[\mathrm{Pt}(\mathrm{en}) \mathrm{Cl}_2\right]$ $(B)$ $\left[\mathrm{Pt}(\mathrm{en})_2\right] \mathrm{Cl}_2$
$(C)$ $\left[\mathrm{Pt}(\text { en })_2 \mathrm{Cl}_2\right] \mathrm{Cl}_2$ $(D)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
(use $R =0.083 L\,bar\,mol ^{-1}\, K ^{-1}$ )
Statement $I$: The boiling point of hydrides of Group $16$ elements follow the order
$\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{Te}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{~S}$.
Statement $II$: On the basis of molecular mass, $\mathrm{H}_2 \mathrm{O}$ is expected to have lower boiling point than the othe members of the group but due to the presence of extensive $\mathrm{H}$-bonding in $\mathrm{H}_2 \mathrm{O}$, it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
