MCQ
Which of the following order is correct ?
- A$He > Ne > Ar > Kr$ (Adsorption tendency)
- B$He > Ne > Ar > Kr$ (Critical temperature)
- C$He > Ne > Ar > Kr$ (Boiling point)
- ✓$Xe > Kr > Ar > Ne$ (Polarisation)
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$I.$ $CH_3 -CH = CH_2$ $\xrightarrow[{{\text{(CC}}{{\text{l}}_4}{\text{ )}}}]{{{\text{C}}{{\text{l}}_2}}}$ $\begin{array}{*{20}{c}}
{Cl\,\,}\\
{\,\,|\,\,\,\,\,}\\
{C{H_3} - CH - C{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl}
\end{array}$
$II.$ $\begin{array}{*{20}{c}}
O\\
{||}\\
{C{H_3} - C - C{H_3}}
\end{array}$ $\xrightarrow[{{}^\Theta OH}]{{{\text{HCN}}}}$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,OH}\\
|\\
{C{H_3} - C - C{H_3}}\\
|\\
{\,\,\,\,\,CN}
\end{array}$
$III.$ $CH_3-CH_2-CH_3$ $\xrightarrow[{hv}]{{C{l_2}}}$ $\begin{array}{*{20}{c}}
{Cl\,\,}\\
{|\,\,\,\,}\\
{C{H_3} - CH - C{H_3}}
\end{array}$