- A$C{O_2}$
- ✓$COC{l_2}$
- C$C{O_2},\;C{l_2}$
- DNone of these
$CHC{{l}_{3}}+O\to COC{{l}_{2}}+HCl$
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Assertion $A :$- Carbon forms two important oxides $- CO$ and $CO _2 . CO$ is neutral whereas $CO _2$ is acidic in nature.
Reason $R :$- $CO _2$ can combine with water in a limited way to form carbonic acid, while $CO$ is sparingly soluble in water.
In the light of the above statements, choose the most appropriate answer from the options given below :-
$\mathop I\limits_{({C_3}{H_6}C{l_2})} \xrightarrow{{KOH(aq)}}II\xrightarrow[{(ii){H_2}O/{H^ + }}]{{(i)C{H_3}MgBr}}III\xrightarrow{{Anhy.ZnC{l_2} + Conc.HCl}}$ given turbidity immediately
${C_6}{H_5}C{H_2}CH(OH)CH{(C{H_3})_2}\xrightarrow{{conc.\,{H_2}S{O_4}}}\,?$