- A$CHC{l_3} + {(C{H_3})_2}CO$
- B${(C{H_3})_2}CO + {C_6}{H_5}N{H_2}$
- C$CHC{l_3} + {C_6}{H_6}$
- ✓${(C{H_3})_2}CO + C{S_2}$
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$Time (sec)$ Rate $(mol\, L^{-1} sec.^{-1})$
$0$ $1.60 \times 10^{-2}$
$10$ $1.60 \times 10^{-2}$
$20$ $1.60 \times 10^{-2}$
$30$ $1.60 \times 10^{-2}$
From the above data, the order of reaction is
$A$. All group $16$ elements form oxides of general formula $\mathrm{EO}_2$ and $\mathrm{EO}_3$ where $\mathrm{E}=\mathrm{S}, \mathrm{Se}, \mathrm{Te}$ and Po. Both the types of oxides are acidic in nature.
$B$. $\mathrm{TeO}_2$ is an oxidising agent while $\mathrm{SO}_2$ is reducing in nature.
$C$. The reducing property decreases from $\mathrm{H}_2 \mathrm{~S}$ to $\mathrm{H}_2 \mathrm{Te}$ down the group.
$D$. The ozone molecule contains five lone pairs of electrons.
Choose the correct answer from the options given below: