Unit 7 The P Block Elements (Intext Questions-10)
Intext Questions
7.18 Why does $\mathrm{O_3}$ act as a powerful oxidising agent?
Answer Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is very reactive. $\underset{\text{Ozone}}{\mathrm{O_3}} \xrightarrow{\Delta} \underset{\text{Oxygen}}{\mathrm{O_2}}+\underset{\text{Nasant oxygen}}{[\mathrm{O}]}$ Therefore, ozone acts as a powerful oxidising agent.Show Answer
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Answer
Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with potassium iodide solution buffered with a borate buffer ( $\mathrm{pH}$ 9.2), iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate using starch as an indicator. The reactions involved in the process are given below.
$$ \begin{aligned} & \underset{\text { Iodide }}{2 \mathrm{I}^{-}}+\mathrm{H_2} \mathrm{O}+ \underset{\text { Ozone }}{\mathrm{O_3}} \longrightarrow 2 \mathrm{OH}^{-}+\underset{\text { Iodine }}{\mathrm{I_2}}+\mathrm{O_2} \\ & \mathrm{I_2}+ \underset{\substack{\text { Sodium } \\ \text { thiosulphate}}}{2 \mathrm{Na_2} \mathrm{~S_2} \mathrm{O_3}} \longrightarrow \underset{\substack{\text { Sodium } \\ \text { tetrathionate }}}{\mathrm{Na_2} \mathrm{~S_4} \mathrm{O_6}}+2 \mathrm{NaI} \\ & \quad \end{aligned} $$
