question_answer 80)

  (i) Draw the gas phase and solid phase structure of\[{{H}_{2}}{{O}_{2}}\]. (ii)\[{{H}_{2}}{{O}_{2}}\]is a better oxidising agent than water. Explain.

Answer:

  (i) Structure of \[{{H}_{2}}{{O}_{2}}\] is slightly different in gas phase and solid phase.                 \[{{r}_{1}}=147.5pm,\]\[{{r}_{2}}=95pm\]                 \[{{\theta }_{1}}={{111.5}^{{}^\circ }},\,\,\,\,\,\,{{\theta }_{2}}={{94.8}^{{}^\circ }}\]                   \[{{r}_{1}}=98.5pm,\]\[{{r}_{2}}=145.8pm.\]                 \[{{\theta }_{1}}={{90.2}^{{}^\circ }},\]\[{{\theta }_{2}}={{101.2}^{{}^\circ }}\] (ii) is a better oxidising agent than water. \[{{H}_{2}}{{O}_{2}}\]acts as an oxidising agent in acid as well as in alkaline medium. \[{{H}_{2}}{{O}_{2}}+2{{H}^{+}}+2e\xrightarrow{Acid}2{{H}_{2}}O\]\[{{E}^{{}^\circ }}=1.77V\] \[{{H}_{2}}{{O}_{2}}+O{{H}^{-}}+2e\xrightarrow{Alkaline}3O{{H}^{-}}\] \[{{E}^{{}^\circ }}=0.88V\] Oxidation state of oxygen changes from -1 to -2. Oxidising nature of \[{{H}_{2}}{{O}_{2}}\] can be interpreted on account of possession of labile oxygen. \[{{H}_{2}}{{O}_{3}}\to {{H}_{2}}O+O\] When water acts as an oxidising agent, it is reduced to\[{{H}_{2}}\].Water reacts with number of active metals whose electrode potential is less than - 0.83 V.                                 \[\underset{\text{Reductant}}{\mathop{2Na+}}\,\underset{Oxidant}{\mathop{2{{H}_{2}}O}}\,\to 2NaOH+{{H}_{2}}\]