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NCERT Exemplar · Q10

Q.Identify disproportionation reaction

(i) CH4 + 2O2 → CO2 + 2H2O
(ii) CH4 + 4Cl2 → CCl4 + 4HCl
(iii) 2F2 + 2OH^- → 2F^- + OF2 + H2O
(iv) 2NO2 + 2OH^- → NO2^- + NO3^- + H2O
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A disproportionation reaction is one where the same element is simultaneously oxidised and reduced. In option (iv), nitrogen in NO2NO_2 (oxidation state +4) changes to NO2−NO_2^- (+3) and NO3−NO_3^- (+5) — so it is the disproportionation reaction.

The key to spotting a disproportionation reaction is to track the oxidation states of each element on both sides. If a single element appears in two different products with one higher and one lower oxidation state than in the reactant, you have disproportionation.

Let’s check each option step by step.

  1. Option (i): CH4+2O2→CO2+2H2OCH_4 + 2O_2 \rightarrow CO_2 + 2H_2O

    Carbon in CH4CH_4 has oxidation state −4-4; in CO2CO_2 it is +4+4 — only oxidation. Oxygen in O2O_2 is 00; in H2OH_2O and CO2CO_2 it is −2-2 — only reduction. No element appears in two different oxidation states in the products. Not disproportionation.

  2. Option (ii): CH4+4Cl2→CCl4+4HClCH_4 + 4Cl_2 \rightarrow CCl_4 + 4HCl

    Carbon goes from −4-4 to +4+4 (oxidation). Chlorine goes from 00 to −1-1 (reduction). Again, each element changes to a single new state. Not disproportionation.

  3. Option (iii): 2F2+2OH−→2F−+OF2+H2O2F_2 + 2OH^- \rightarrow 2F^- + OF_2 + H_2O

    Fluorine in F2F_2 is 00. In F−F^- it is −1-1 (reduction). In OF2OF_2, oxygen is −2-2 and fluorine is +1+1 (oxidation of fluorine from 00 to +1+1). So fluorine is both reduced and oxidised — this looks like disproportionation of fluorine. …

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