Chemistry · Electronic concepts of oxidation and reduction, redox reactions, oxidation number, rules for assigning oxidation number and balancing of redox reactions
Out of the following redox reactions III.N IIII.PCl disproportionation is not sh
Out of the following redox reactions \( \boldsymbol{I} \cdot \boldsymbol{N} \boldsymbol{H}_{4} \boldsymbol{N} \boldsymbol{O}_{3} \stackrel{\Delta}{\rightarrow} \boldsymbol{N}_{2} \boldsymbol{O}+\boldsymbol{2} \boldsymbol{H}_{2} \boldsymbol{O} \) III.N \( \boldsymbol{H}_{4} \boldsymbol{N} \boldsymbol{O}_{2} \stackrel{\Delta}{\rightarrow} \boldsymbol{N}_{2}+\mathbf{2} \boldsymbol{H}_{2} \boldsymbol{O} \) IIII.PCl\( _{5} \stackrel{\Delta}{\rightarrow} P C l_{3}+C l_{2} \) disproportionation is not shown in:
- A. I and II
- B. II and III
- C. I and III
- D. I, II and III
Step-by-step solution
Disproportionation requires the same element in one oxidation state to be simultaneously oxidized and reduced. In I, N in -3 and +5 states combine to form N2O (N at +1), which is comproportionation. In II, N in -3 and +3 combine to form N2 (N at 0), also comproportionation. In III, P in +5 is reduced to +3, and Cl in -1 is oxidized to 0; no element undergoes both oxidation and reduction. Thus, none of the reactions show disproportionation.
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