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AIIMS 2010
Exam
AIIMS
Year
2010
Subject
Chemistry
Chapter
p Block Elements (Group 15, 16, 17 & 18)

AIIMS 2010 · Chemistry · p Block Elements (Group 15, 16, 17 & 18)

Bromine is added to cold dilute aqueous solution of NaOH . The...

Bromine is added to cold dilute aqueous solution of NaOH\mathrm{NaOH}. The mixture is boiled. Which of the following statements is not true?

Options

  1. ADuring the reaction bromine is present in four different oxidation states.
  2. BThe greatest difference between the various oxidation states of bromine is 5 .
  3. COn acidification of the final mixture bromine is formed.
  4. DDisproportionation of bromine occurs during the reaction.

Correct answer

B. The greatest difference between the various oxidation states of bromine is 5 .

Step-by-step solution

2NaOH(dilute) +Br2⟶ cold NaBrO+NaBr+H2O3NaBrO Sodium hypobromide ⟶300K3NaBrO\begin{aligned}& 2 \mathrm{NaOH}_{\text{(dilute) }}+\mathrm{Br}_{2} \\ & \stackrel{\text{ cold }}{\longrightarrow} \mathrm{NaBrO}+\mathrm{NaBr}+\mathrm{H}_{2} \mathrm{O}\\ & \underset{\text{ Sodium hypobromide }}{3 \mathrm{NaBrO}} \stackrel{300 \mathrm{ K}}{\longrightarrow} 3 \mathrm{NaBrO} \end{aligned}On acidification, the finalmixturegivesbromine.final mixture gives bromine.5NaBrO+NaBrO3+6HCl⟶6NaCl+3Br2+3H2O\begin{aligned}& 5 \mathrm{NaBrO}+\mathrm{NaBrO}_{3}+6 \mathrm{HCl} \longrightarrow\\ & 6 \mathrm{NaCl}+3 \mathrm{Br}_{2}+3 \mathrm{H}_{2} \mathrm{O} \end{aligned}Thus,duringThus, during the reaction,bromineispresentinfourdifferentoxidationstatesi.e.,zeroinBr2,+1inNaBrOreaction, bromine is present in four different oxidation states i.e., zero in \mathrm{Br}_2,+1 in \mathrm{NaBrO},-1 in NaBr\mathrm{NaBr} and +5inNaBrO3.Thegreatestdifferencebetweenvariousoxidationstatesofbromineis6andnot5.Onacidificationofthefinalmixture,Br2isformedanddisproportionationofBr2occursduringthereactiongivingBrO−,Br−andBrO3−+5 in \mathrm{NaBrO}_{3}. The greatest difference between various oxidation states of bromine is 6 and not 5. On acidification of the final mixture, \mathrm{Br}_{2} is formed and disproportionation of \mathrm{Br}_{2} occurs during the reaction giving \mathrm{BrO}^{-}, \mathrm{Br}^{-} and \mathrm{BrO}_{3}^{-} ions.
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