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D and F Block Elements question

2023 · 30 Jan · Shift 1 · Q8
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D and F Block Elements question

2023 · 30 Jan · Shift 1 · Q8

JEE MainChemistryD and F Block ElementsMCQ+4 / −1
During the qualitative analysis of SO32−\mathrm{SO}_{3}^{2-}SO32−​ using dilute H2SO4,SO2\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{2}H2​SO4​,SO2​ gas is evolved which turns K2Cr2O7\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}K2​Cr2​O7​ solution (acidified with dilute H2SO4\mathrm{H}_{2} \mathrm{SO}_{4}H2​SO4​) :
  1. A
    blue
  2. B
    black
  3. C
    red
  4. D
    green
View written solutionFree

Correct answer: D

  1. Reaction of sulfite with dilute sulfuric acid

When SO32− \mathrm{SO_3^{2-}}SO32−​ is treated with dilute H2SO4\mathrm{H_2SO_4}H2​SO4​, sulfur dioxide gas is evolved:

SO32−+2H+→SO2+H2O\mathrm{SO_3^{2-} + 2H^+ \rightarrow SO_2 + H_2O}SO32−​+2H+→SO2​+H2​O
  1. Nature of SO2\mathrm{SO_2}SO2​

SO2\mathrm{SO_2}SO2​ is a reducing agent.

  1. Action on acidified potassium dichromate

Acidified K2Cr2O7\mathrm{K_2Cr_2O_7}K2​Cr2​O7​ contains dichromate ion, Cr2O72−\mathrm{Cr_2O_7^{2-}}Cr2​O72−​, in which chromium is in the +6+6+6 oxidation state. It is orange in color.

SO2\mathrm{SO_2}SO2​ reduces Cr(VI)\mathrm{Cr(VI)}Cr(VI) to Cr(III)\mathrm{Cr(III)}Cr(III):

Cr2O72−+3SO2+2H+→2Cr3++3SO42−+H2O\mathrm{Cr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O}Cr2​O72−​+3SO2​+2H+→2Cr3++3SO42−​+H2​O
  1. Color change

Cr3+\mathrm{Cr^{3+}}Cr3+ ion is green in solution.

Hence, acidified K2Cr2O7\mathrm{K_2Cr_2O_7}K2​Cr2​O7​ turns green.

  1. Option check
  • A: blue →\rightarrow→ incorrect
  • B: black →\rightarrow→ incorrect
  • C: red →\rightarrow→ incorrect
  • D: green →\rightarrow→ correct

Therefore, the correct answer is D.

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