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D and F Block Elements question

2025 · 23 Jan · Shift 2 · Q19
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D and F Block Elements question

2025 · 23 Jan · Shift 2 · Q19

JEE MainChemistryD and F Block ElementsMCQ+4 / −1
Consider the following reactions K2Cr2O7→−H2OKOH[ A]→−H2OH2SO4[ B]+K2SO4\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{KOH}}[\mathrm{~A}] \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{H}_2 \mathrm{SO}_4}[\mathrm{~B}]+\mathrm{K}_2 \mathrm{SO}_4K2​Cr2​O7​KOH−H2​O​[ A]H2​SO4​−H2​O​[ B]+K2​SO4​ The products [A][A][A] and [B][B][B], respectively are :
  1. A
    K2Cr(OH)6\mathrm{K}_2 \mathrm{Cr}(\mathrm{OH})_6K2​Cr(OH)6​ and Cr2O3\mathrm{Cr}_2 \mathrm{O}_3Cr2​O3​
  2. B
    K2CrO4\mathrm{K}_2 \mathrm{CrO}_4K2​CrO4​ and Cr2O3\mathrm{Cr}_2 \mathrm{O}_3Cr2​O3​
  3. C
    K2CrO4\mathrm{K}_2 \mathrm{CrO}_4K2​CrO4​ and K2Cr2O7\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7K2​Cr2​O7​
  4. D
    K2CrO4\mathrm{K}_2 \mathrm{CrO}_4K2​CrO4​ and CrO\mathrm{CrO}CrO
View written solutionFree

Correct answer: C

  1. Identify the first reaction

Potassium dichromate and potassium chromate are interconvertible depending on the medium:

Cr2O72−+2OH−⇌2CrO42−+H2O\mathrm{Cr_2O_7^{2-} + 2OH^- \rightleftharpoons 2CrO_4^{2-} + H_2O}Cr2​O72−​+2OH−⇌2CrO42−​+H2​O

So, when K2Cr2O7\mathrm{K_2Cr_2O_7}K2​Cr2​O7​ is treated with KOH\mathrm{KOH}KOH, the equilibrium shifts toward chromate.

Thus,

K2Cr2O7+2KOH→2K2CrO4+H2O\mathrm{K_2Cr_2O_7 + 2KOH \rightarrow 2K_2CrO_4 + H_2O}K2​Cr2​O7​+2KOH→2K2​CrO4​+H2​O

Hence,

[A]=K2CrO4[A] = \mathrm{K_2CrO_4}[A]=K2​CrO4​

  1. Identify the second reaction

Acidification of chromate converts it back to dichromate:

2CrO42−+2H+⇌Cr2O72−+H2O\mathrm{2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O}2CrO42−​+2H+⇌Cr2​O72−​+H2​O

In potassium salt form with sulfuric acid:

2K2CrO4+H2SO4→K2Cr2O7+K2SO4+H2O\mathrm{2K_2CrO_4 + H_2SO_4 \rightarrow K_2Cr_2O_7 + K_2SO_4 + H_2O}2K2​CrO4​+H2​SO4​→K2​Cr2​O7​+K2​SO4​+H2​O

Therefore,

[B]=K2Cr2O7[B] = \mathrm{K_2Cr_2O_7}[B]=K2​Cr2​O7​

  1. Match with the options
  • Option A: K2Cr(OH)6\mathrm{K_2Cr(OH)_6}K2​Cr(OH)6​ and Cr2O3\mathrm{Cr_2O_3}Cr2​O3​ — incorrect
  • Option B: K2CrO4\mathrm{K_2CrO_4}K2​CrO4​ and Cr2O3\mathrm{Cr_2O_3}Cr2​O3​ — incorrect
  • Option C: K2CrO4\mathrm{K_2CrO_4}K2​CrO4​ and K2Cr2O7\mathrm{K_2Cr_2O_7}K2​Cr2​O7​ — correct
  • Option D: K2CrO4\mathrm{K_2CrO_4}K2​CrO4​ and CrO\mathrm{CrO}CrO — incorrect
  1. Final answer
[A]=K2CrO4,[B]=K2Cr2O7[A] = \mathrm{K_2CrO_4}, \qquad [B] = \mathrm{K_2Cr_2O_7}[A]=K2​CrO4​,[B]=K2​Cr2​O7​

So the correct option is C.

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