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NEET CHEMISTRYThe d-and f-Block ElementsMedium

Question

Acidified K2Cr2O7K_2Cr_2O_7 solution turns green when Na2SO3Na_2SO_3 is added to it. This is due to the formation of:

A

CrO42−CrO_4^{2-}

B

Cr2(SO3)3Cr_2(SO_3)_3

C

CrSO4CrSO_4

D

Cr2(SO4)3Cr_2(SO_4)_3

Step-by-Step Solution

Potassium dichromate (K2Cr2O7K_2Cr_2O_7) is a strong oxidising agent in acidic medium. It oxidises sulphite ions (SO32−SO_3^{2-}) from sodium sulphite into sulphate ions (SO42−SO_4^{2-}). In this process, the orange-coloured dichromate ion (Cr2O72−Cr_2O_7^{2-}) is reduced to the green-coloured chromium(III) ion (Cr3+Cr^{3+}).

The chemical equation is: Cr2O72−+3SO32−+8H+→2Cr3++3SO42−+4H2OCr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O

Since the reaction medium is acidified (usually with sulphuric acid), the chromium(III) ions exist as chromium(III) sulphate, Cr2(SO4)3Cr_2(SO_4)_3 .

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from The d-and f-Block Elements. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThe d-and f-Block Elementsacidifiedsolutionformation

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