| ⇦ | 
| ⇨ |
The change in oxidation state of sulphur when sulphur dioxide is passed in an acidified K₂Cr₂O₇ solution is from
Options
(a) 4 to 0
(b) 4 to 2
(c) 4 to 6
(d) 6 to 4
Correct Answer:
4 to 6
Explanation:
Acidified K₂Cr₂O₇ solution oxidised SO₂ into Cr₂(SO₄)₃.
3SO₂ + K₂Cr₂O₇ + H₂O → K₂SO₄ + Cr₂(SO₄)₃ + H₂O
Oxidation of S =+4
Oxidation of (SO₄) = +6
Related Questions: - The change in optical rotation (with time) of freshy prepared solution of sugar
- Which gas is formed when CaC₂ reacts with H₂O
- Which of the following reaction is endothermic
- If an electron has spin quantum number of +1/2 and a magnetic quantum number
- Hydrogen can be fused to form helium at
Topics: Hydrogen and Redox Reactions
(174)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The change in optical rotation (with time) of freshy prepared solution of sugar
- Which gas is formed when CaC₂ reacts with H₂O
- Which of the following reaction is endothermic
- If an electron has spin quantum number of +1/2 and a magnetic quantum number
- Hydrogen can be fused to form helium at
Topics: Hydrogen and Redox Reactions (174)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply