| ⇦ |
| ⇨ |
Oxidation number of chromium in Na₂Cr₂O₇ is
Options
(a) 2
(b) 4
(c) 3
(d) 6
Correct Answer:
6
Explanation:
Oxidation number of chromium in Na₂Cr₂O₇ is +6. Oxidation number of Na = +1. Oxidation number of O = -2. Therefore 2 + 2x -2 * 7 = 0.⇒ 2x – 12 = 0 , ⇒ x = +6.
Related Questions: - The type of hybridisation of boron in diboron is
- Each unit cell of NaCl consists of 6 chlorine atoms and
- Glycerol reacts with KHSO₄ to form
- In Kjeldahl’s mathod, the nitrogen present is estimated as
- How many coulombs are required for the oxidation of 1 mole H₂O to O₂
Topics: D and F Block Elements
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The type of hybridisation of boron in diboron is
- Each unit cell of NaCl consists of 6 chlorine atoms and
- Glycerol reacts with KHSO₄ to form
- In Kjeldahl’s mathod, the nitrogen present is estimated as
- How many coulombs are required for the oxidation of 1 mole H₂O to O₂
Topics: D and F Block Elements (91)
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