⇦ | ![]() | ⇨ |
Decreasing order of stability of O₂ , O₂⁻ , O₂⁺ and O₂²⁻ is
Options
(a) O₂²⁻ > O₂⁻ > O₂ >O₂⁺
(b) O₂> O₂⁺ > O₂²⁻ > O₂⁻
(c) O₂⁻ > O₂²⁻ > O₂⁺ >O₂
(d) O₂⁺ > O₂ > O₂⁻ > O₂²⁻
Correct Answer:
O₂⁺ > O₂ > O₂⁻ > O₂²⁻
Explanation:
According to molecular orbital theory as bond order decreases stability of the molecule decreases
Bond order =1/2 (N(b)-N(a))
Bond order for O₂⁺=1/2 (10-5)=2.5
Bond order for O₂=1/2 (10-6)=2
Bond order for O₂⁻=1/2 (10-7)=1.5
Bond order for O₂²⁻=1/2 (10-8)=1.0
hence the correct order is
O₂⁺ > O₂ > O₂⁻> O₂²⁻
Related Questions:
- As the nuclear charge increases from neon to calcium the orbital energy
- Which kind of isomerism is exhibited by octahedral [Co(NH₃)₄Br₂]Cl
- What is formed when formaldehyde reacts with KOH
- Assuming complete ionisation same moles of which of the following compounds
- Which of the following transition element shows the highest oxidation state
Topics: Chemical Bonding and Molecular Structure
(86)
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