⇦ | ![]() | ⇨ |
Four diatomic species are listed below. Identilfy the correct order in which the bond order is increasing in them:
Options
(a) NO < O₂⁻ < C₂²⁻ < He₂⁺
(b) O₂⁻ < NO < C₂²⁻ < He₂⁺
(c) C₂²⁻ < He₂⁺ < O₂⁻ < NO
(d) He₂⁺ < O₂⁻ < NO < C₂²⁻
Correct Answer:
He₂⁺ < O₂⁻ < NO < C₂²⁻
Explanation:
Bond order = Nₙ – Nₘ / 2
He₂⁺ = σ(1s)² σ*(1s) B.O. = 0.5
O₂⁻ = KKσ(2s)² σ*(2s)² σ(2pₖ)²
π(2pᵢ)² π(2pⱼ)² π*(2pᵢ)² π*(2pⱼ)¹ B.O. = 1.5
NO = KKσ(2s)² σ*(2s)²π(2pᵢ)²
π(2pⱼ)² σ(2pₖ)² π*(2pᵢ)¹ B.O. = 2.5
C₂²⁻ = KKσ(2s)² σ*(2s)²π(2pᵢ)² π(2pⱼ)² σ(2pₖ)² B.O. = 3.0
Related Questions:
- In the Kjeldahl’s method for estimation of nitrogen present in a soil sample
- The chemical substance used for the treatments of
- The role of cryolite in the metallurgy of extraction of aluminium is to
- Which of the following C-H bonds has lowest bond dissociation energy
- The acidic, basic or amphoteric nature of Mn₂O₇, V₂O₅ and CrO are respectively
Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
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