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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: - Which one of the following is true in electrolytic refining
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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
- Which one of the following is true in electrolytic refining
- Which one of the following electrolytes has the same value of van’t Hoff’s factor
- Acetaldehyde can not show
- An increase in pressure would favour which of the following reaction?
- The solubility of CuBr is 2 ˣ 10⁻⁴ mol/L at 25⁰C. The Ksp value for CuBr is
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
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