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NEET CHEMISTRYOrganic Chemistry – Some Basic Principles and Techniques Medium

Question

The pair of species with the same bond order is:

A

O₂²⁻, B₂

B

O₂⁺, NO⁺

C

NO, CO

D

N₂, O₂

Step-by-Step Solution

  1. Formula: Bond Order=12(Nb−Na)\text{Bond Order} = \frac{1}{2} (N_b - N_a), where NbN_b is the number of bonding electrons and NaN_a is the number of antibonding electrons.
  2. Analyze Option A (O22−,B2O_2^{2-}, B_2):
  • O22−O_2^{2-} (Peroxide Ion): Total electrons = 16+2=1816 + 2 = 18. Configuration: σ1s2σ1s∗2σ2s2σ2s∗2σ2pz2(π2px2=π2py2)(π2px∗2=π2py∗2)\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}). Nb=10,Na=8N_b = 10, N_a = 8. Bond Order=0.5(10−8)=1\text{Bond Order} = 0.5(10 - 8) = 1.
  • B2B_2 (Boron Molecule): Total electrons = 1010. Configuration: σ1s2σ1s∗2σ2s2σ2s∗2(π2px1=π2py1)\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} (\pi_{2p_x}^1 = \pi_{2p_y}^1). Nb=6,Na=4N_b = 6, N_a = 4. Bond Order=0.5(6−4)=1\text{Bond Order} = 0.5(6 - 4) = 1.
  • Result: Both have a Bond Order of 1.
  1. Analyze other options:
  • O2+O_2^+ (BO = 2.5) vs NO+NO^+ (BO = 3).
  • NONO (BO = 2.5) vs COCO (BO = 3).
  • N2N_2 (BO = 3) vs O2O_2 (BO = 2).

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Organic Chemistry – Some Basic Principles and Techniques . Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYOrganic Chemistry – Some Basic Principles and Techniques species

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