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NEET CHEMISTRYChemical Bonding and Molecular StructureEasy

Question

The bond order of 1.5 is shown by:

A

O₂⁺

B

O₂⁻

C

O₂²⁻

D

O₂

Step-by-Step Solution

  1. Recall Bond Order Formula: Bond Order=12(Nb−Na)\text{Bond Order} = \frac{1}{2} (N_b - N_a) where NbN_b is the number of electrons in bonding molecular orbitals and NaN_a is the number of electrons in antibonding molecular orbitals.
  2. Analyze O2O_2 (16 electrons):
  • Configuration: σ1s2σ1s∗2σ2s2σ2s∗2σ2pz2(π2px2=π2py2)(π2px∗1=π2py∗1)\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}^{*1} = \pi_{2p_y}^{*1})
  • Nb=10,Na=6N_b = 10, N_a = 6. Bond Order = 12(10−6)=2\frac{1}{2}(10-6) = 2.
  1. Analyze O2+O_2^+ (15 electrons):
  • Remove 1 electron from antibonding π∗\pi^*. Nb=10,Na=5N_b = 10, N_a = 5.
  • Bond Order = 12(10−5)=2.5\frac{1}{2}(10-5) = 2.5.
  1. Analyze O2−O_2^- (Superoxide, 17 electrons):
  • Add 1 electron to antibonding π∗\pi^*. Nb=10,Na=7N_b = 10, N_a = 7.
  • Bond Order = 12(10−7)=1.5\frac{1}{2}(10-7) = 1.5.
  1. Analyze O22−O_2^{2-} (Peroxide, 18 electrons):
  • Add 2 electrons to antibonding π∗\pi^*. Nb=10,Na=8N_b = 10, N_a = 8.
  • Bond Order = 12(10−8)=1\frac{1}{2}(10-8) = 1.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Chemical Bonding and Molecular Structure. 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.

CHEMISTRYChemical Bonding and Molecular Structure

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