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NEET CHEMISTRYEasy

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.
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