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NEET CHEMISTRYMedium

The correct order of increasing bond order among the given is:

A

O₂⁻ < NO < C₂²⁻ < He₂⁺

B

NO < C₂²⁻ < O₂⁻ < He₂⁺

C

C₂²⁻ < He₂⁺ < NO < O₂⁻

D

He₂⁺ < O₂⁻ < NO < C₂²⁻

Step-by-Step Solution

  1. Formula: Bond Order (B.O.) = 12(Nb−Na)\frac{1}{2} (N_b - N_a).

  2. Analyze He₂⁺:

  • Total electrons: 2+2−1=32 + 2 - 1 = 3.
  • Configuration: σ1s2σ1s∗1\sigma_{1s}^2 \sigma_{1s}^{*1}.
  • B.O. = 12(2−1)=0.5\frac{1}{2} (2 - 1) = 0.5.
  1. Analyze O₂⁻ (Superoxide ion):
  • Total electrons: 8+8+1=178 + 8 + 1 = 17.
  • Configuration: σ1s2σ1s∗2σ2s2σ2s∗2σ2pz2(π2px2=π2py2)(π2px∗2=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}^{*2} = \pi_{2p_y}^{*1}).
  • Nb=10,Na=7N_b = 10, N_a = 7.
  • B.O. = 12(10−7)=1.5\frac{1}{2} (10 - 7) = 1.5.
  1. Analyze NO:
  • Total electrons: 7+8=157 + 8 = 15.
  • Similar to O2+O_2^+. Configuration ends in π∗\pi^* having 1 electron.
  • Nb=10,Na=5N_b = 10, N_a = 5.
  • B.O. = 12(10−5)=2.5\frac{1}{2} (10 - 5) = 2.5.
  1. Analyze C₂²⁻ (Acetylide ion):
  • Total electrons: 6+6+2=146 + 6 + 2 = 14. Isoelectronic with N2N_2.
  • Configuration: σ1s2σ1s∗2σ2s2σ2s∗2(π2px2=π2py2)σ2pz2\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} (\pi_{2p_x}^2 = \pi_{2p_y}^2) \sigma_{2p_z}^2.
  • Nb=10,Na=4N_b = 10, N_a = 4.
  • B.O. = 12(10−4)=3.0\frac{1}{2} (10 - 4) = 3.0.
  1. Conclusion: Increasing order is 0.5<1.5<2.5<3.00.5 < 1.5 < 2.5 < 3.0, which corresponds to He2+<O2−<NO<C22−He_2^+ < O_2^- < NO < C_2^{2-}.
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