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NEET CHEMISTRYEasy

A paramagnetic molecule among the following is:

A

O₂⁻

B

CN⁻

C

NO⁺

D

CO

Step-by-Step Solution

  1. Define Paramagnetism: Paramagnetism arises due to the presence of unpaired electrons in an atom, ion, or molecule [Class 12 Chemistry, Unit 4, Sec 4.3.9].
  2. Analyze Electron Counts:
  • O2−O_2^- (Superoxide Ion): Total electrons = 8(O)+8(O)+1(charge)=178 (O) + 8 (O) + 1 (charge) = 17. An odd number of electrons ensures at least one is unpaired. According to MOT, the configuration is 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}). It has 1 unpaired electron. Thus, it is Paramagnetic.
  • CN−CN^- (Cyanide Ion): Total electrons = 6(C)+7(N)+1=146 (C) + 7 (N) + 1 = 14. All electrons are paired (sigma1s2...sigma2pz2\\sigma_{1s}^2 ... \\sigma_{2p_z}^2). Diamagnetic.
  • NO+NO^+ (Nitrosonium Ion): Total electrons = 7(N)+8(O)−1=147 (N) + 8 (O) - 1 = 14. Isoelectronic with CN−CN^- and N2N_2. All electrons paired. Diamagnetic.
  • COCO (Carbon Monoxide): Total electrons = 6(C)+8(O)=146 (C) + 8 (O) = 14. Isoelectronic with N2N_2. All electrons paired. Diamagnetic.
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