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NEET CHEMISTRYEasy

Which of the following is a paramagnetic compound?

A

N₂

B

H₂

C

Li₂

D

O₂

Step-by-Step Solution

  1. Definition: A substance is paramagnetic if it contains unpaired electrons. It is diamagnetic if all electrons are paired.
  2. Analyze N2N_2 (14 electrons): Configuration is 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. All electrons are paired. Diamagnetic.
  3. Analyze H2H_2 (2 electrons): Configuration is σ1s2\sigma_{1s}^2. All electrons paired. Diamagnetic.
  4. Analyze Li2Li_2 (6 electrons): Configuration is σ1s2σ1s∗2σ2s2\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2. All electrons paired. Diamagnetic.
  5. Analyze O2O_2 (16 electrons): Configuration is 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}). The last two electrons occupy the degenerate antibonding π∗\pi^* orbitals singly (Hund's Rule). Therefore, it has 2 unpaired electrons and is Paramagnetic.
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