NEET UG 2019 — Chemistry Inorganic Chemistry

Key Idea If all the electron in a molecule are paired, the molecule is diamagnetic and if there are unpaired electrons in a molecule, it is paramagnetic. Molecular orbital configuration of given molecules are as follows: (1) $\mathrm{N}_2(\mathrm{Z}=14)-\sigma \mathrm{ls}^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \sigma{ }^* 2 \mathrm{~s}^2, \pi 2 \mathrm{p}_{\mathrm{x}}^2$N2​(Z=14)−σls2,σ∗1 s2,σ2 s2,σ∗2 s2,π2px2​ $=\pi 2 \mathrm{p}_{\mathrm{y}}^2, \sigma 2 \mathrm{p}_{\mathrm{z}}^2$=π2py2​,σ2pz2​ It is a diamagnetic molecule due to absence of unpaired electron. (2) $\mathrm{H}_2(\mathrm{Z}=2)-1 \sigma^2$H2​(Z=2)−1σ2 It is a diamagnetic molecule. (3) $\mathrm{Li}_2(\mathrm{Z}=6)-\sigma \mathrm{ls}^2, \sigma{ }^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2$Li2​(Z=6)−σls2,σ∗1 s2,σ2 s2 It is a diamagnetic molecule. (4) $\mathrm{O}_2(\mathrm{Z}=16)-\sigma \mathrm{ls}^2, \sigma{ }^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \sigma * 2 \mathrm{~s}^2$O2​(Z=16)−σls2,σ∗1 s2,σ2 s2,σ∗2 s2, $$\sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2, \pi * 2 p_x^1=\pi * 2 p_y^1$$σ2pz2​,π2px2​=π2py2​,π∗2px1​=π∗2py1​ It is a paramagnetic molecule due to presence of unpaired electrons in each orbital of degenerate levels. Thus, option (4) is correct.