Chemistry - CHEMICAL BONDING Question with Solution | TestHub

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) ${\mathrm{N}}_2(\mathrm{Z}=14)-\sigma {\mathrm{ls}}^2,{\sigma }^{\ast }1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\sigma {}^{\ast }2{\mathrm{s}}^2,\pi 2{\mathrm{p}}_{\mathrm{x}}^2$ $=\pi 2{\mathrm{p}}_{\mathrm{y}}^2,\sigma 2{\mathrm{p}}_{\mathrm{z}}^2$ It is a diamagnetic molecule due to absence of unpaired electron. (2) ${\mathrm{H}}_2(\mathrm{Z}=2)-1{\sigma }^2$ It is a diamagnetic molecule. (3) ${\mathrm{Li}}_2(\mathrm{Z}=6)-\sigma {\mathrm{ls}}^2,\sigma {}^{\ast }1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2$ It is a diamagnetic molecule. (4) ${\mathrm{O}}_2(\mathrm{Z}=16)-\sigma {\mathrm{ls}}^2,\sigma {}^{\ast }1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\sigma \ast 2{\mathrm{s}}^2$, $$\sigma 2p_z^2,\pi 2p_x^2=\pi 2p_y^2,\pi \ast 2p_x^1=\pi \ast 2p_y^1$$ It is a paramagnetic molecule due to presence of unpaired electrons in each orbital of degenerate levels. Thus, option (4) is correct.