Statement-I : $\mathrm{O}_{2}$ molecule is diamagnetic while $\mathrm{C}_{2}$ molecule is paramagnet

Question

Statement-I : $\mathrm{O}_{2}$ molecule is diamagnetic while $\mathrm{C}_{2}$ molecule is paramagnetic in nature.\ Statement-II : Bond order of $\mathrm{O}_{2}$ molecule is 1.5 and that of $\mathrm{C}_{2}$ molecule is 2.5.

Professor PC Thomas Classes · Answer

(c) : $\mathrm{O}_{2}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 sight)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 sight)^{2}\left(\sigma 2 p_{z}{ }^{2}ight)$\ $\left(\pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}ight)\left(\pi^{\star} 2 p_{x}^{1}=\pi^{\star} 2 p_{y}^{1}ight)$ 
Bond order $=\frac{1}{2}\left(N_{b}-N_{a}ight)=\frac{1}{2}(10-6)=2$\ It contains two unpaired electrons, hence it is paramagnetic in nature.\ $\mathrm{C}_{2}:(\sigma 1 s)^{2}(\sigma * 1 s)^{2}(\sigma 2 s)^{2}(\sigma * 2 s)^{2}\left(\pi 2 p_{x}{ }^{2}=\pi 2 p_{y}{ }^{2}ight)$\ Bond order $=\frac{1}{2}(8-4)=2$\ All electrons are paired hence it is diamagnetic in nature.\