Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Bond order of the molecules : $${\mathrm{O}}_2^{-}:KK(\sigma 2s{)}^2{\left({\sigma }_2\ast \right)}^2{\left(\sigma 2p_{\mathrm{z}}\right)}^2{\left(\pi 2p_{\mathrm{x}}\right)}^2{\left(\pi 2p_{\mathrm{y}}\right)}^2{\left({\pi }^{\ast }2p_x\right)}^2{\left({\pi }^{\ast }2p_y\right)}^1$$ Bond order of ${\mathrm{O}}_2^{-}=\frac{1}{2}(8-5)=1.5$ ${\mathrm{He}}_2^{+}:(\sigma 1s{)}^2{\left({\sigma }^{\ast }1s\right)}^1$ Bond order of ${\mathrm{He}}_2^{+}=\frac{1}{2}(2-1)=0.5$ $${\mathrm{C}}_2:KK(\sigma 2s{)}^2{\left({\sigma }^{\ast }2s\right)}^2{\left(\pi 2p_{\mathrm{x}}\right)}^2{\left(\pi 2p_{\mathrm{y}}\right)}^2$$ Bond order $=\frac{1}{2}(6-2)=2.0$ $$\text{ NO : }KK(\sigma 2s{)}^2{\left({\sigma }^{\ast }2s\right)}^2{\left(\sigma 2p_{\mathrm{z}}\right)}^2{\left(\pi 2^2{p}_{\mathrm{x}}\right)}^2{\left(\pi 2^2{p}_{\mathrm{y}}\right)}^2{\left(\pi \ast 2p_{\mathrm{x}}\right)}^1$$ Bond order $=\frac{1}{2}(8-3)=2.5$