Chemistry - CHEMICAL BONDING Question with Solution | TestHub

(i) ${\mathrm{Ne}}_2\Rightarrow \left(\sigma 1{\mathrm{s}}^2\right)\left({\sigma }^{\ast }1{\mathrm{s}}^2\right)\left(\sigma 2{\mathrm{s}}^2\right)\left({\sigma }^{\ast }2{\mathrm{s}}^2\right)\left(\sigma 2{\mathrm{p}}_{\mathrm{z}}^2\right)\left(\pi 2{\mathrm{p}}_{\mathrm{x}}^2=\pi 2{\mathrm{p}}_{\mathrm{y}}^2\right)\left({\pi }^{\ast }2{\mathrm{p}}_{\mathrm{x}}^2={\pi }^{\ast }2{\mathrm{p}}_{\mathrm{y}}^2\right)\left({\sigma }^{\ast }2{\mathrm{p}}_{\mathrm{z}}^2\right)$ B.O. $=\frac{6-6}{2}=0$ (ii) ${\mathrm{F}}_2\Rightarrow \left(\sigma 1{\mathrm{s}}^2\right)\left({\sigma }^{\ast }1{\mathrm{s}}^2\right)\left(\sigma 2{\mathrm{s}}^2\right)\left({\sigma }^{\ast }2{\mathrm{s}}^2\right)\left(\sigma 2{\mathrm{p}}_{\mathrm{z}}^2\right)\left(\pi 2{\mathrm{p}}_{\mathrm{x}}^2=\pi 2{\mathrm{p}}_{\mathrm{y}}^2\right)\left({\pi }^{\ast }2{\mathrm{p}}_{\mathrm{x}}^2={\pi }^{\ast }2{\mathrm{p}}_{\mathrm{y}}^2\right)$ (iii) ${\mathrm{O}}_2^{\oplus }\Rightarrow \left(\sigma 1s^2\right)\left({\sigma }^{\ast }1s^2\right)\left(\sigma 2s^2\right)\left({\sigma }^{\ast }2s^2\right)\left(\sigma 2p_z^2\right)\left(\pi 2p_x^2=\pi 2p_y^2\right)\left({\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y\right)$ B.O. $=\frac{6-1}{2}=2.5$ ${\mathrm{O}}_2\Rightarrow \left(\sigma 1s^2\right)\left({\sigma }^{\ast }1s^2\right)\left(\sigma 2s^2\right)\left({\sigma }^{\ast }2s^2\right)\left(\sigma 2p_{\mathrm{z}}^2\right)\left(\pi 2p_{\mathrm{x}}^2=\pi 2p_{\mathrm{y}}^2\right)\left({\pi }^{\ast }2p_{\mathrm{x}}^1={\pi }^{\ast }2p_{\mathrm{y}}^1\right)$ B.O. $\frac{6-2}{2}=2$ (Bond order increases, Bond strength increases) (iv) Size of atom increases, Bond length increases Size of Li > B So, Bond length of ${\mathrm{Li}}_2>{\mathrm{B}}_2$