Chemistry - Complete s-block Question with Solution | TestHub

Key Idea Molecules with zero bond order, do not exist. (a) ${\mathrm{Be}}_2^{+}(4+4-1=7)=$ $$\begin{array}{rl}& \sigma 1{\mathrm{s}}^2,\sigma 1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^1\\ & \mathrm{BO}=\frac{4-3}{2}=0.5\end{array}$$ (b) ${\mathrm{Be}}_2(4+4=8)=\sigma 1{\mathrm{s}}^2,\sigma 1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2$ $$\mathrm{BO}=\frac{4-4}{2}=0$$ (c) ${\mathrm{B}}_2(5+5=10)$ $$\begin{array}{c}=\sigma 1{\mathrm{s}}^2,\sigma 1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\pi 2p_{\mathrm{x}}^1\approx \pi 2{\mathrm{p}}_{\mathrm{y}}^1\\ \mathrm{BO}=\frac{6-4}{2}=1\end{array}$$ (d) $$\begin{array}{rl}& {\mathrm{Li}}_2(3+3=6)\\ & =\sigma 1{\mathrm{s}}^2,\sigma 1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2\\ & =\frac{4-2}{2}=1\end{array}$$ $\mathrm{BO}=\frac{4-2}{2}=1$ Thus, ${\mathrm{Be}}_2$ does not exist under normal conditions.