4. Chemical bonding and molecular structure — Chemistry STD 11 Science — Question

(b) $\mathrm{Li}_{2}-<\mathrm{Li}_{2}^{+}<\mathrm{Li}_{2}$

Explanation:

$\mathrm{Li}_{2}(3+3=6)=\sigma 1 \mathrm{s}^{2}, \quad \hat{\sigma} 1 \mathrm{s}^{2}, \sigma 2 \mathrm{s}^{2}$

Bond order $=\frac{N_{b}-N_{a}}{2}=\frac{4-2}{2}=1$

$\mathrm{Li}_{2}^{+}(3+3-1=5)=\sigma 1 \mathrm{s}^{2}, \dot{\sigma}^{2} 1 s^{2}, \cdot \sigma 2 s^{1}$

Bond order $=\frac{3-2}{2}=\frac{1}{2}=0.5$

Bond order $=\frac{4-3}{2}=\frac{1}{2}=0.5$

Stability order is $\mathrm{Li}_{2}>\mathrm{Li}_{2}^{+}>\mathrm{Li}_{2}^{-}$ (because Li $_{2}^{-}$ have more number of electrons in antibonding orbitals which destabilises the species).