1. Write the molecular orbital configuration for lithium dimer species

Lithium has atomic configuration: $Li: 1s^2 2s^1$

For bonding in $Li_2$, only the valence $2s$ electrons are considered. The relevant molecular orbitals are: $\sigma(2s),\ \sigma^*(2s)$

2. Now fill electrons for each species

(i) $Li_2$

Total valence electrons $= 2$

Configuration: $\sigma(2s)^2$

Bond order: $\text{B.O.} = \frac{N_b - N_a}{2} = \frac{2-0}{2}=1$

(ii) $Li_2^+$

Total valence electrons $= 1$

Configuration: $\sigma(2s)^1$

Bond order: $\text{B.O.} = \frac{1-0}{2}=\frac{1}{2}$

(iii) $Li_2^-$

Total valence electrons $= 3$

Configuration: $\sigma(2s)^2\sigma^*(2s)^1$

Bond order: $\text{B.O.} = \frac{2-1}{2}=\frac{1}{2}$

3. Compare stability

Greater bond order generally means greater stability.

So clearly, $Li_2 \text{ is more stable than } Li_2^+ \text{ and } Li_2^-$

Now compare $Li_2^+$ and $Li_2^-$. Both have bond order $\frac12$, so we compare them by electron placement:

• In $Li_2^+$, there is no electron in antibonding orbital.
• In $Li_2^-$, one electron is present in antibonding orbital $\sigma^*(2s)$, which reduces stability.

Hence, $Li_2^- < Li_2^+ < Li_2$

4. Match with options

This corresponds to Option C.