Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The stability of a diatomic species is directly related to its bond order. A higher bond order indicates greater stability.

1. Li₂: 6 electrons, configuration $(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2$. Bond order = $(4-2)/2=1$.

2. Li₂⁺: 5 electrons, configuration $(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^1$. Bond order = $(3-2)/2=0.5$.

3. Li₂⁻: 7 electrons, configuration $(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^1$. Bond order = $(4-3)/2=0.5$.

Since Li₂ has a bond order of 1, and Li₂⁺ and Li₂⁻ both have a bond order of 0.5, Li₂ is the most stable. To differentiate between Li₂⁺ and Li₂⁻, we consider the electron in the antibonding orbital. Li₂⁻ has one electron in the ${\sigma }^{\ast }2s$ antibonding orbital, which destabilizes the molecule more than the removal of an electron from the bonding orbital in Li₂⁺. Therefore, Li₂⁺ is more stable than Li₂⁻.

The order of increasing stability is Li₂⁻ < Li₂⁺ < Li₂.

The final answer is $\boxed{B}$.