Question

Stability of the species $ Li_2 , Li_2^- , \, \, and \, \, Li_2^+ $ increases in the order of

• A. $ Li < Li_2^+ < Li_2^- $
• B. $ Li_2^- < Li_2^+ < Li_2 $
• C. $ Li_2 < Li_2^- < Li_2^+ $
• D. $ Li_2^- < Li_2 < Li_2^+ $

Answer 1

The Correct Option is B

To determine the stability order of the diatomic species $ Li_2 , Li_2^- , \, \, and \, \, Li_2^+ $, we need to understand the concept of molecular orbital theory and how the number of electrons affects stability.

1. According to molecular orbital (MO) theory, the stability of a molecule is determined by the bond order, which is calculated as: \[\text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}\]
2. Let's consider each species:
   • $Li_2$: Lithium (Li) atomic number is 3, hence each Li atom contributes 3 electrons. The electronic configuration of Li is 1s^2\, 2s^1. In $Li_2$, there are a total of 6 electrons. The electrons fill the molecular orbitals as follows:
     \text{MO: } \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2}
     Bond Order = \frac{4-2}{2} = 1
   • $Li_2^+$: This species has one less electron than $Li_2$. So, it has 5 electrons. The electron configuration will be:
     \text{MO: } \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{1}
     Bond Order = \frac{3-2}{2} = 0.5
   • $Li_2^-$: This species has one more electron than $Li_2$. So, it has 7 electrons. The electron configuration will be:
     \text{MO: } \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*1}
     Bond Order = \frac{4-3}{2} = 0.5
3. Conclusion: Based on the bond order calculation, we can determine the relative stability of these species:
   • $Li_2^-$: Bond Order = 0.5
   • $Li_2^+$: Bond Order = 0.5
   • $Li_2$: Bond Order = 1
   Therefore, the stability order is: $ Li_2^- < Li_2^+ < Li_2 $.

Hence, the correct answer is $ Li_2^- < Li_2^+ < Li_2 $.