Stability of the species $Li_2$, $Li_2^-$ and $Li_2^+$ increases in the order of:
A$Li_2 < Li_2^+ < Li_2^-$
B$Li_2^- < Li_2^+ < Li_2$
C$Li_2 < Li_2^- < Li_2^+$
D$Li_2^- < Li_2 < Li_2^+$
Answer & Solution
Correct answer: B. $Li_2^- < Li_2^+ < Li_2$
Use molecular orbital theory. For $Li_2$, the valence configuration is $(\sigma 2s)^2$, so bond order $=\frac{2-0}{2}=1$. For $Li_2^+$, one electron is removed from the bonding orbital, so bond order becomes $\frac{1-0}{2}=0.5$. For $Li_2^-$, the extra electron enters the antibonding $\sigma^{*}2s$ orbital, so bond order is also $\frac{2-1}{2}=0.5$, but it is less stable than $Li_2^+$ because the added antibonding electron increases repulsion and weakens the bond more effectively. Therefore stability increases as $Li_2^- < Li_2^+ < Li_2$.
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