Stability of the species Li_2, Li_2^-, and Li | vedclass

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(B) The stability of a molecule is directly proportional to its bond order. If bond orders are equal, the species with more electrons in antibonding o

Vedclass · Accepted Answer

$Li_2^- < Li_2^+ < Li_2$

Vedclass · Answer

(B) The stability of a molecule is directly proportional to its bond order. If bond orders are equal, the species with more electrons in antibonding orbitals is less stable.$Li_2$ ($6$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2$. Bond order $= \frac{4 - 2}{2} = 1.0$.$Li_2^+$ ($5$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^1$. Bond order $= \frac{3 - 2}{2} = 0.5$.$Li_2^-$ ($7$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^1$. Bond order $= \frac{4 - 3}{2} = 0.5$.Comparing $Li_2^+$ and $Li_2^-$, both have a bond order of $0.5$, but $Li_2^-$ has more electrons in the antibonding orbital $(\sigma^* 2s^1)$, making it less stable than $Li_2^+$.Therefore, the stability order is $Li_2^- < Li_2^+ < Li_2$.

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

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