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

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Question

(B) The stability of a molecular species is directly proportional to its bond order. The electronic configuration of $Li_2$ ($6$ electrons) is $\sigma

Vedclass · Accepted Answer

$Li_2^- < Li_2^+ < Li_2$

Vedclass · Answer

(B) The stability of a molecular species is directly proportional to its bond order. The electronic configuration of $Li_2$ ($6$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2$. Bond order $= (4-2)/2 = 1$. The electronic configuration of $Li_2^+$ ($5$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^1$. Bond order $= (3-2)/2 = 0.5$. The electronic configuration of $Li_2^-$ ($7$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^1$. Bond order $= (4-3)/2 = 0.5$. Comparing $Li_2^+$ and $Li_2^-$,$Li_2^+$ is more stable than $Li_2^-$ because $Li_2^-$ has an electron in the antibonding $\sigma^* 2s$ orbital,which decreases stability. Thus,the order of stability 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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