Arrange the following species in order of increasing stability: $L i_{2}, L i_{2}^{+}, L i_{2}^{-}$

L i_{2}^{+} < L i_{2}^{-} < L i_{2}

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L i_{2}^{-} < L i_{2} < L i_{2}^{+}

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L i_{2} < L i_{2}^{+} = L i_{2}^{-}

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L i_{2}^{+} = L i_{2}^{-} < L i_{2}

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Solution

The correct option is D $L i_{2}^{+} = L i_{2}^{-} < L i_{2}$
The correct order of stability is $L i_{2}^{+} = L i_{2}^{-} < L i_{2}$
$L i_{2}$ molecule has MO electronic configuration of $K K (2 s)^{2}$ . Its bond order is $\frac{N_{b} - N_{a}}{2} = \frac{2 - 0}{2} = 2$
$L i_{2}^{+}$ molecule has MO electronic configuration of $K K (2 s)^{1}$ . Its bond order is $\frac{N_{b} - N_{a}}{2} = \frac{1 - 0}{2} = 0.5$
$L i_{2}^{-}$ molecule has MO electronic configuration of $K K (2 s)^{2} (2 s^{*})^{1}$ . Its bond order is $\frac{N_{b} - N_{a}}{2} = \frac{2 - 1}{2} = 0.5$
Higher is the bond order, higher is the stability.

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Stability of the species $L i_{2}$ , ${L i_{2}}^{-}$ and ${L i_{2}}^{+}$ increases in the order of

L i_{2}

L i_{2}^{-}

L i_{2}^{+}

L i_{2}

L i_{2}^{-}

L i_{2}^{+}

H_{2}, L i_{2}, B_{2}

Correct order of radius of the $1^{s t}$ orbit of $H$ , $H e^{+}, L i^{2 +}, B e^{3 +}$ is:

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