Which one is correct for stability order of given diatomic molecules?

H_{2} > B_{2} > {L i}_{2}

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H_{2} > {L i}_{2} > B_{2}

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B_{2} > {L i}_{2} > H_{2}

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{L i}_{2} > B_{2} > H_{2}

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Solution

The correct option is A $H_{2} > B_{2} > {L i}_{2}$
$H_{2} > B_{2} > {L i}_{2}$ is correct order of stability.
$H_{2}$ has no antibonding electron and thus most stable. Also ${L i}_{2}$ and $B_{2}$ have $2$ and four antibonding electrons, however small size of $B_{2}$ leads to more stability than ${L i}_{2}$ .
$H_{2}$ ${L i}_{2}$ $B_{2}$
B.E $438$ $110$ $290$ $k J {m o l}^{- 1}$

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