$H_{2}, L i_{2}, B_{2}$ each has bond order equal to 1, the order of their stability is :

H_{2} = L i_{2} = B_{2}

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H_{2} > L i_{2} > B_{2}

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H_{2} > B_{2} > L i_{2}

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B_{2} > L i_{2} < H_{2}

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Solution

The correct option is C $H_{2} > B_{2} > L i_{2}$
$H_{2}, L i_{2}, B_{2}$ are not equal stable. $L i$ atom is much larger in size than $H - a t o m$ . The bond length in $L i_{2}$ is much larger than bond length in $H_{2}$ . Moreover, $L i_{2}$ molecule has two electrons in the antibonding molecular orbital while $H_{2}$ has no electrons in the antibonding molecular orbital. Thus, $L i_{2}$ is less stable than $H_{2}$ (Bond energy of $L i_{2} = 110 k J m o l^{- 1} +$ Bond energy of $H_{2} = 438 k J m o l^{- 1})$ . Boron atom is smaller than lithium atom but larger than hydrogen atom.The $B_{2}$ is more stable than $L i_{2}$ but less stable than $H_{2}$ (Bond energy of $B_{2} = 290 k J m o l^{- 1}$ ).

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