Question

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

• A. $H_{2}=Li_{2}=B_{2}$
• B. $H_{2}>Li_{2}>B_{2}$
• C. $H_{2}>B_{2}>Li_{2}$
• D. $B_{2}>Li_{2}>H_{2}$

Answer 1

Answer: (C)

$H _{2}, Li _{2}, B _{2}$ are not equally stable. $Li$ atom is much larger in size than $H$-atom. The bond order in $Li _{2}$ is much larger than bond length in $H _{2}$. Moreover, $Li _{2}$ molecule has two electrons in the antibonding molecular orbital while $H _{2}$ has no electrons in the antiobinding molecular orbital. Thus, $L i_{2}$ is less sable than $ H _{2}\left(\right.$ Bond energy of $Li _{2}=110 \,kJ \,mol^{-2}+$ Bond energy of $ H _{2}=438 \,k$. $\left. mol ^{-1}\right)$
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
$H _{2}\left(\right.$ Bond energy of $\left. B _{2}=290 \,kJ\, mol ^{-1}\right)$