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, $Li _{2}$ is less sable than

$H _{2}$ (Bond energy of $Li _{2}=110 kJ mol ^{-2}+$ Bond energy of $H _{2}=438 \,kJ\,mol^{-1}$

Boron atom is smaller than lithium atom but larger than hydrogen atom. The $B _{2}$ is more stable than $Li _{2}$ but less stable than $H _{2}\left(\right.$ Bond energy of $\left. B _{2}=290 kJ mol ^{-1}\right)$