Question

$H_2,L{i}_2,B_2$ each has bond order equal to 1, the order of their stability is :

• A. $H_2=L{i}_2=B_2$
• B. $H_2>L{i}_2>B_2$
• C. $H_2>B_2>L{i}_2$
• D. $B_2>L{i}_2<H_2$

Answer 1

The correct option is C $H_2>B_2>L{i}_2$

$H_2,L{i}_2,B_2$ are not equal stable. $Li$ atom is much larger in size than $H-atom$. 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 $L{i}_2=110kJmo{l}^{-1}+$ Bond energy of $H_2=438kJmo{l}^{-1})$. 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 $B_2=290kJmo{l}^{-1}$).