Question

Which one is correct for stability order of given diatomic molecules?
A. $H_2>B_2>L{i}_2$
B. $H_2>L{i}_2>B_2$
C. $B_2>L{i}_2>H_2$
D. $L{i}_2>B_2>H_2$

Answer 1

Hint: To determine the stability of the given diatomic molecules we will use the MOT theory. According to which atomic orbitals combine to form molecular orbital. The molecular orbitals are of three types, bonding, non-bonding and antibonding. Stability is directly proportional to the electron density in bonding orbital and inversely proportional to the electron density in antibonding orbitals.

Complete answer:
Atoms react to form molecules. As the molecules form, atomic orbitals combine to form molecular orbitals. Then the total electrons of the molecules get arranged in these molecular orbitals. The diagram which shows the formation of molecular orbital, the arrangement of the molecular orbitals and electrons in these molecular orbitals is known as the molecular orbital diagram.
Two atomic orbital combines to form two molecular orbitals. Out of these two molecular orbital one is known as bonding which has low energy and the second is known as antibonding which has high energy. s-and one p-orbital forms sigma bond and the remaining two p-orbitals form pi bonding.
Due to high energy, the electron in antibonding decreases the stability of the molecules.
So, we will write the molecular electronic configuration of each given atom to determine the antibonding orbitals.
$H_2={\left(1s\sigma \right)}^2{\left(1s{\sigma }^{\ast }\right)}^0$
$L{i}_2={\left(1s\sigma \right)}^2{\left(1s{\sigma }^{\ast }\right)}^2{\left(2s\sigma \right)}^2{\left(2s{\sigma }^{\ast }\right)}^0$
$$B_2={\left(1s\sigma \right)}^2{\left(1s{\sigma }^{\ast }\right)}^2{\left(2s\sigma \right)}^2{\left(2s{\sigma }^{\ast }\right)}^2{\left(2p_x\pi \gg 2p_y\pi \right)}^2\left(2p_Z\sigma \right){\left(2p_x{\pi }^{\ast }\gg 2p_y{\pi }^{\ast }\right)}^2\left(2p_Z{\sigma }^{\ast }\right)$$
So, from the above molecular electronic configuration we can say the $H_2$has no antibonding electrons, $L{i}_2$ has two antibonding electrons and $$B_2$$has four antibonding electrons.
As the stability is inversely proportional to the antibonding electrons so, the stability order should be$H_2>L{i}_2>B_2$ but the boron atom is of small size so, due to high effective nuclear charge it is more stable than lithium molecule so, the correct stability order is, $H_2>B_2>L{i}_2$.

Therefore, option (A) $H_2>B_2>L{i}_2$ is correct.

Note: The MOT is used to determine the many physical and chemical properties of the molecule such as bond order, stability, magnetic property, etc. stability is also directly proportional to the bond order which is determined as, Bonding electron – Anti bonding electrons / $2$ .

Here, the given species have the same bond order so the stability is compared on the basis of antibonding electrons. The highest occupied molecular orbital is known as HOMO and the lowest unoccupied molecular orbital is known as LUMO.