Question

According to molecular orbital theory, which of the following is true with respect to $Li_2^ +$ and $Li_2^ -$?
(A) Both are unstable
(B) $Li_2^ +$ is unstable and $Li_2^ -$ is stable
(C) $Li_2^ +$ is stable and $Li_2^ -$ is unstable.
(D) Both are stable.

Answer 1

Calculate the number of electrons individually in both $Li_2^ +$ and $Li_2^ -$. Then, write their electronic configuration according to molecular orbital theory. Lesser the number of electrons in the antibonding orbitals, greater is the stability of the molecule.

Complete step by step solution:
According to molecular orbital theory (MOT), from the electronic configuration of the molecule, it is possible to get information about the stability of the molecule as discussed below:
-If the number of electrons in the bonding orbitals are greater than those in antibonding electrons, then the molecule is stable.
-If the number of electrons in the bonding orbitals are less than those in antibonding electrons, then the molecule is unstable.
Li contains 3 electrons and thus, two lithium molecules will have 6 electrons. Consequently, $Li_2^ +$ will contain 5 electrons. Similarly, $Li_2^ -$ will have 7 electrons.
Now, for writing the electronic configuration of $Li_2^ +$ and $Li_2^ -$ according to molecular orbital theory, you must know the increasing order of various molecular orbitals and it is: $\sigma 1s < \sigma *1s < \sigma 2s < \sigma *2s < \pi 2{p_x} = \pi 2{p_y} < 2\sigma {p_z} < \pi *2{p_x} = \pi *2{p_y} < \sigma *2{p_z}$
So, electronic configuration of $Li_2^ +$ having 6 electrons: $\sigma 1{s^2},\sigma *1{s^2},\sigma 2{s^1}$
Electronic configuration of $Li_2^ -$ having 7 electrons: $\sigma 1{s^2},\sigma *1{s^2},\sigma 2{s^2},\sigma *2{s^1}$
Thus, we can see, in the $Li_2^ +$ molecule there are three electrons in the bonding orbitals which is more than two electrons in the antibonding orbital. Therefore, $Li_2^ +$ is stable according to MOT.
And, in $Li_2^ -$, there are 4 electrons in the bonding orbitals which is more than three electrons in the antibonding orbitals. Thus, $Li_2^ -$ is stable according to MOT.
Therefore in conclusion, Both $Li_2^ +$ and $Li_2^ -$ are stable.

Hence, option (D) is correct.

Note: According to MOT, bond order = $\dfrac{{{N_b} - {N_a}}}{2}$
Where ${N_b}$ are the bonding electrons and ${N_a}$ are the antibonding electrons.
A positive bond order (i.e, ${N_b}$ $>$ ${N_a}$) means a stable molecule while a negative bond order (i.e, ${N_b}$ $<$ ${N_a}$) means an unstable molecule.