Question

According to molecular orbital theory, which of the following is true with respect to $L{i}_2^{+}$ and $L{i}_2^{-}$?
(A) Both are unstable
(B) $L{i}_2^{+}$ is unstable and $L{i}_2^{-}$ is stable
(C) $L{i}_2^{+}$ is stable and $L{i}_2^{-}$ is unstable.
(D) Both are stable.

Answer 1

Hint: Calculate the number of electrons individually in both $L{i}_2^{+}$ and $L{i}_2^{-}$. 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, $L{i}_2^{+}$ will contain 5 electrons. Similarly, $L{i}_2^{-}$ will have 7 electrons.
Now, for writing the electronic configuration of $L{i}_2^{+}$ and $L{i}_2^{-}$ according to molecular orbital theory, you must know the increasing order of various molecular orbitals and it is: $\sigma 1s<\sigma \ast 1s<\sigma 2s<\sigma \ast 2s<\pi 2p_x=\pi 2p_y<2\sigma p_z<\pi \ast 2p_x=\pi \ast 2p_y<\sigma \ast 2p_z$
So, electronic configuration of $L{i}_2^{+}$ having 6 electrons: $$\sigma 1s^2,\sigma \ast 1s^2,\sigma 2s^1$$
Electronic configuration of $L{i}_2^{-}$ having 7 electrons: $\sigma 1s^2,\sigma \ast 1s^2,\sigma 2s^2,\sigma \ast 2s^1$
Thus, we can see, in the $L{i}_2^{+}$ molecule there are three electrons in the bonding orbitals which is more than two electrons in the antibonding orbital. Therefore, $L{i}_2^{+}$ is stable according to MOT.
And, in $L{i}_2^{-}$, there are 4 electrons in the bonding orbitals which is more than three electrons in the antibonding orbitals. Thus, $L{i}_2^{-}$ is stable according to MOT.
Therefore in conclusion, Both $L{i}_2^{+}$ and $L{i}_2^{-}$ are stable.

Hence, option (D) is correct.

Note: According to MOT, bond order = ${\displaystyle \frac{N_b-N_a}{2}}$
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.