Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: According to molecular orbital theory, which of the following is true with respect to ${\text{Li}}_2^{+}$ and ${\text{Li}}_2^{-}$?

Molecular Orbital Theory (MOT) is a powerful tool that allows us to predict the existence and relative stability of molecules and ions. In this problem, we are tasked with comparing the stability of two lithium diatomic ions: ${\text{Li}}_2^{+}$ and ${\text{Li}}_2^{-}$.

To do this, we must construct their electronic configurations and calculate their bond orders. But what happens when the bond orders are identical? Let's dive into the fascinating world of molecular orbitals to find out.

First, let's determine the total number of electrons for each species. A neutral lithium atom ($\text{Li}$) has an atomic number of $Z=3$, meaning it has 3 electrons.

For the ${\text{Li}}_2^{+}$ ion, we have two lithium atoms, which would normally give us $3+3=6$ electrons. However, the positive charge indicates the loss of one electron. Therefore, the total number of electrons in ${\text{Li}}_2^{+}$ is $6-1=5$.

For the ${\text{Li}}_2^{-}$ ion, we again start with 6 electrons from the two lithium atoms. The negative charge indicates the gain of one extra electron. Thus, the total number of electrons in ${\text{Li}}_2^{-}$ is $6+1=7$.

Now, we apply the Aufbau principle to fill the molecular orbitals in order of increasing energy: $\sigma 1s<{\sigma }^{\ast }1s<\sigma 2s<{\sigma }^{\ast }2s$.

For ${\text{Li}}_2^{+}$ (5 electrons), the electronic configuration is: $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^1$

The bond order (BO) is calculated using the formula: where $N_b$ is the number of bonding electrons and $N_a$ is the number of anti-bonding electrons.

For ${\text{Li}}_2^{+}$, we have 3 bonding electrons (in $\sigma 1s$ and $\sigma 2s$) and 2 anti-bonding electrons (in ${\sigma }^{\ast }1s$).

Now, let's look at ${\text{Li}}_2^{-}$ (7 electrons). Its electronic configuration is: $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^1$

For ${\text{Li}}_2^{-}$, we have 4 bonding electrons (in $\sigma 1s$ and $\sigma 2s$) and 3 anti-bonding electrons (in ${\sigma }^{\ast }1s$ and ${\sigma }^{\ast }2s$).

Both ions have a bond order of $0.5$. A positive bond order indicates that both species can theoretically exist. However, the question asks us to compare their relative stabilities.

Here is the crucial catch: When two species have the exact same bond order, their stability is determined by the number of electrons residing in the anti-bonding orbitals ($N_a$). Anti-bonding electrons inherently destabilize the molecule because they occupy higher energy states that counteract the attractive forces of the bonding electrons.

Let's compare the anti-bonding electrons: - For ${\text{Li}}_2^{+}$, $N_a=2$. - For ${\text{Li}}_2^{-}$, $N_a=3$.

Since ${\text{Li}}_2^{-}$ has more anti-bonding electrons, it experiences greater internal repulsion and destabilization. Therefore, ${\text{Li}}_2^{+}$ is more stable than ${\text{Li}}_2^{-}$.

This makes option (d) the correct answer. Always remember this tie-breaker rule; it is a classic trap in competitive exams!