Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Stability of the species ${\text{Li}}_2$, ${\text{Li}}_2^{-}$ and ${\text{Li}}_2^{+}$ increases in the order of

Decoding Molecular Orbitals

When atoms combine to form molecules, their atomic orbitals merge to create molecular orbitals. Some of these new orbitals are bonding (they hold the atoms together), while others are antibonding (they push the atoms apart). To determine how stable a molecule is, we use a simple yet powerful tool called the Bond Order.

The formula for Bond Order is: where $N_b$ is the number of electrons in bonding orbitals, and $N_a$ is the number of electrons in antibonding orbitals.

Let's apply this to our three lithium species: ${\text{Li}}_2$, ${\text{Li}}_2^{+}$, and ${\text{Li}}_2^{-}$.

A neutral lithium atom has an atomic number of 3, meaning it has 3 electrons. Therefore, the diatomic molecule ${\text{Li}}_2$ has a total of $3+3=6$ electrons.

Filling these into the molecular orbitals from lowest energy to highest, we get the electronic configuration:

Counting the electrons, we have $N_b=4$ (from the $\sigma 1s$ and $\sigma 2s$ orbitals) and $N_a=2$ (from the ${\sigma }^{\ast }1s$ orbital).

Plugging these into our formula:

A bond order of 1 indicates a stable, single covalent bond.

Now, let's look at the cation, ${\text{Li}}_2^{+}$. The positive charge indicates the loss of one electron, leaving us with $6-1=5$ electrons. This electron is removed from the highest occupied molecular orbital, which is the $\sigma 2s$ orbital.

The new configuration is:

Now, $N_b=3$ and $N_a=2$. Calculating the bond order:

The bond order has decreased, meaning the cation is less stable than the neutral molecule.

Finally, consider the anion, ${\text{Li}}_2^{-}$. The negative charge means we have gained an extra electron, giving us $6+1=7$ electrons. This extra electron must go into the next available empty orbital, which is the antibonding ${\sigma }^{\ast }2s$ orbital.

The configuration becomes:

Here, $N_b=4$ and $N_a=3$. Let's calculate the bond order:

We have a situation! Both ${\text{Li}}_2^{+}$ and ${\text{Li}}_2^{-}$ have a bond order of $0.5$. How do we determine which one is more stable?

This is where a crucial rule of Molecular Orbital Theory comes into play: When bond orders are equal, the species with more electrons in antibonding orbitals is less stable.

Antibonding electrons actively destabilize the molecule by increasing electron-electron repulsion and pulling electron density away from the internuclear region. Since ${\text{Li}}_2^{-}$ has 3 antibonding electrons compared to ${\text{Li}}_2^{+}$'s 2 antibonding electrons, ${\text{Li}}_2^{-}$ experiences more internal repulsion.

Therefore, ${\text{Li}}_2^{-}$ is the least stable of the three. The correct increasing order of stability is: