Stability of the species $$Li_2$$, $$Li_2^-$$ and $$Li_2^+$$ increases in the order of

For comparing the stability of homonuclear di-atomic species we use Molecular Orbital (MO) theory. According to this theory the stability is directly related to the bond order; the larger the bond order, the stronger and hence the more stable the molecule.

First we recall the expression for bond order:

$$\text{Bond order } (B.O.) \;=\;\dfrac{N_b - N_a}{2}$$

Here $$N_b$$ is the total number of electrons present in bonding molecular orbitals and $$N_a$$ is the total number of electrons present in antibonding molecular orbitals.

For species formed from lithium atoms we consider the following sequence of molecular orbitals formed from the 1s and 2s atomic orbitals:

$$\sigma(1s),\;\;\sigma^{\ast}(1s),\;\;\sigma(2s),\;\;\sigma^{\ast}(2s)$$

The 1s orbitals are inner (core) orbitals; nevertheless we must include them while counting electrons because they still contribute to the bond order expression.

Now we analyse each species one by one.

(i) The neutral molecule $$Li_2$$

Each lithium atom contributes three electrons, so the total number of electrons is

$$2 \times 3 = 6$$

Filling the molecular orbitals in ascending order of energy we obtain the configuration

$$\sigma(1s)^2\;\sigma^{\ast}(1s)^2\;\sigma(2s)^2$$

Counting electrons, we have

Bonding electrons $$N_b = 2 \;( \sigma 1s) + 2 \;(\sigma 2s) = 4$$

Antibonding electrons $$N_a = 2 \;(\sigma^{\ast} 1s) = 2$$

Hence

$$B.O.(Li_2) = \dfrac{4 - 2}{2} = 1$$

(ii) The anion $$Li_2^-$$

This species possesses one extra electron, so the total number of electrons is

$$6 + 1 = 7$$

The additional electron enters the next available orbital, namely $$\sigma^{\ast}(2s)$$, giving the configuration

$$\sigma(1s)^2\;\sigma^{\ast}(1s)^2\;\sigma(2s)^2\;\sigma^{\ast}(2s)^1$$

Now

Bonding electrons $$N_b = 2 \;(\sigma 1s) + 2 \;(\sigma 2s) = 4$$

Antibonding electrons $$N_a = 2 \;(\sigma^{\ast} 1s) + 1 \;(\sigma^{\ast} 2s) = 3$$

So

$$B.O.(Li_2^-) = \dfrac{4 - 3}{2} = 0.5$$

(iii) The cation $$Li_2^+$$

Here one electron is removed from the neutral molecule, leaving

$$6 - 1 = 5$$ electrons.

The electron is taken from the highest occupied bonding MO, $$\sigma(2s)$$, so the configuration becomes

$$\sigma(1s)^2\;\sigma^{\ast}(1s)^2\;\sigma(2s)^1$$

Counting electrons gives

Bonding electrons $$N_b = 2 \;(\sigma 1s) + 1 \;(\sigma 2s) = 3$$

Antibonding electrons $$N_a = 2 \;(\sigma^{\ast} 1s) = 2$$

Hence

$$B.O.(Li_2^+) = \dfrac{3 - 2}{2} = 0.5$$

We see that $$Li_2$$ has the highest bond order $$1$$, while both $$Li_2^-$$ and $$Li_2^+$$ possess a smaller bond order of $$0.5$$. Between the two charged species we must still decide which is less stable.

Observe that in $$Li_2^-$$ the extra electron occupies the high-energy antibonding orbital $$\sigma^{\ast}(2s)$$, whereas in $$Li_2^+$$ the deficiency is in the lower-energy bonding orbital $$\sigma(2s)$$. Introducing an electron into a high-energy antibonding level destabilises a molecule more than merely removing an electron from a lower-energy bonding level. Therefore, although the algebraic bond orders are equal, the actual energy of $$Li_2^-$$ is higher (and the molecule consequently less stable) than that of $$Li_2^+$$.

Combining all the above results, stability increases in the sequence

$$Li_2^- \;<\; Li_2^+ \;<\; Li_2$$

Hence, the correct answer is Option D.