1. Electronic configuration of Li atom

Lithium has atomic number $3$.

$Li = 1s^2\,2s^1$

In molecular orbital theory, the core $1s$ electrons are usually ignored for bond order comparison, because they fill both bonding and antibonding MOs equally and cancel out.

So we focus on the valence $2s$ electrons.

2. MO scheme for $Li_2$ species

For lithium molecules, the relevant valence molecular orbitals are:

$\sigma(2s),\ \sigma^*(2s)$

Bond order is given by

$\text{Bond order} = \frac{N_b - N_a}{2}$

where:

• $N_b$ = number of electrons in bonding MOs
• $N_a$ = number of electrons in antibonding MOs

A species is generally stable if bond order $>0$.

3. For $Li_2^+$

Each Li contributes one valence electron, so $Li_2$ would have $2$ valence electrons. Since it is $Li_2^+$, one electron is removed.

Thus valence electrons in $Li_2^+$:

$2-1=1$

Electronic configuration:

$\sigma(2s)^1$

So,

• bonding electrons $N_b = 1$
• antibonding electrons $N_a = 0$

Therefore,

$\text{Bond order} = \frac{1-0}{2} = \frac{1}{2}$

Since bond order is positive, $Li_2^+$ is stable.

4. For $Li_2^-$

$Li_2$ has $2$ valence electrons, and one extra electron is added for $Li_2^-$.

Thus valence electrons in $Li_2^-$:

$2+1=3$

Electronic configuration:

$\sigma(2s)^2\,\sigma^*(2s)^1$

So,

• bonding electrons $N_b = 2$
• antibonding electrons $N_a = 1$

Therefore,

$\text{Bond order} = \frac{2-1}{2} = \frac{1}{2}$

Since bond order is positive, $Li_2^-$ is also stable.

5. Evaluate options

• A: $Li_2^+$ unstable and $Li_2^-$ stable $\rightarrow$ False
• B: $Li_2^+$ stable and $Li_2^-$ unstable $\rightarrow$ False
• C: Both are stable $\rightarrow$ True
• D: Both are unstable $\rightarrow$ False

6. Final answer

Both $Li_2^+$ and $Li_2^-$ have bond order $\frac{1}{2}$, so both are stable.