# Why "Li"_2^+ is more stable than "Li"_2 ?

## According to J.D Lee, compounds with fraction bond number are unstable-- Li2+ BOND ORDER = 0.5 Li2 BOND ORDER =1 Hence Li2+ must be unstable than Li2 but then why Li2 is more stable than Li2+. Please explain reasons.

Aug 3, 2018

No... it's the other way around. ${\text{Li}}_{2}$ is more stable than ${\text{Li}}_{2}^{+}$, because the bond is (hypothetically) stronger (probably gas-phase).

Here we consider the molecular orbital diagram (MO) of ${\text{Li}}_{2}$:

The bond order can be calculated in a simple manner. Just take electrons that are in each MO, and

• for each electron in a bonding MO, it adds $0.5$ to the bond order, because more bonding character strengthens the bond...
• for each electron in an antibonding MO, it subtracts $0.5$ from the bond order, because more antibonding character weakens the bond...

Hence, the bond order of ${\text{Li}}_{2}$ is

$\frac{1}{2} + \frac{1}{2} - \frac{1}{2} - \frac{1}{2} + \frac{1}{2} = 1$,

indicating a single sigma bond (because they are in the ${\sigma}_{2 s}$ MO, a sigma orbital).

On the other hand, ${\text{Li}}_{2}^{+}$ has one less electron, and it must be from the ${\sigma}_{2 s}$, so it loses $0.5$ of a bond order. Thus,

color(blue)("Bond order" ("Li"_2^+) = 0.5)

And so, it has, hypothetically, half of a sigma bond. Clearly, half of a bond is less stable than one entire bond of the same type.