Question

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.

Answer 1

No... it's the other way around. `"Li"_2` is more stable than `"Li"_2^(+)`, because the bond is (hypothetically) stronger (probably gas-phase).

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Here we consider the molecular orbital diagram (MO) of `"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 `"Li"_2` is

`1/2 + 1/2 - 1/2 - 1/2 + 1/2 = 1`,

indicating a single sigma bond (because they are in the `sigma_(2s)` MO, a sigma orbital).

On the other hand, `"Li"_2^(+)` has one less electron, and it must be from the `sigma_(2s)`, 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.