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

Question

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.

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Answer 1 · 2018-08-03T21:25:35

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

Here we consider the molecular orbital diagram (MO) of $"Li"_2$ :

![http://wps.prenhall.com/]( useruploads.socratic.org )

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.