1. Criterion from Molecular Orbital Theory

A diatomic species exists if its bond order is positive.

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

where:

• $N_b$ = number of electrons in bonding molecular orbitals
• $N_a$ = number of electrons in antibonding molecular orbitals

If bond order $>0$, the species can exist. If bond order $=0$, it does not exist.

2. Option A: ${O_2}^{2-}$

$O_2$ has 16 electrons, so ${O_2}^{2-}$ has 18 electrons.

For oxygen, filling of MOs gives bond order of neutral $O_2$ as 2. Adding 2 electrons fills antibonding $\pi^*$ orbitals further, reducing bond order by 1.

So,

$\text{Bond order of } {O_2}^{2-} = 1$

Since bond order is positive, ${O_2}^{2-}$ exists.

3. Option B: $Be_2$

Each Be atom has 4 electrons, so $Be_2$ has 8 electrons.

Electronic configuration in MOs:

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

Bonding electrons = 4 in relevant filled bonding orbitals ($\sigma 1s$, $\sigma 2s$) Antibonding electrons = 4 in relevant filled antibonding orbitals ($\sigma^*1s$, $\sigma^*2s$)

Thus,

$\text{Bond order} = \frac{4-4}{2} = 0$

So $Be_2$ does not exist.

4. Option C: $He_2^-$

Each He has 2 electrons, so $He_2^-$ has $2+2+1=5$ electrons.

MO configuration:

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

(Equivalently, considering only 1s-derived MOs for helium, the extra electron gives a positive bond order overall.)

Using bond order counting:

• bonding electrons = 3
• antibonding electrons = 2

$\text{Bond order} = \frac{3-2}{2} = \frac{1}{2}$

So $He_2^-$ exists.

5. Option D: $He_2^+$

$He_2^+$ has $2+2-1=3$ electrons.

MO configuration:

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

Thus,

• bonding electrons = 2
• antibonding electrons = 1

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

So $He_2^+$ exists.

6. Conclusion

The only species with bond order zero is:

$\boxed{Be_2}$

So the species that does not exist is Option B.