1. Determine diamagnetism using Molecular Orbital (MO) theory

A species is:

• Diamagnetic if all electrons are paired.
• Paramagnetic if it has one or more unpaired electrons.

For hydrogen and helium molecular species, the relevant MO order is:

$\sigma_{1s},\; \sigma_{1s}^*$

We now examine each option.

2. Option A: $He_2^+$

Each He atom has 2 electrons, so:

$He_2^+ : 2+2-1 = 3 \text{ electrons from the }1s\text{ orbitals?}$

Actually total electrons:

$He_2^+ = 4 - 1 = 3 \text{ electrons?}$

This is incorrect counting for the whole species because each He has 2 electrons, so neutral $He_2$ has:

$2+2=4 \text{ electrons}$

Thus,

$He_2^+ = 4-1 = 3 \text{ electrons in molecular orbitals built from }1s$

Filling MOs:

$\sigma_{1s}^2\, \sigma_{1s}^{*1}$

There is one unpaired electron, so $He_2^+$ is paramagnetic.

3. Option B: $H_2$

Each H atom has 1 electron, so total electrons:

$H_2 = 1+1 = 2$

Filling MOs:

$\sigma_{1s}^2$

All electrons are paired, so $H_2$ is diamagnetic.

4. Option C: $H_2^+$

Total electrons:

$H_2^+ = 2-1 = 1$

Filling MOs:

$\sigma_{1s}^1$

There is one unpaired electron, so $H_2^+$ is paramagnetic.

5. Option D: $H_2^-$

Total electrons:

$H_2^- = 2+1 = 3$

Filling MOs:

$\sigma_{1s}^2\,\sigma_{1s}^{*1}$

There is one unpaired electron, so $H_2^-$ is paramagnetic.

6. Conclusion

Only $H_2$ has all electrons paired.

Therefore, the diamagnetic species is:

$\boxed{H_2}$

So the correct option is B.

7. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

They match.