1. Use MO theory for 1s combinations

For species formed from H and He using only $1s$ orbitals, the molecular orbital order is:

$\sigma_{1s} < \sigma_{1s}^*$

Bond order is given by:

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

where:

• $N_b$ = number of electrons in bonding MO
• $N_a$ = number of electrons in antibonding MO

A molecule/species is considered viable if bond order $>0$. If bond order $=0$, it is not viable.

2. Check each option

Option A: ${\rm He}_2^{2+}$

Each He atom has 2 electrons, so neutral ${\rm He}_2$ would have 4 electrons.

For ${\rm He}_2^{2+}$, total electrons:

$4 - 2 = 2$

Electronic configuration:

$\sigma_{1s}^2$

So,

$N_b = 2, \quad N_a = 0$

Bond order:

$\frac{2-0}{2} = 1$

So ${\rm He}_2^{2+}$ is viable.

Option B: ${\rm He}_2^{+}$

Total electrons:

$4 - 1 = 3$

Electronic configuration:

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

So,

$N_b = 2, \quad N_a = 1$

Bond order:

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

So ${\rm He}_2^{+}$ is viable.

Option C: ${\rm H}_2^{-}$

Neutral ${\rm H}_2$ has 2 electrons.

For ${\rm H}_2^{-}$, total electrons:

$2 + 1 = 3$

Electronic configuration:

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

So,

$N_b = 2, \quad N_a = 1$

Bond order:

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

So ${\rm H}_2^{-}$ is viable.

Option D: ${\rm H}_2^{2-}$

Total electrons:

$2 + 2 = 4$

Electronic configuration:

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

So,

$N_b = 2, \quad N_a = 2$

Bond order:

$\frac{2-2}{2} = 0$

Hence ${\rm H}_2^{2-}$ is not a viable molecule.

3. Conclusion

The species with bond order zero is:

$\boxed{{\rm H}_2^{2-}}$

So the correct option is D.