In which of the following pairs of molecules/ions, both the species are not likely to exist?

We start with the molecular-orbital (MO) theory for homonuclear diatomic species that are made only from the 1s atomic orbitals. In this very first MO set we have just two orbitals:

$$\sigma_{1s}\;(\text{bonding}), \qquad \sigma_{1s}^{\*}\;(\text{antibonding}).$$

The bond order is obtained from the standard formula

$$\text{Bond order} \;=\;\dfrac{N_b-N_a}{2},$$

where $$N_b$$ is the number of electrons in bonding orbitals and $$N_a$$ is the number of electrons in antibonding orbitals. A bond order that is zero or negative means the species cannot exist, whereas a positive bond order means the species can, at least in principle, exist.

Let us evaluate the bond order for every species that appears in the four options.

1. $$H_2^{2+}$$ (0 electron)

Because two positive charges remove both electrons, we have

$$N_b = 0,\; N_a = 0 \;\Longrightarrow\; \text{Bond order}= \dfrac{0-0}{2}=0.$$

So $$H_2^{2+}$$ is not expected to exist.

2. $$H_2$$ (2 electrons)

Configuration: $$\sigma_{1s}^2$$

$$N_b = 2,\; N_a = 0 \;\Longrightarrow\; \text{Bond order}= \dfrac{2-0}{2}=1 >0.$$

This molecule exists (ordinary hydrogen).

3. $$H_2^{+}$$ (1 electron)

Configuration: $$\sigma_{1s}^1$$

$$N_b = 1,\; N_a = 0 \;\Longrightarrow\; \text{Bond order}= \dfrac{1-0}{2}=0.5 >0.$$

Hence $$H_2^{+}$$ can exist.

4. $$H_2^{-}$$ (3 electrons)

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

$$N_b = 2,\; N_a = 1 \;\Longrightarrow\; \text{Bond order}= \dfrac{2-1}{2}=0.5 >0.$$

Therefore $$H_2^{-}$$ is also, in principle, possible.

5. $$He_2$$ (4 electrons)

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

$$N_b = 2,\; N_a = 2 \;\Longrightarrow\; \text{Bond order}= \dfrac{2-2}{2}=0.$$

So $$He_2$$ is not expected to exist.

6. $$He_2^{2+}$$ (2 electrons)

The double positive charge removes two electrons, leaving 2 electrons in total:

Configuration: $$\sigma_{1s}^2$$

$$N_b = 2,\; N_a = 0 \;\Longrightarrow\; \text{Bond order}= \dfrac{2-0}{2}=1 >0.$$

Thus $$He_2^{2+}$$ should exist.

7. $$He_2^{2-}$$ (6 electrons)

After the basic 1s set is filled (4 electrons), the extra two electrons enter the next higher bonding MO, $$\sigma_{2s}$$. The order up to that point is

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

This gives

$$N_b = 2 (\sigma_{1s}) + 2 (\sigma_{2s}) = 4,\qquad N_a = 2 (\sigma_{1s}^{\*})$$

$$\Longrightarrow\; \text{Bond order}= \dfrac{4-2}{2}=1 >0.$$

So $$He_2^{2-}$$ can also exist according to MO theory.

Now we examine each option to see where both species have bond order zero.

A. $$H_2^{2+}$$ (bond order 0) and $$He_2$$ (bond order 0)  ⟹  both non-existent.

B. $$H_2^{-}$$ (bond order 0.5) and $$He_2^{2+}$$ (bond order 1)  ⟹  both exist.

C. $$H_2^{+}$$ (bond order 0.5) and $$He_2^{2-}$$ (bond order 1)  ⟹  both exist.

D. $$H_2$$ (bond order 1) and $$He_2^{2-}$$ (bond order 1)  ⟹  both exist.

Only option A contains a pair for which neither species is expected to exist.

Hence, the correct answer is Option A.