According to molecular orbital theory, which of the following molecule will not be available?

According to molecular orbital (MO) theory we first combine the two atomic $$1s$$ orbitals of hydrogen or helium to obtain two molecular orbitals: a lower-energy bonding orbital $$\sigma(1s)$$ and a higher-energy antibonding orbital $$\sigma^{\*}(1s)$$.

The total number of electrons of the species is distributed in these two orbitals in the order of increasing energy, following the Pauli principle and Hund’s rule whenever necessary.

Before we start filling electrons, we recall the bond-order formula that MO theory gives us:

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

where $$N_b$$ is the number of electrons in all bonding molecular orbitals and $$N_a$$ is the number of electrons in all antibonding molecular orbitals. A positive value of bond order means the molecule/ion is stable enough to be isolated, while a zero or negative value means it will not exist under normal conditions.

Now we examine every option one by one and compute its bond order step by step.

Option A : $$H_2^{2-}$$

Neutral $$H_2$$ possesses 2 electrons. The charge $$2-$$ adds 2 more electrons, so

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

We fill the MOs:

$$\sigma(1s)\,2e^-;\; \sigma^{\*}(1s)\,2e^-$$

Thus $$N_b=2,\; N_a=2$$ and

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

Because the bond order is zero, no net bond is left, so $$H_2^{2-}$$ cannot exist.

Option B : $$He_2^{2+}$$

Neutral $$He_2$$ would have $$2\times 2=4$$ electrons. The $$2+$$ charge removes 2 electrons, leaving

$$\text{Total electrons}=4-2=2$$

Electron filling:

$$\sigma(1s)\,2e^-;\; \sigma^{\*}(1s)\,0e^-$$

So $$N_b=2,\; N_a=0$$

$$\text{Bond order}=\frac{2-0}{2}=1$$

A positive bond order of 1 shows that $$He_2^{2+}$$ is stable enough to be detectable.

Option C : $$He_2^{+}$$

Again start from 4 electrons of $$He_2$$ and remove 1 electron for the $$+$$ charge:

$$\text{Total electrons}=4-1=3$$

Electron filling:

$$\sigma(1s)\,2e^-;\; \sigma^{\*}(1s)\,1e^-$$

Thus $$N_b=2,\; N_a=1$$

$$\text{Bond order}=\frac{2-1}{2}=0.5$$

Since the bond order is positive, $$He_2^{+}$$ can exist, although it is weaker than $$He_2^{2+}$$.

Option D : $$H_2^{-}$$

Neutral $$H_2$$ has 2 electrons; the $$-$$ charge adds 1 more:

$$\text{Total electrons}=2+1=3$$

Electron filling:

$$\sigma(1s)\,2e^-;\; \sigma^{\*}(1s)\,1e^-$$

So $$N_b=2,\; N_a=1$$

$$\text{Bond order}=\frac{2-1}{2}=0.5$$

This positive value indicates that $$H_2^{-}$$ is also a realizable species.

Among all the given species, only $$H_2^{2-}$$ has a bond order of zero, meaning it will not be available in practice.

Hence, the correct answer is Option A.