The total number of electrons in all bonding molecular orbitals of $$O_2^{2-}$$ is _________ (Round off to the nearest integer)

For molecular-orbital (MO) treatment we start with the electronic details of the species.

Each oxygen atom has atomic number 8, so a neutral $$\mathrm{O_2}$$ molecule contains $$2 \times 8 = 16$$ electrons. The dianion $$\mathrm{O_2^{2-}}$$ possesses two extra electrons. Hence the total electron count is

$$16 + 2 = 18.$$

Next we recall the accepted MO energy order for homonuclear diatomics of oxygen and beyond (i.e. having $$Z \ge 8$$):

$$ \sigma_{1s}\,<\,\sigma_{1s}^\ast\,<\, \sigma_{2s}\,<\,\sigma_{2s}^\ast\,<\, \sigma_{2p_z}\,<\, \pi_{2p_x} = \pi_{2p_y}\,<\, \pi_{2p_x}^\ast = \pi_{2p_y}^\ast\,<\, \sigma_{2p_z}^\ast . $$

We now place the 18 electrons, two per orbital in the sequence shown.

1. $$\sigma_{1s}$$ receives $$2$$ electrons   $$\Rightarrow$$  running total $$=2$$.

2. $$\sigma_{1s}^\ast$$ receives $$2$$ electrons   $$\Rightarrow$$  running total $$=4$$.

3. $$\sigma_{2s}$$ receives $$2$$ electrons   $$\Rightarrow$$  running total $$=6$$.

4. $$\sigma_{2s}^\ast$$ receives $$2$$ electrons   $$\Rightarrow$$  running total $$=8$$.

5. $$\sigma_{2p_z}$$ receives $$2$$ electrons   $$\Rightarrow$$  running total $$=10$$.

6. $$\pi_{2p_x}$$ receives $$2$$ electrons.

   $$\pi_{2p_y}$$ receives $$2$$ electrons   $$\Rightarrow$$  running total $$=14$$.

7. The remaining $$18-14 = 4$$ electrons must enter the lowest antibonding set:

   $$\pi_{2p_x}^\ast$$ receives $$2$$ electrons.

   $$\pi_{2p_y}^\ast$$ receives $$2$$ electrons   $$\Rightarrow$$  running total $$=18$$, filling is complete.

Our goal is to count only the electrons present in bonding MOs. These bonding orbitals are

$$\sigma_{1s},\;\sigma_{2s},\;\sigma_{2p_z},\;\pi_{2p_x},\;\pi_{2p_y}.$$

The electron population in them is therefore

$$ 2 \;(\sigma_{1s}) + 2 \;(\sigma_{2s}) + 2 \;(\sigma_{2p_z}) + 2 \;(\pi_{2p_x}) + 2 \;(\pi_{2p_y}) = 10. $$

So the total number of electrons occupying bonding molecular orbitals in $$\mathrm{O_2^{2-}}$$ is $$10$$.

Hence, the correct answer is Option C.