Number of molecules/species from the following having one unpaired electron is ______. $$O_2,\ O_2^{-1},\ NO,\ CN^{-1},\ O_2^{2-}$$

We need to count how many of the given species have exactly one unpaired electron, using Molecular Orbital Theory (MOT).

Key Concept: In MOT, electrons fill molecular orbitals in order of increasing energy. For diatomic molecules of second-period elements, the filling order depends on whether the species has 14 or fewer electrons (where $$\sigma_{2p}$$ is higher than $$\pi_{2p}$$) or more than 14 electrons (where $$\sigma_{2p}$$ is lower than $$\pi_{2p}$$).

1. $$O_2$$ (16 electrons):

Electronic configuration: $$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^*_{2p})^2$$

The last two electrons go into the two degenerate $$\pi^*_{2p}$$ orbitals. By Hund's rule, they occupy different orbitals with parallel spins. This gives 2 unpaired electrons.

Since we need exactly 1 unpaired electron, $$O_2$$ does not qualify.

2. $$O_2^-$$ (17 electrons):

Compared to $$O_2$$, this has one additional electron: $$(\pi^*_{2p})^3$$

The two degenerate $$\pi^*_{2p}$$ orbitals now hold 3 electrons: one orbital has 2 (paired), the other has 1 (unpaired). This gives 1 unpaired electron. This qualifies.

3. $$NO$$ (15 electrons):

Electronic configuration: $$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^*_{2p})^1$$

The 15th electron enters one of the $$\pi^*_{2p}$$ orbitals, giving 1 unpaired electron. This qualifies.

4. $$CN^-$$ (14 electrons):

$$CN^-$$ is isoelectronic with $$N_2$$ and $$CO$$ (all have 14 electrons). For species with 14 or fewer electrons, the MO ordering has $$\pi_{2p}$$ below $$\sigma_{2p}$$:

$$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4 (\sigma_{2p})^2$$

All electrons are paired, giving 0 unpaired electrons. Does not qualify.

5. $$O_2^{2-}$$ (18 electrons):

Compared to $$O_2$$, this has two additional electrons: $$(\pi^*_{2p})^4$$

Both $$\pi^*_{2p}$$ orbitals are now completely filled (2 electrons each), so all electrons are paired. This gives 0 unpaired electrons. Does not qualify.

Counting the species with exactly one unpaired electron:

$$O_2^-$$ and $$NO$$ — that is 2 species.

The answer is 2.