1. Idea: A species is paramagnetic if it has one or more unpaired electrons.

2. We use Molecular Orbital (MO) theory.

For lighter homonuclear diatomics up to $\mathrm{N}_2$: $\sigma(2s) < \sigma^*(2s) < \pi(2p_x)=\pi(2p_y) < \sigma(2p_z)$

For oxygen species: $\sigma(2s) < \sigma^*(2s) < \sigma(2p_z) < \pi(2p_x)=\pi(2p_y) < \pi^*(2p_x)=\pi^*(2p_y)$

Now examine each species.

3. $\mathrm{B}_2$

Each B has 5 electrons, so total electrons $=10$. Core $1s$ electrons are paired and do not affect paramagnetism discussion; valence filling gives: $\sigma(2s)^2\,\sigma^*(2s)^2\,\pi(2p_x)^1\,\pi(2p_y)^1$ There are two unpaired electrons.

So, $\mathrm{B}_2$ is paramagnetic.

4. $\mathrm{Li}_2$

Each Li has 3 electrons, total $=6$. Valence configuration: $\sigma(2s)^2$ All electrons are paired.

So, $\mathrm{Li}_2$ is diamagnetic.

5. $\mathrm{C}_2$

Each C has 6 electrons, total $=12$. Valence MO filling: $\sigma(2s)^2\,\sigma^*(2s)^2\,\pi(2p_x)^2\,\pi(2p_y)^2$ All electrons are paired.

So, $\mathrm{C}_2$ is diamagnetic.

6. $\mathrm{C}_2^{-}$

$\mathrm{C}_2$ has 12 electrons, so $\mathrm{C}_2^{-}$ has 13 electrons. After filling up to $\sigma(2s)^2\,\sigma^*(2s)^2\,\pi(2p_x)^2\,\pi(2p_y)^2$ one extra electron goes into: $\sigma(2p_z)^1$ This gives one unpaired electron.

So, $\mathrm{C}_2^{-}$ is paramagnetic.

7. $\mathrm{O}_2^{2-}$

$\mathrm{O}_2$ has 16 electrons, so $\mathrm{O}_2^{2-}$ has 18 electrons. For oxygen species, filling is: $\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2\,\pi(2p_x)^2\,\pi(2p_y)^2\,\pi^*(2p_x)^2\,\pi^*(2p_y)^2$ All electrons are paired.

So, $\mathrm{O}_2^{2-}$ is diamagnetic.

8. $\mathrm{O}_2^{+}$

$\mathrm{O}_2$ has 16 electrons, so $\mathrm{O}_2^{+}$ has 15 electrons. For $\mathrm{O}_2$, the last two electrons occupy separate $\pi^*$ orbitals. Removing one electron gives: $\pi^*(2p)^3$ Hence there is one unpaired electron.

So, $\mathrm{O}_2^{+}$ is paramagnetic.

9. $\mathrm{He}_2^{+}$

Each He has 2 electrons, so $\mathrm{He}_2$ would have 4; thus $\mathrm{He}_2^{+}$ has 3 electrons. MO configuration: $\sigma(1s)^2\,\sigma^*(1s)^1$ There is one unpaired electron.

So, $\mathrm{He}_2^{+}$ is paramagnetic.

10. Count paramagnetic species

Paramagnetic species are:

• $\mathrm{B}_2$
• $\mathrm{C}_2^{-}$
• $\mathrm{O}_2^{+}$
• $\mathrm{He}_2^{+}$

Total number: $4$

11. Comparison with stored answer

Derived answer $=4$, and stored correct answer $=4$. So they agree.