The number of paramagnetic species among the following is $$B_2, Li_2, C_2, C_2^-, O_2^{2-}, O_2^+$$ and $$He_2^+$$

A paramagnetic species has one or more unpaired electrons. We use Molecular Orbital Theory (MOT) to determine the electronic configuration and count unpaired electrons for each species.

Key MOT orbital filling order:

For $$B_2$$, $$C_2$$, $$N_2$$ (up to 14 electrons): $$\sigma_{1s} \lt \sigma^*_{1s} \lt \sigma_{2s} \lt \sigma^*_{2s} \lt \pi_{2p_x} = \pi_{2p_y} \lt \sigma_{2p_z} \lt \pi^*_{2p_x} = \pi^*_{2p_y} \lt \sigma^*_{2p_z}$$

For $$O_2$$ and heavier (15+ electrons): $$\sigma_{1s} \lt \sigma^*_{1s} \lt \sigma_{2s} \lt \sigma^*_{2s} \lt \sigma_{2p_z} \lt \pi_{2p_x} = \pi_{2p_y} \lt \pi^*_{2p_x} = \pi^*_{2p_y} \lt \sigma^*_{2p_z}$$

1. $$B_2$$ (total electrons = 5 + 5 = 10):

Configuration: $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{2}\, \sigma_{2s}^2\, {\sigma^*_{2s}}^{2}\, \pi_{2p_x}^1\, \pi_{2p_y}^1$$

The last two electrons enter the two degenerate $$\pi_{2p}$$ orbitals with parallel spins (Hund's rule), giving 2 unpaired electrons.

Bond order = $$\f\frac{6-4}{2} = 1$$. Paramagnetic.

2. $$Li_2$$ (total electrons = 3 + 3 = 6):

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

All electrons are paired. 0 unpaired electrons.

Bond order = $$\f\frac{4-2}{2} = 1$$. Diamagnetic.

3. $$C_2$$ (total electrons = 6 + 6 = 12):

Configuration: $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{2}\, \sigma_{2s}^2\, {\sigma^*_{2s}}^{2}\, \pi_{2p_x}^2\, \pi_{2p_y}^2$$

Both $$\pi_{2p}$$ orbitals are completely filled. 0 unpaired electrons.

Bond order = $$\f\frac{8-4}{2} = 2$$. Diamagnetic.

4. $$C_2^-$$ (total electrons = 12 + 1 = 13):

Configuration: $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{2}\, \sigma_{2s}^2\, {\sigma^*_{2s}}^{2}\, \pi_{2p_x}^2\, \pi_{2p_y}^2\, \sigma_{2p_z}^1$$

The 13th electron enters the $$\sigma_{2p_z}$$ orbital (which comes after $$\pi_{2p}$$ in the filling order for lighter elements). This gives 1 unpaired electron.

Bond order = $$\f\frac{9-4}{2} = 2.5$$. Paramagnetic.

5. $$O_2^{2-}$$ (total electrons = 16 + 2 = 18):

Configuration (using O-type ordering): $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{2}\, \sigma_{2s}^2\, {\sigma^*_{2s}}^{2}\, \sigma_{2p_z}^2\, \pi_{2p_x}^2\, \pi_{2p_y}^2\, {\pi^*_{2p_x}}^{2}\, {\pi^*_{2p_y}}^{2}$$

Both antibonding $$\pi^*$$ orbitals are completely filled. 0 unpaired electrons.

Bond order = $$\f\frac{10-8}{2} = 1$$. Diamagnetic.

6. $$O_2^+$$ (total electrons = 16 - 1 = 15):

Configuration: $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{2}\, \sigma_{2s}^2\, {\sigma^*_{2s}}^{2}\, \sigma_{2p_z}^2\, \pi_{2p_x}^2\, \pi_{2p_y}^2\, {\pi^*_{2p_x}}^{1}$$

One electron is removed from the antibonding $$\pi^*$$ orbital, leaving 1 unpaired electron.

Bond order = $$\f\frac{10-5}{2} = 2.5$$. Paramagnetic.

7. $$He_2^+$$ (total electrons = 2 + 2 - 1 = 3):

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

The antibonding orbital has one electron. 1 unpaired electron.

Bond order = $$\f\frac{2-1}{2} = 0.5$$. Paramagnetic.

Summary:

The paramagnetic species are: $$B_2$$, $$C_2^-$$, $$O_2^+$$, and $$He_2^+$$.

Therefore, the number of paramagnetic species is 4.