We need to count the species which satisfy both conditions:

1. Paramagnetic
2. Bond order = 1

We examine each species using molecular orbital (MO) theory.

1. $\mathrm{H}_2$

For $\mathrm{H}_2$, total electrons $=2$.

MO filling: $\sigma_{1s}^2$

Bond order: $\text{B.O.} = \frac{N_b-N_a}{2} = \frac{2-0}{2}=1$

All electrons are paired, so it is diamagnetic.

So, $\mathrm{H}_2$ does not satisfy both conditions.

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

Total electrons $= 2+2-1 = 3$ electrons.

MO filling: $\sigma_{1s}^2\,\sigma_{1s}^{*1}$

Bond order: $\text{B.O.} = \frac{2-1}{2} = \frac{1}{2}$

It is paramagnetic (one unpaired electron), but bond order is not 1.

So, $\mathrm{He}_2^{+}$ does not qualify.

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

For $\mathrm{O}_2$, bond order is 2. Removing one electron gives $\mathrm{O}_2^{+}$.

$\mathrm{O}_2$ has MO configuration ending in: $\pi_{2p_x}^{*1}\pi_{2p_y}^{*1}$

In $\mathrm{O}_2^{+}$, one antibonding electron is removed, so bond order increases by $\frac{1}{2}$: $\text{B.O.} = 2.5$

It is paramagnetic, but bond order is not 1.

So, $\mathrm{O}_2^{+}$ does not qualify.

4. $\mathrm{N}_2^{2-}$

$\mathrm{N}_2$ has 14 electrons; $\mathrm{N}_2^{2-}$ has 16 electrons.

For $\mathrm{N}_2$, bond order is 3. Adding two electrons goes into antibonding $\pi^*$ orbitals, decreasing bond order by 1.

Thus, $\text{B.O.} = 3-1=2$

The two added electrons occupy degenerate orbitals singly first, so it is paramagnetic.

But bond order is not 1.

So, $\mathrm{N}_2^{2-}$ does not qualify.

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

$\mathrm{O}_2$ has bond order 2. Adding two electrons to antibonding orbitals decreases bond order by 1: $\text{B.O.} = 2-1=1$

MO in peroxide ion has all electrons paired in $\pi^*$ orbitals. So it is diamagnetic.

Thus, $\mathrm{O}_2^{2-}$ has bond order 1 but is not paramagnetic.

So, it does not qualify.

6. $\mathrm{F}_2$

$\mathrm{F}_2$ has 18 valence electrons overall pattern like oxygen-family diatomics.

Bond order: $\text{B.O.} = 1$

But all electrons are paired, so it is diamagnetic.

So, $\mathrm{F}_2$ does not qualify.

7. $\mathrm{Ne}_2^{+}$

$\mathrm{Ne}_2$ would have bond order 0. Removing one electron from antibonding orbital gives: $\text{B.O.} = 0.5$

It is paramagnetic, but bond order is not 1.

So, $\mathrm{Ne}_2^{+}$ does not qualify.

8. $\mathrm{B}_2$

Each B has 5 electrons, so total electrons $=10$.

For $\mathrm{B}_2$, MO order for lighter molecules is: $\sigma_{1s},\ \sigma_{1s}^*,\ \sigma_{2s},\ \sigma_{2s}^*,\ \pi_{2p_x}=\pi_{2p_y},\ \sigma_{2p_z}$

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

Bond order: $\text{B.O.} = \frac{(2+2)-(2+2)+2}{2}?$

More directly from valence MOs:

• Bonding electrons in $\sigma_{2s}, \pi_{2p_x}, \pi_{2p_y}$ = $2+2=4$
• Antibonding electrons in $\sigma_{2s}^*$ = $2$

So, $\text{B.O.} = \frac{4-2}{2}=1$

There are two unpaired electrons in the two $\pi$ orbitals, so $\mathrm{B}_2$ is paramagnetic.

Thus, $\mathrm{B}_2$ does qualify.

Final count

Only $\mathrm{B}_2$ is both paramagnetic and has bond order equal to 1.

Therefore, the required number of species is: $\boxed{1}$

Comparison with stored answer

Stored correct answer = 1

Our derived answer = 1

So, the answer agrees with the stored correct answer.