The number of species from the following which are paramagnetic and with bond order equal to one is _______
$$H_2, He_2^+, O_2^+, N_2^{2-}, O_2^{2-}, F_2, Ne_2^+, B_2$$

We need to find the number of species that are paramagnetic AND have bond order = 1 from: $$H_2, He_2^+, O_2^+, N_2^{2-}, O_2^{2-}, F_2, Ne_2^+, B_2$$.

Analyze each species.

$$H_2$$: $$\sigma_{1s}^2$$. Bond order = 1. Diamagnetic (all paired). No.

$$He_2^+$$: $$\sigma_{1s}^2 {\sigma^*_{1s}}^{1}$$. Bond order = 0.5. No (BO ≠ 1).

$$O_2^+$$: Bond order = 2.5. No.

$$N_2^{2-}$$: 16 electrons. Same as $$O_2$$. Bond order = 2. No.

$$O_2^{2-}$$: 18 electrons. $$\sigma_{1s}^2{\sigma^*_{1s}}^{2}\sigma_{2s}^2{\sigma^*_{2s}}^{2}\sigma_{2p}^2\pi_{2p}^4{\pi^*_{2p}}^{4}$$. Bond order = (10-8)/2 = 1. All electrons paired. Diamagnetic. No.

$$F_2$$: 18 electrons. Bond order = 1. Diamagnetic. No.

$$Ne_2^+$$: 19 electrons. Bond order = (10-9)/2 = 0.5. No.

$$B_2$$: 10 electrons. $$\sigma_{1s}^2{\sigma^*_{1s}}^{2}\sigma_{2s}^2{\sigma^*_{2s}}^{2}\pi_{2p}^1\pi_{2p}^1$$. Bond order = (6-4)/2 = 1. Two unpaired electrons in $$\pi_{2p}$$. Paramagnetic with bond order 1. Yes!

Count.

Only $$B_2$$ satisfies both conditions = 1 species.

Conclusion.

The number is 1.

Therefore, the answer is 1.