How many of the following species are paramagnetic?
`N_(2)^(+), O_(2)^(-), B_(2), C_(2), CN^(-), NO^(+), CO, H_(2)^(-), He_(2)^(+)`

Explanation

To determine how many of the given species are paramagnetic, we need to analyze the number of valence electrons in each species and check if they have unpaired electrons. A species is considered paramagnetic if it has one or more unpaired electrons.

### Step-by-Step Solution:

1. **N₂⁺ (Nitrogen Dimer Cation)**
- Nitrogen (N) has 5 valence electrons.
- For N₂, there are 10 valence electrons (5 from each nitrogen).
- N₂⁺ has one less electron: 10 - 1 = 9 electrons.
- The electron configuration shows that there will be unpaired electrons.
- **Conclusion**: N₂⁺ is paramagnetic.

2. **O₂⁻ (Oxygen Dimer Anion)**
- Oxygen (O) has 6 valence electrons.
- For O₂, there are 12 valence electrons (6 from each oxygen).
- O₂⁻ has one extra electron: 12 + 1 = 13 electrons.
- The electron configuration shows that there will be unpaired electrons.
- **Conclusion**: O₂⁻ is paramagnetic.

3. **B₂ (Boron Dimer)**
- Boron (B) has 3 valence electrons.
- For B₂, there are 6 valence electrons (3 from each boron).
- The electron configuration shows that all electrons are paired.
- **Conclusion**: B₂ is not paramagnetic.

4. **C₂ (Carbon Dimer)**
- Carbon (C) has 4 valence electrons.
- For C₂, there are 8 valence electrons (4 from each carbon).
- The electron configuration shows that all electrons are paired.
- **Conclusion**: C₂ is not paramagnetic.

5. **CN⁻ (Cyanide Anion)**
- Carbon has 4 valence electrons and nitrogen has 5.
- Total for CN: 4 + 5 = 9 valence electrons.
- CN⁻ has one extra electron: 9 + 1 = 10 electrons.
- The electron configuration shows that all electrons are paired.
- **Conclusion**: CN⁻ is not paramagnetic.

6. **NO⁺ (Nitric Oxide Cation)**
- Nitrogen has 5 valence electrons and oxygen has 6.
- Total for NO: 5 + 6 = 11 valence electrons.
- NO⁺ has one less electron: 11 - 1 = 10 electrons.
- The electron configuration shows that there will be unpaired electrons.
- **Conclusion**: NO⁺ is paramagnetic.

7. **CO (Carbon Monoxide)**
- Carbon has 4 valence electrons and oxygen has 6.
- Total for CO: 4 + 6 = 10 valence electrons.
- The electron configuration shows that all electrons are paired.
- **Conclusion**: CO is not paramagnetic.

8. **H₂⁻ (Hydrogen Dimer Anion)**
- Each hydrogen has 1 valence electron.
- For H₂, there are 2 valence electrons.
- H₂⁻ has one extra electron: 2 + 1 = 3 electrons.
- The electron configuration shows that there will be unpaired electrons.
- **Conclusion**: H₂⁻ is paramagnetic.

9. **He₂⁺ (Helium Dimer Cation)**
- Each helium has 2 valence electrons.
- For He₂, there are 4 valence electrons.
- He₂⁺ has one less electron: 4 - 1 = 3 electrons.
- The electron configuration shows that there will be unpaired electrons.
- **Conclusion**: He₂⁺ is paramagnetic.

### Summary of Results:
- Paramagnetic species: N₂⁺, O₂⁻, NO⁺, H₂⁻, He₂⁺
- Total paramagnetic species = 5

### Final Answer:
**5 species are paramagnetic.**