How many of the following species have bond order = 1 and are paramagnetic as well? He₂²⁺; O₂^2-; Ne₂²⁺; F₂; B₂; H₂; O₂²⁺

The correct answer is 1.

To determine which of the given species have a bond order of 1 and are paramagnetic, we can use Molecular Orbital (MO) Theory. A species is paramagnetic if it has one or more unpaired electrons. The bond order is calculated using the formula:

$\text{Bond Order}=\frac{N_b-N_a}{2}$

where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.

Let us analyze each species provided in the question:

1. ${\text{He}}_2^{2+}$:
Total electrons = 4 - 2 = 2 electrons.
MO configuration: ${\left({\sigma }_{1s}\right)}^2$
Bond order = $\frac{2-0}{2}=1$.
Since all electrons are paired, it is diamagnetic.

2. ${\text{O}}_2^{2-}$ (Peroxide ion):
Total electrons = 16 + 2 = 18 electrons.
MO configuration: ${\left({\sigma }_{1s}\right)}^2{\left({\sigma }_{1s}^{*}\right)}^2{\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\sigma }_{2p_z}\right)}^2{\left({\pi }_{2p_x}\right)}^2{\left({\pi }_{2p_y}\right)}^2{\left({\pi }_{2p_x}^{*}\right)}^2{\left({\pi }_{2p_y}^{*}\right)}^2$
Bond order = $\frac{10-8}{2}=1$.
Since all electrons are paired, it is diamagnetic.

3. ${\text{Ne}}_2^{2+}$:
Total electrons = 20 - 2 = 18 electrons.
Bond order = $\frac{10-8}{2}=1$.
It has no unpaired electrons, so it is diamagnetic.

4. ${\text{F}}_2$:
Total electrons = 18 electrons.
Bond order = $\frac{10-8}{2}=1$.
It has no unpaired electrons, so it is diamagnetic.

5. ${\text{B}}_2$:
Total electrons = 10 electrons.
MO configuration: ${\left({\sigma }_{1s}\right)}^2{\left({\sigma }_{1s}^{*}\right)}^2{\left({\sigma }_{2s}\right)}^2{\left({\sigma }_{2s}^{*}\right)}^2{\left({\pi }_{2p_x}\right)}^1{\left({\pi }_{2p_y}\right)}^1$
Bond order = $\frac{6-4}{2}=1$.
Since it contains two unpaired electrons in the degenerate ${\pi }_{2p}$ orbitals, it is paramagnetic.

6. ${\text{H}}_2$:
Total electrons = 2 electrons.
Bond order = $\frac{2-0}{2}=1$.
All electrons are paired, so it is diamagnetic.

7. ${\text{O}}_2^{2+}$:
Total electrons = 16 - 2 = 14 electrons.
Bond order = $\frac{10-4}{2}=3$.
It has a bond order of 3, so it does not meet the requirement of having a bond order of 1.

Conclusion:
Only 1 species (${\text{B}}_2$) has a bond order equal to 1 and is paramagnetic.