The number of species from the following which are paramagnetic and with bond order equal to one is H₂, He₂^+, O₂^+, N₂²⁻, O₂²⁻, F₂, Ne₂^+, B₂$$\mathrm {H}_2, \mathrm{He}_2^+, \mathrm{O}_2^+, \mathrm{N}_2^{2-}, \mathrm{O}_2^{2-}, \mathrm{F}_2, \mathrm{Ne}_2^+, \mathrm{B}_2$$

Core Logic

Using Molecular Orbital (MO) Theory, we evaluate the bond order (BO = (Nb - Nₐ)/(2)$BO = \frac{N_b - N_a}{2}$) and magnetic nature (unpaired electrons = paramagnetic, all paired = diamagnetic) for each species:

Step 1: Final Selection

We need the species that satisfies BOTH conditions:

• Paramagnetic
• Bond Order = 1

Looking at the table, B₂$B_2$ is the only molecule that is paramagnetic (it has 2 unpaired electrons in degenerate π₂ₚ$\pi_{2p}$ orbitals) and has a bond order of 1.

Total number of such species = 1.

Pattern Recognition

B₂$B_2$ (10 electrons) and O₂$O_2$ (16 electrons) are the classic exceptions in MO theory that are paramagnetic despite having an even number of electrons. B₂$B_2$ has BO = 1, and O₂$O_2$ has BO = 2.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure