To determine which species is paramagnetic, we need to analyze their molecular orbital (MO) configurations. Paramagnetic species have unpaired electrons, while diamagnetic species have all electrons paired. Since 2s-2p mixing is not operative, the energy order of molecular orbitals follows the standard sequence: σ1s < σ*1s < σ2s < σ*2s < σ2p_z < π2p_x = π2p_y < π*2p_x = π*2p_y < σ*2p_z.

Let's evaluate each option step by step:

Step 1: Determine the total number of electrons for each molecule.

B₂: Boron atomic number is 5, so B₂ has 10 electrons.

Be₂: Beryllium atomic number is 4, so Be₂ has 8 electrons.

C₂: Carbon atomic number is 6, so C₂ has 12 electrons.

N₂: Nitrogen atomic number is 7, so N₂ has 14 electrons.

Step 2: Write the MO electron configuration for each without 2s-2p mixing.

The order is: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x⁰ π*2p_y⁰ σ*2p_z⁰

We fill electrons according to Aufbau principle, Hund's rule, and Pauli exclusion principle.

For B₂ (10 electrons):

Configuration: σ1s² σ*1s² σ2s² σ*2s² (σ2p_z² π2p_x¹ π2p_y¹)

Since σ2p_z is lower than π2p, we fill σ2p_z first. After σ*2s² (4 electrons used), next 6 electrons: σ2p_z takes 2 electrons, then π2p orbitals take 4 electrons, but only 4 electrons remain, so π2p_x and π2p_y get 2 electrons each? Wait, actually, after σ*2s², we have 6 electrons to place. σ2p_z orbital can hold 2 electrons, then π2p orbitals (degenerate) can hold up to 4 electrons. So for 6 electrons: σ2p_z² (2 electrons), then π2p_x² π2p_y² (4 electrons). All electrons are paired. But this would be diamagnetic. However, experimentally B₂ is paramagnetic due to 2s-2p mixing, but here mixing is not operative, so it should be diamagnetic. Let's confirm the configuration:

Total electrons: 10. After core (σ1s² σ*1s²) ignore for valence, valence electrons: 6 (from 2s and 2p). σ2s² σ*2s² uses 4 electrons, then σ2p_z² uses 2 electrons, then π2p orbitals: but only 0 electrons left? No, 6 valence electrons: σ2s² σ*2s² (4 electrons), then σ2p_z² (2 electrons) — all paired. So B₂ should be diamagnetic without mixing.

Correction: Valence electrons only consider 2s and 2p. For B₂, total valence electrons = 3+3=6.

So configuration: σ2s² σ*2s² σ2p_z² π2p_x⁰ π2p_y⁰. All paired, diamagnetic.

For Be₂ (8 electrons, valence electrons=4):

Configuration: σ2s² σ*2s². All paired, diamagnetic. Also, Be₂ does not exist stably.

For C₂ (12 electrons, valence electrons=8):

Configuration: σ2s² σ*2s² σ2p_z² π2p_x² π2p_y⁰? Wait, 8 valence electrons: after σ2s² σ*2s² (4 electrons), σ2p_z² (2 electrons, total 6), then π2p orbitals: 2 electrons left, which go into π2p_x and π2p_y. Since they are degenerate, Hund's rule applies: π2p_x¹ π2p_y¹ — two unpaired electrons. So paramagnetic.

For N₂ (14 electrons, valence electrons=10):

Configuration: σ2s² σ*2s² σ2p_z² π2p_x² π2p_y². All paired, diamagnetic.

Therefore, without 2s-2p mixing, C₂ is paramagnetic due to two unpaired electrons in the π orbitals.

Final Answer: C₂ is the paramagnetic species.

Related Topics and Formulae

Molecular Orbital Theory: Describes the electronic structure of molecules using molecular orbitals formed by the combination of atomic orbitals. Key points:

• Bonding orbitals (σ, π) lower energy, antibonding orbitals (σ*, π*) higher energy.
• Electron configuration determines magnetic property (paramagnetic with unpaired electrons, diamagnetic with all paired).
• Bond order = (Number of bonding electrons - Number of antibonding electrons)/2.

Standard MO Energy Order (without mixing):

For B₂, C₂, etc., without 2s-2p mixing: σ1s < σ*1s < σ2s < σ*2s < σ2p_z < π2p_x = π2p_y < π*2p_x = π*2p_y < σ*2p_z

With mixing, the order changes, but here it is not operative.