Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is :

Unlocking Magnetic Mysteries

Molecular Orbital (MO) Theory is one of the most elegant frameworks in chemistry, allowing us to predict the magnetic properties and bond orders of diatomic molecules with stunning accuracy. But what happens when we tweak the fundamental rules? This JEE Advanced problem does exactly that by asking us to imagine a universe where $2s-2p$ mixing is NOT operative.

Let's dive deep into the quantum mechanics of this hypothetical scenario and see how it flips our standard understanding of molecules like $B_2$ and $C_2$.

The "No Mixing" Hypothetical

In standard MO theory, the energy levels of homonuclear diatomic molecules depend heavily on the atomic number. For elements like Oxygen ($O_2$) and Fluorine ($F_2$), the energy gap between the $2s$ and $2p$ atomic orbitals is large. Because they are far apart in energy, they do not interact or "mix" significantly. As a result, the ${\sigma }_{2p_z}$ molecular orbital forms at a lower energy than the degenerate ${\pi }_{2p_x}$ and ${\pi }_{2p_y}$ orbitals.

However, for lighter elements like Boron ($B_2$), Carbon ($C_2$), and Nitrogen ($N_2$), the $2s$ and $2p$ orbitals are closer in energy. They undergo $2s-2p$ mixing, which pushes the ${\sigma }_{2p_z}$ orbital higher in energy, placing it above the ${\pi }_{2p}$ orbitals.

But this question explicitly commands us: Assume $2s-2p$ mixing is NOT operative.

This means we must use the energy sequence typically reserved for $O_2$ and $F_2$ for all the given options:

The Trap of $B_2$

Let's test the molecules one by one by filling their electrons into our new, unmixed energy diagram.

1. Beryllium ($B{e}_2$) A Beryllium atom has $4$ electrons, so $B{e}_2$ has $8$ electrons in total. Filling them in order: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}$ All electrons are perfectly paired. Thus, $B{e}_2$ is diamagnetic.

2. Boron ($B_2$) Boron has $5$ electrons, making $10$ electrons for the $B_2$ molecule. Let's add two more electrons to the $B{e}_2$ configuration. Normally (with mixing), these two electrons would singly occupy the ${\pi }_{2p_x}$ and ${\pi }_{2p_y}$ orbitals, making $B_2$ paramagnetic. But without mixing, the ${\sigma }_{2p_z}$ orbital is lower in energy! ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2{\mathbf{p}}_{\mathbf{z}}}^2$ Both electrons pair up in the ${\sigma }_{2p_z}$ orbital. In this hypothetical scenario, $B_2$ becomes diamagnetic. This is a classic trap designed to catch students who rely purely on memory rather than applying the given constraints.

Why $C_2$ Becomes Paramagnetic

3. Carbon ($C_2$) Carbon has $6$ electrons, so $C_2$ has $12$ electrons. We need to place $4$ electrons into the $2p$ molecular orbitals. Normally (with mixing), the ${\pi }_{2p}$ orbitals are lower, so all $4$ electrons pair up in ${\pi }_{2p_x}$ and ${\pi }_{2p_y}$, making $C_2$ diamagnetic.

But let's see what happens without mixing: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2{\mathbf{p}}_{\mathbf{z}}}^2{\pi }_{2{\mathbf{p}}_{\mathbf{x}}}^1{\pi }_{2{\mathbf{p}}_{\mathbf{y}}}^1$

The first two $p$-electrons fill the ${\sigma }_{2p_z}$ orbital. The remaining two electrons must enter the degenerate ${\pi }_{2p_x}$ and ${\pi }_{2p_y}$ orbitals. According to Hund's Rule of Maximum Multiplicity, they will occupy these orbitals singly before pairing up.

We now have two unpaired electrons! Therefore, without $2s-2p$ mixing, $C_2$ is paramagnetic.

4. Nitrogen ($N_2$) Just to be thorough, $N_2$ has $14$ electrons. The configuration becomes: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2{\pi }_{2p_x}^2{\pi }_{2p_y}^2$ All orbitals are completely filled and paired. $N_2$ remains diamagnetic, regardless of whether mixing occurs or not.

By carefully applying the "no mixing" constraint, we observed a fascinating role reversal: $B_2$ lost its paramagnetism, while $C_2$ gained it. This problem is a beautiful reminder that in chemistry, the "rules" are deeply tied to underlying physical interactions. When you change the physics, you change the chemistry.