Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Among ${\text{H}}_2$, ${\text{He}}_2^{+}$, ${\text{Li}}_2$, ${\text{Be}}_2$, ${\text{B}}_2$, ${\text{C}}_2$, ${\text{N}}_2$, ${\text{O}}_2^{-}$, and ${\text{F}}_2$, the number of diamagnetic species is - (Atomic number : H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8, f = 9)

Imagine you are a detective, and your suspects are a lineup of diatomic molecules. Your mission? To find out which ones are diamagnetic.

In the quantum world, electrons love company. When all electrons in a molecule are perfectly paired up, their magnetic fields cancel out, making the molecule diamagnetic. But if even a single electron is left flying solo, it acts like a tiny magnet, and the whole molecule becomes paramagnetic.

To solve this case, we need our master tool: Molecular Orbital Theory.

Molecular Orbital Theory

Molecular Orbital Theory (MOT) tells us exactly how electrons are distributed when two atoms bond. But there is a twist! The energy sequence of these orbitals isn't the same for everyone.

For lighter elements up to Nitrogen ($Z\le 7$), the $2s$ and $2p$ orbitals are close in energy. This causes s-p mixing, which pushes the ${\sigma }_{2p_z}$ orbital higher in energy than the ${\pi }_{2p}$ orbitals.

However, for heavier elements like Oxygen and Fluorine ($Z>7$), the energy gap is too large. There is no s-p mixing, so the ${\sigma }_{2p_z}$ orbital stays lower in energy. Keeping these two sequences in mind is the key to cracking this problem.

Up to Beryllium

Let's start interrogating our suspects by filling their electrons into the molecular orbitals.

Hydrogen ($H_2$) has 2 electrons. Both comfortably pair up in the lowest ${\sigma }_{1s}$ orbital. No unpaired electrons here, so it is diamagnetic.

Helium ion ($H{e}_2^{+}$) has 3 electrons. The first two pair up in ${\sigma }_{1s}$, but the third one is forced into the antibonding ${\sigma }_{1s}^{\ast }$ orbital. It is all alone, making it paramagnetic.

Lithium ($L{i}_2$) has 6 electrons. Following the sequence, the last two electrons pair up in the ${\sigma }_{2s}$ orbital. It is diamagnetic.

Beryllium ($B{e}_2$) has 8 electrons. The last two fill the antibonding ${\sigma }_{2s}^{\ast }$ orbital. Everything is paired, so theoretically, it is diamagnetic.

Boron to Nitrogen

Now things get interesting.

Boron ($B_2$) has 10 electrons. After filling 8 electrons, the next two enter the degenerate ${\pi }_{2p_x}$ and ${\pi }_{2p_y}$ orbitals. According to Hund's rule, electrons prefer to occupy degenerate orbitals singly before pairing up. So, we get two unpaired electrons! This makes $B_2$ paramagnetic.

Carbon ($C_2$) has 12 electrons. Those two $\pi$ orbitals now get fully filled with 4 electrons. Everyone has a partner, making it diamagnetic.

Nitrogen ($N_2$) has 14 electrons. The next two electrons fill the ${\sigma }_{2p_z}$ orbital. Again, all electrons are paired. Nitrogen is highly stable and diamagnetic.

Oxygen and Fluorine

For the heavier elements, we switch to the second energy sequence without s-p mixing.

Superoxide ion ($O_2^{-}$) has 17 electrons. After filling 14 electrons, the remaining 3 go into the antibonding ${\pi }^{\ast }$ orbitals. One orbital gets a pair, but the other gets only one electron. That single unpaired electron makes it paramagnetic.

Finally, Fluorine ($F_2$) has 18 electrons. The ${\pi }^{\ast }$ orbitals are now completely filled with 4 electrons. All paired up! Fluorine is diamagnetic.

Let's tally up our diamagnetic suspects: $H_2$, $L{i}_2$, $B{e}_2$, $C_2$, $N_2$, and $F_2$. That gives us a total count of 6.

But wait, there is a catch! Beryllium ($B{e}_2$) has an equal number of bonding and antibonding electrons, giving it a bond order of zero. This means it practically doesn't exist as a stable molecule. If you exclude it from the lineup, the count drops to 5.

Because of this ambiguity, the JEE officially accepted both 5 and 6 as correct answers. However, based purely on the theoretical electronic configuration, the count is 6. Case closed!