Bond Order and Magnetism of B2 Violating Hunds Rule

To determine the bond order and magnetic nature of the diatomic boron molecule ($\text{B}_2$) when Hund's rule is violated, we use Molecular Orbital (MO) Theory.

1. Total Number of Electrons

The atomic number of Boron ($\text{B}$) is $5$. Therefore, the total number of electrons in a neutral $\text{B}_2$ molecule is:
$\text{Total electrons} = 2 \times 5 = 10$

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2. Molecular Orbital Configuration

For homonuclear diatomic molecules with 14 or fewer electrons (such as $\text{B}_2$), the energy level diagram of molecular orbitals in increasing order of energy is:
$\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < (\pi_{2p_x} = \pi_{2p_y}) < \sigma_{2p_z} < (\pi^*_{2p_x} = \pi^*_{2p_y}) < \sigma^*_{2p_z}$

Filling the first 8 electrons into the lowest energy molecular orbitals gives:
\sigma_{1s}^2 \, \sigma^*_{1s}^2 \, \sigma_{2s}^2 \, \sigma^*_{2s}^2

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3. Effect of Violating Hund's Rule

Hund's rule states that pairing of electrons in degenerate orbitals (orbitals with equal energy, such as $\pi_{2p_x}$ and $\pi_{2p_y}$) does not occur until each orbital is singly occupied with parallel spins.

• Under normal conditions (Hund's rule followed): The remaining 2 electrons enter $\pi_{2p_x}$ and $\pi_{2p_y}$ singly:
  $\pi_{2p_x}^1 \, \pi_{2p_y}^1$
  This results in two unpaired electrons, making $\text{B}_2$ paramagnetic.

• When Hund's rule is violated: The remaining 2 electrons pair up in a single orbital before filling the other degenerate orbital:
  $\pi_{2p_x}^2 \, \pi_{2p_y}^0 \quad \text{or} \quad \pi_{2p_x}^0 \, \pi_{2p_y}^2$
  Since all electrons are fully paired, the molecule contains zero unpaired electrons, making $\text{B}_2$ diamagnetic.

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4. Calculation of Bond Order

The bond order is defined as:
$\text{Bond Order} = \frac{N_b - N_a}{2}$

where:

• $N_b$ = number of bonding electrons $= 2 \, (\sigma_{1s}) + 2 \, (\sigma_{2s}) + 2 \, (\pi_{2p_x}) = 6$
• $N_a$ = number of antibonding electrons $= 2 \, (\sigma^*_{1s}) + 2 \, (\sigma^*_{2s}) = 4$

Substituting these values:
$\text{Bond Order} = \frac{6 - 4}{2} = 1$

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Conclusion

Assuming Hund's rule is violated, the diatomic molecule $\text{B}_2$ has a bond order of 1 and is diamagnetic.

Correct Option: A