Step 1: Determine the total number of electrons in the $B_2$ molecule.

Boron (B) has an atomic number of 5. Therefore, a single boron atom has 5 electrons. The diatomic molecule $B_2$ will have a total of $5 + 5 = 10$ electrons.

Step 2: Write the molecular orbital (MO) energy level order for $B_2$.

For diatomic molecules of elements up to and including Nitrogen ($N_2$), the order of increasing energy of molecular orbitals is: $\sigma1s < \sigma^*1s < \sigma2s < \sigma^*2s < (\pi2p_x = \pi2p_y) < \sigma2p_z < (\pi^*2p_x = \pi^*2p_y) < \sigma^*2p_z$ Note that the $\pi2p_x$ and $\pi2p_y$ orbitals are degenerate (have the same energy).

Step 3: Fill the molecular orbitals with the 10 electrons, violating Hund's rule.

Hund's rule of maximum multiplicity states that when filling degenerate orbitals, electrons are placed one in each orbital before any orbital is doubly occupied. The question specifies that we must assume Hund's rule is violated. This means that when filling degenerate orbitals, electrons will pair up in one orbital before occupying the next empty degenerate orbital.

Let's fill the MOs with the 10 electrons:

1. The first 2 electrons go into the $\sigma1s$ orbital: $(\sigma1s)^2$.
2. The next 2 electrons go into the $\sigma^*1s$ orbital: $(\sigma^*1s)^2$.
3. The next 2 electrons go into the $\sigma2s$ orbital: $(\sigma2s)^2$.
4. The next 2 electrons go into the $\sigma^*2s$ orbital: $(\sigma^*2s)^2$. So far, we have used 8 electrons. We have 2 electrons remaining.
5. The next available orbitals are the degenerate $\pi2p_x$ and $\pi2p_y$ orbitals. Since Hund's rule is violated, both remaining electrons will pair up in one of these orbitals, for example, the $\pi2p_x$ orbital. This gives $(\pi2p_x)^2 (\pi2p_y)^0$.

The complete electronic configuration for $B_2$ with Hund's rule violated is: $(\sigma1s)^2 (\sigma^*1s)^2 (\sigma2s)^2 (\sigma^*2s)^2 (\pi2p_x)^2$

Step 4: Calculate the bond order.

The bond order is calculated using the formula: $\text{Bond Order} = \frac{1}{2} (N_b - N_a)$ where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in anti-bonding molecular orbitals.

From our configuration:

• Bonding electrons ($N_b$) are in $\sigma1s$, $\sigma2s$, and $\pi2p_x$ orbitals. So, $N_b = 2 + 2 + 2 = 6$.
• Anti-bonding electrons ($N_a$) are in $\sigma^*1s$ and $\sigma^*2s$ orbitals. So, $N_a = 2 + 2 = 4$.

Plugging these values into the formula: $\text{Bond Order} = \frac{1}{2} (6 - 4) = \frac{1}{2} (2) = 1$

Step 5: Determine the magnetic nature.

The magnetic nature of a molecule is determined by the presence of unpaired electrons.

• If there are unpaired electrons, the molecule is paramagnetic.
• If all electrons are paired, the molecule is diamagnetic.

Looking at the electronic configuration, $(\sigma1s)^2 (\sigma^*1s)^2 (\sigma2s)^2 (\sigma^*2s)^2 (\pi2p_x)^2$, we can see that all orbitals are filled with pairs of electrons. There are no unpaired electrons.

Therefore, the $B_2$ molecule, under this condition, is diamagnetic.

Conclusion

Based on our calculations, if Hund's rule is violated, the $B_2$ molecule has a bond order of 1 and is diamagnetic. This corresponds to option A.