What is the bond order of B2- ? is it paramagnetic or diamagnetic?
The Correct Answer and Explanation is :
To determine the bond order of the (\text{B}_2^-) ion and whether it is paramagnetic or diamagnetic, let’s break it down step by step.
Step 1: Molecular Orbital Theory and Electron Configuration
Molecular orbital theory helps us understand the bonding in molecules. For the (\text{B}_2) molecule, the electron configuration is based on the combination of atomic orbitals from each boron atom.
- Boron (B) has an atomic number of 5, which means it has 5 electrons. For two boron atoms, there are a total of 10 electrons.
- The molecular orbitals for diatomic molecules like (\text{B}_2) are filled in the following order (from lowest to highest energy):
- (\sigma_{1s}), (\sigma^{1s}), (\sigma{2s}), (\sigma^{2s}), (\pi{2p}), (\pi^{2p}), (\sigma{2p}), and (\sigma^_{2p}).
The bonding and antibonding orbitals are filled as:
- (\sigma_{1s}) and (\sigma^*_{1s}) (2 electrons in each, paired).
- (\sigma_{2s}) and (\sigma^*_{2s}) (2 electrons in each, paired).
- (\pi_{2p}) (2 electrons, 1 in each (\pi) orbital).
- (\pi^*_{2p}) (1 electron).
For (\text{B}_2^-), the negative charge means an additional electron, making it have 11 electrons in total.
Step 2: Bond Order Calculation
The bond order is calculated using the formula:
[
\text{Bond order} = \frac{1}{2} (\text{Number of bonding electrons} – \text{Number of antibonding electrons})
]
For (\text{B}_2^-), the electron configuration is:
- Bonding electrons: (\sigma_{1s}^2, \sigma_{2s}^2, \pi_{2p}^4) (8 electrons).
- Antibonding electrons: (\sigma^{2s}^2, \pi^{2p}^1) (3 electrons).
Thus, the bond order is:
[
\text{Bond order} = \frac{1}{2} (8 – 3) = \frac{1}{2} (5) = 2.5
]
Step 3: Magnetic Properties
To determine whether (\text{B}2^-) is paramagnetic or diamagnetic, we look at the electron configuration. Since there is an unpaired electron in the (\pi^*{2p}) orbital, (\text{B}_2^-) is paramagnetic.
Conclusion
- Bond order of (\text{B}_2^-): 2.5
- Magnetic properties: Paramagnetic
Let me generate an image to visualize this molecular orbital diagram.
Here is the molecular orbital diagram for the (\text{B}2^-) ion, showing the electron configuration and the bond order calculation. As you can see, the ion has a bond order of 2.5, and since it has an unpaired electron in the antibonding (\pi^*{2p}) orbital, it is paramagnetic. Let me know if you need further explanations or adjustments!
