Using Molecular Orbital Theory, Draw The Molecular Orbital Energy Diagram For B2 (A) Draw The Atomic Orbital Diagram For Each Boron Atom, (B) Draw And Label The Molecular Orbital Diagram In The Middle, Using Dashed Lines To Show Which Atomic Orbitals Combine, (C) Fill In All Electrons. The Energy Levels For The Molecular Orbitals Are: Lowest To Highest
The Correct Answer and Explanation is :
To construct the molecular orbital (MO) energy diagram for the ( \text{B}_2 ) molecule using Molecular Orbital Theory, we follow several steps, taking into account the electronic configuration of boron and how its atomic orbitals combine to form molecular orbitals.
Step 1: Atomic Orbital Diagram for Each Boron Atom
Each boron atom has an atomic number of 5, which means its electron configuration is ( 1s^2 2s^2 2p^1 ). In a molecular orbital diagram, the valence electrons (those in the ( 2s ) and ( 2p ) orbitals) participate in bonding.
- The atomic orbitals of each boron atom are:
- ( 2s ) (2 electrons)
- ( 2p ) (1 electron)
Step 2: Molecular Orbital Diagram
For the ( \text{B}_2 ) molecule, we have two boron atoms, so we consider the atomic orbitals from both atoms combining to form molecular orbitals.
- The atomic orbitals involved in bonding are the ( 2s ) and ( 2p ) orbitals.
- The ( 2s ) orbitals combine to form a ( \sigma_{2s} ) bonding orbital and a ( \sigma^*_{2s} ) antibonding orbital.
- The ( 2p ) orbitals combine to form a ( \sigma_{2p_z} ) bonding orbital and a ( \pi_{2p_x} ) and ( \pi_{2p_y} ) degenerate bonding orbitals. The antibonding orbitals are ( \pi^{2p_x} ), ( \pi^{2p_y} ), and ( \sigma^*_{2p_z} ).
The energy diagram for the molecular orbitals of ( \text{B}_2 ) is shown as follows:
- ( \sigma_{2s} ) is the lowest energy orbital.
- ( \sigma^*_{2s} ) is the next higher energy.
- ( \sigma_{2p_z} ) is higher than ( \sigma^*_{2s} ).
- ( \pi_{2p_x} ) and ( \pi_{2p_y} ) are at higher energy.
- ( \pi^{2p_x} ) and ( \pi^{2p_y} ) are higher than ( \pi_{2p_x} ).
- ( \sigma^*_{2p_z} ) is the highest energy orbital.
Step 3: Filling in Electrons
- Each boron atom contributes 5 electrons, so for ( \text{B}_2 ), we have a total of 10 electrons to distribute.
- The order of filling the molecular orbitals (from lowest to highest energy) is as follows:
- 2 electrons in ( \sigma_{2s} )
- 2 electrons in ( \sigma^*_{2s} )
- 4 electrons in ( \pi_{2p_x} ) and ( \pi_{2p_y} )
- 2 electrons in ( \sigma_{2p_z} )
Thus, the molecular orbital diagram for ( \text{B}_2 ) will have:
- ( 2 ) electrons in ( \sigma_{2s} )
- ( 2 ) electrons in ( \sigma^*_{2s} )
- ( 4 ) electrons in ( \pi_{2p_x} ) and ( \pi_{2p_y} )
- ( 2 ) electrons in ( \sigma_{2p_z} )
Conclusion
The ( \text{B}_2 ) molecule, with a total of 10 valence electrons, has a bond order of 1, as calculated by the formula:
[
\text{Bond Order} = \frac{1}{2} \left( \text{(number of electrons in bonding MOs)} – \text{(number of electrons in antibonding MOs)} \right)
]
This shows that the ( \text{B}_2 ) molecule has a single bond.