Consider The Molecular Orbital Diagram For The H2, Shown Below. What Is The Bond Order For H2? 1s 015 Н H
The Correct Answer and Explanation is :
Bond Order of H₂:
The molecular orbital (MO) diagram for ( H_2 ) shows the interaction of the 1s atomic orbitals from two hydrogen atoms. The bond order can be calculated using the formula:
[
\text{Bond Order} = \frac{\text{Number of electrons in bonding orbitals} – \text{Number of electrons in antibonding orbitals}}{2}
]
For ( H_2 ), the key details are:
- Each hydrogen atom contributes one electron from its 1s orbital, resulting in two total electrons.
- These electrons are placed in the molecular orbitals, starting with the lower-energy bonding orbital (( \sigma_{1s} )) and then filling the antibonding orbital (( \sigma_{1s}^* )) if additional electrons are available.
- In ( H_2 ), both electrons occupy the ( \sigma_{1s} ) bonding orbital.
Calculation:
- Number of electrons in bonding orbitals (( \sigma_{1s} )) = 2
- Number of electrons in antibonding orbitals (( \sigma_{1s}^* )) = 0
[
\text{Bond Order} = \frac{2 – 0}{2} = 1
]
Thus, the bond order for ( H_2 ) is 1.
Explanation:
The bond order indicates the strength and stability of the bond between atoms. A bond order of 1 means that ( H_2 ) has a single covalent bond, formed by the sharing of electrons between the two hydrogen nuclei. The ( \sigma_{1s} ) bonding orbital is stabilized because the wavefunctions of the two 1s orbitals constructively interfere, resulting in an increase in electron density between the nuclei. This stabilizes the molecule and creates an attractive force that holds the two hydrogen atoms together.
The antibonding orbital (( \sigma_{1s}^* )), on the other hand, is higher in energy due to destructive interference of the wavefunctions, reducing electron density between the nuclei. Since ( H_2 ) has no electrons in this orbital, there is no destabilization.
This bond order of 1 corresponds to a strong, stable bond in ( H_2 ), making it a simple yet significant diatomic molecule.