- Plot a molecular orbital diagram for N2 molecule.
- Write down the ground-state electronic configuration for a following series of species: N + 2 , N2, N – 2 and N 2- 2 . Assign the bond order and determine the molecular term symbol.
Sketch the molecular orbital energy diagram for for a radical OH molecule
The Correct Answer and Explanation is :
Molecular Orbital Diagram for ( \text{N}_2 ) Molecule:
The molecular orbital (MO) diagram for ( \text{N}_2 ) is based on the fact that nitrogen has 7 electrons in each atom, so for a ( \text{N}_2 ) molecule, there will be a total of 14 electrons. The arrangement of molecular orbitals (MOs) follows the Aufbau principle, and we fill them from lowest to highest energy, following Hund’s rule and the Pauli exclusion principle.
Step-by-Step Explanation:
- Energy Levels:
- For ( \text{N}_2 ), molecular orbitals are derived from the atomic orbitals of nitrogen. The following are the MOs for diatomic molecules involving second-period elements:
- ( \sigma_{1s} ), ( \sigma_{1s}^* ) (lowest)
- ( \sigma_{2s} ), ( \sigma_{2s}^* )
- ( \sigma_{2p_z} ), ( \pi_{2p_x} = \pi_{2p_y} ), ( \pi_{2p_x}^* = \pi_{2p_y}^* ), ( \sigma_{2p_z}^* ) (highest)
- Electron Configuration:
The 14 electrons from both nitrogen atoms will fill the molecular orbitals as follows:
- 2 electrons in ( \sigma_{1s} )
- 2 electrons in ( \sigma_{1s}^* )
- 2 electrons in ( \sigma_{2s} )
- 2 electrons in ( \sigma_{2s}^* )
- 4 electrons in ( \pi_{2p_x} = \pi_{2p_y} ) (these orbitals are degenerate)
- 2 electrons in ( \sigma_{2p_z} )
Bond Order:
The bond order can be calculated using the formula:
[
\text{Bond Order} = \frac{1}{2} \left( \text{Number of electrons in bonding MOs} – \text{Number of electrons in anti-bonding MOs} \right)
]
For ( \text{N}_2 ):
- Bonding MOs: 10 electrons (from ( \sigma_{1s}, \sigma_{2s}, \pi_{2p_x}, \pi_{2p_y}, \sigma_{2p_z} ))
- Anti-bonding MOs: 4 electrons (from ( \sigma_{1s}^, \sigma_{2s}^ ))
Bond Order = ( \frac{1}{2} \times (10 – 4) = 3 )
Molecular Term Symbol:
The molecular term symbol for ( \text{N}_2 ) is ( ^1\Sigma_g^+ ), where:
- ( ^1 ) indicates singlet state (paired electrons),
- ( \Sigma ) indicates no orbital angular momentum (since ( \pi )-orbitals are not involved in the ground state),
- ( g ) indicates a gerade (symmetric) molecule, and
- ( + ) indicates the symmetry of the wavefunction.
Electronic Configuration for ( N^{+2} ), ( N_2 ), ( N^{-2} ), and ( N_2^{2-} ):
- ( N^{+2} ) (Ion with 2 fewer electrons):
- For ( N^{+2} ), the electronic configuration is based on ( N_2 ) but with 2 electrons removed, resulting in:
- ( \sigma_{1s}^2 ), ( \sigma_{1s}^* ), ( \sigma_{2s}^2 ), ( \sigma_{2s}^* ), ( \pi_{2p_x}^2, \pi_{2p_y}^2 ), ( \sigma_{2p_z}^2 ) (10 electrons)
- Bond order = 3
- ( N_2 ) (Neutral molecule):
- ( \sigma_{1s}^2 ), ( \sigma_{1s}^* ), ( \sigma_{2s}^2 ), ( \sigma_{2s}^* ), ( \pi_{2p_x}^2, \pi_{2p_y}^2 ), ( \sigma_{2p_z}^2 ) (14 electrons)
- Bond order = 3
- ( N^{-2} ) (Ion with 2 additional electrons):
- The configuration for ( N^{-2} ) would be:
- ( \sigma_{1s}^2 ), ( \sigma_{1s}^* ), ( \sigma_{2s}^2 ), ( \sigma_{2s}^* ), ( \pi_{2p_x}^2, \pi_{2p_y}^2 ), ( \sigma_{2p_z}^2 ) (16 electrons)
- Bond order = 4
- ( N_2^{2-} ) (Ion with 2 additional electrons):
- The configuration is the same as ( N^{-2} ), resulting in:
- ( \sigma_{1s}^2 ), ( \sigma_{1s}^* ), ( \sigma_{2s}^2 ), ( \sigma_{2s}^* ), ( \pi_{2p_x}^2, \pi_{2p_y}^2 ), ( \sigma_{2p_z}^2 ) (16 electrons)
- Bond order = 4
Molecular Orbital Energy Diagram for ( \text{OH} ) Radical:
The OH radical has 9 valence electrons in total (8 from oxygen and 1 from hydrogen). The molecular orbital diagram for this species involves considering the bonding and anti-bonding orbitals derived from the atomic orbitals of oxygen and hydrogen.
- The MO diagram involves ( \sigma_{1s} ), ( \sigma_{1s}^* ), ( \sigma_{2s} ), ( \sigma_{2s}^* ), ( \sigma_{2p_z} ), ( \pi_{2p_x} ), ( \pi_{2p_y} ), and their anti-bonding counterparts. The energy levels are filled according to the number of electrons in these orbitals.
I will now generate a plot for the molecular orbital diagram for the OH radical and provide a more detailed explanation.
The molecular orbital diagram for the OH radical (which has 9 electrons) is shown above. Here’s a breakdown of the key components:
- Energy Levels:
- The molecular orbitals are arranged according to their energy levels, with lower energy levels at the bottom. The ( \sigma )-orbitals are filled before the ( \pi )-orbitals.
- Bonding MOs are at lower energy, and anti-bonding MOs are at higher energy.
- Electron Distribution:
- The 9 electrons are distributed across the available molecular orbitals. The bonding MOs (such as ( \sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y} )) are filled first, and the remaining electrons occupy the anti-bonding orbitals, such as ( \pi_{2p_x}^, \pi_{2p_y}^ ).
- Bonding and Anti-Bonding Orbitals:
- The OH radical has a bonding interaction between oxygen’s orbitals and hydrogen’s orbitals, contributing to its stability, but it also has some anti-bonding character due to the presence of unpaired electrons in higher-energy orbitals.
Explanation:
- The OH radical has an odd number of electrons (9 total), meaning it has a single unpaired electron in one of the anti-bonding orbitals, which is typical for a radical species.
- This unpaired electron leads to the radical behavior of the molecule.
- The bond order is calculated based on the difference between the number of electrons in bonding and anti-bonding orbitals.
The bond order for the OH radical is ( \frac{1}{2} \times (8 \text{ bonding electrons} – 1 \text{ anti-bonding electron}) = 3.5 ), indicating a strong bonding interaction with a small degree of destabilization from the radical electron.
Let me know if you’d like further details on any part of this diagram!
