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The MO diagram for diboron (B-B, electron configuration 1σ g 2 1σ u 2 2σ g 2 2σ u 2 1π u 2) requires the introduction of an atomic orbital overlap model for p orbitals. The three dumbbell -shaped p-orbitals have equal energy and are oriented mutually perpendicularly (or orthogonally ).
The qualitative approach of MO analysis uses a molecular orbital diagram to visualize bonding interactions in a molecule. In this type of diagram, the molecular orbitals are represented by horizontal lines; the higher a line the higher the energy of the orbital, and degenerate orbitals are placed on the same level with a space between them.
Molecular orbital diagram of dinitrogen molecule, N 2. There are five bonding orbitals and two antibonding orbitals (marked with an asterisk; orbitals involving the inner 1s electrons not shown), giving a total bond order of three. Atomic nitrogen, also known as active nitrogen, is highly reactive, being a triradical with three unpaired electrons.
Molecular orbital diagram of He 2. Bond order is the number of chemical bonds between a pair of atoms. The bond order of a molecule can be calculated by subtracting the number of electrons in anti-bonding orbitals from the number of bonding orbitals, and the resulting number is then divided by two. A molecule is expected to be stable if it has ...
An initial assumption is that the number of molecular orbitals is equal to the number of atomic orbitals included in the linear expansion. In a sense, n atomic orbitals combine to form n molecular orbitals, which can be numbered i = 1 to n and which may not all be the same. The expression (linear expansion) for the i th molecular orbital would be:
For example, in the case of the F 2 molecule, the F−F bond is formed by the overlap of p z orbitals of the two F atoms, each containing an unpaired electron. Since the nature of the overlapping orbitals are different in H 2 and F 2 molecules, the bond strength and bond lengths differ between H 2 and F 2 molecules.
Triple bonding can be explained in terms of orbital hybridization. In the case of acetylene, each carbon atom has two sp-orbitals and two p-orbitals. The two sp-orbitals are linear, with 180° bond angles, and occupy the x-axis in the cartesian coordinate system. The p-orbitals are perpendicular to the sp-orbitals on the y-axis and the z-axis.
Here the sum extends over π molecular orbitals only, and n i is the number of electrons occupying orbital i with coefficients c ri and c si on atoms r and s respectively. Assuming a bond order contribution of 1 from the sigma component this gives a total bond order (σ + π) of 5/3 = 1.67 for benzene, rather than the commonly cited bond order ...