18 Molecular Orbital Theory II

Devendra Mohan

Contents:

1. Diatomic Molecular Orbitals

2. Energy Level Diagram for Molecular Orbital

3. Electronic Configuration of Molecules In The Mo Concept

4. Assignment

1. DIATOMIC MOLECULAR ORBITALS

The Molecular orbital theory regarding the bonding in ?₂+ and H₂ molecules can be extended to other homo-nuclear diatomic molecules.In general, molecular orbitals are classified in terms of their symmetry about internuclear axis. Since the atoms ? and ? are identical, the mid-point of the ?−? bond is a centre of symmetry for the molecule. Further, If an inversion of a molecular orbital about the centre of symmetry does not change the sign Ψ, it is said to be even and is denoted by the symbol g as a subscript.In case, the sign changes, the orbital is said to be odd and a subscript ? is assigned to the symbol.

It is to note that, ? and ? stand for the German words gerade (means even) and ungerade (mean odd), respectively.

Case –I Combination of two s orbitals

The combination of two 1s orbitals gives rise to the formation of a bonding (1??) and an antibonding (1??∗) orbitals. There is a build up of electron charge between the nuclei leading to the formation of a bond in the case of bonding orbital and a depletion of charge between the nuclei in the case of antibonding orbital. In the above notation the bonding and antibonding orbitals are respectively denoted by (2???) and an antibonding (2??∗_u) molecular orbitals of the same shapes as (2???) and(2??∗_u) orbitals.

Case-II Combination of s and p_x orbitals

A s orbitals may combine with a p orbital if the lobes of the p orbital point along the internuclear axis. A bonding MO result when the adjacent lobes have the same sign and an antibonding MO when the adjacentlobes are of opposite sign. Figure depicts the combination of s and px atomic orbitals.

_{Fig Combination of s and px atomic orbitals}

Case-III Combination of Two p orbitals

The two lobes of each of the three p orbitals have opposite sign. If the internuclear axis is taken as the x-direction, two px atomic combine to give the ??’? 2???? and 2???∗_? as represented in the Figure. Both have symmetry about the bond axis. However, the Combination of two ?? orbitals gives the molecular orbitals 2???? and 2?_??_?∗ as has been shown in the figure. Now because both of these have the same energy these are doubly degenerate. The ?_??_? MO consists of two streamers, one above and one below the nuclei. In this case the bonding orbital is odd and the antibonding orbital is even unlike the earlier ones.

Figure: (a) bonding orbital 2?_??_? and (b) antibonding 2???∗_? molecular orbitals from two ?_? orbital.

Figure: Formation of (a) bonding orbital 2?_??_? , and (b) antibonding orbital 2???∗_? from two ?? orbtials (combination is the same for ?_? orbitals).

It is noteworthy that in the combination of atomic orbitals, the possibility for the combination of 1s and 2s atomic orbitals or s and p orbitals of two identical atoms is not possible because their energies are completely different.

2. Energy Level Diagram for Molecular Orbital

Let the Molecular Orbitals are arranged in the increasing order of energy. The energy levels of molecular orbitals of diatomic molecules resulting from 2p orbitals are somewhat uncertain as the atomic number of the atom increases.

Also, for molecules lighter than O₂, the 2??_? orbitals lie higher than the 2??_? orbitals. But, for molecules heavier than O₂, the 2??_? orbitals are of lower energy than 2??_? orbitals.

Figure depicts the molecular orbital energy level diagram for Homonuclear diatomic molecules. (a) Heavier than O₂ (b) lighter than O₂.

R.S. Mulliken has suggested an alternate terminology for the orbitals 2?? to 2??∗ and its sequence is

??(2??)<??(2??)^∗<??(2?_??)<??(2?_??=2?_??)<??(2???^∗=2?_??∗)<??(2?_??^∗)

The bracket gives the earlier notation. This notation is being used for one-quantum and three-quantum molecular orbitals by writing (K) and (M) before the Mulliken symbol. Thus (?)?? stands for 1?? ,(?)??, for 3??∗, and like that…..

3. Electronic Configuration Of Molecules In The Mo Concept

Electrons can be assigned to the orbitals following the Pauli’s exclusion principle after the molecular orbitals are obtained. The MO’s of the lowest energy are filled first and each MO can get two electrons of opposite spins and in case of degenerate orbitals, Hund’s rule is applied.

Homo-nuclear Diatomic Molecules:

H2: The number of electrons in hydrogen molecule is two. The MO 1?? is formed from 2H (1s) atomic orbitals. As both the electrons are in the lowest bonding molecular orbital, the two electron bond of hydrogen molecule is stronger than the one electron bond of ?⁺₂

2H (1s) ⟹H2[(1??_?)2]

He2: The overlap of two helium 1s orbitals gives the bonding 1??? and antibonding 1???∗ molecular orbitals with two electrons in each. The superposition of two electrons in the bonding orbitals and two in the antibonding orbitals leads to no bonding. Hence He2 molecule does not exist under normal condition.

2H_e (1s²) ⟹He₂[(1???)²(1???∗)]

Li2: Lithium atom has an electronic configuration 1?²2?. Molecular orbitals have to be constructed from 1s as well as the 2s atomic orbital. As the presence of two electrons in the bonding 1??_? orbital and two in the antibonding orbital 1??_?∗ leads to cancellation of the bonding between the two Li atoms, the bonding between the Li atoms result from the pairing of the 2s electrons in the bonding orbital 2??_?.Writing KK for (1???)² (1???∗)², the ground electronic configuration of Li2 is writing as

??₂[??(2??)²]

The configuration of other alkali metals are analogous.

Be₂: Be atom has 4 electrons making a total of 8 electrons for the Be2 molecule. They are distributed as

??2[??(2??_?)2(2??_?∗)²]

As the presence of equal number of bonding and antibonding electrons produces no net bonding effect, the molecule is not likely to exist.

B₂: The 10 electrons in this molecule are distributed as

?2[??(2??_?)²(2??_?∗)²(2??_?)²]

The 2??_? and 2??_? molecular orbitals have nearly the same energy. Therefore, instead of (2??_?)2 , the alternate configuration (2??_?)¹(22??_?)¹ leading to a total spin of one is also possible. These two unpaired electrons per molecule lead to the observed paramagnetism of B2.

C2: Carbon molecule has 12 electrons. The electrons are distributed as

?₂[??(2??_?)²(2??_?∗)²⁽2???)²(2???)²]