Bonding in Some Diatomic Molecules:
Hydrogen Molecule (H2):
The electronic configuration of H is 1s1. Thus the formation of the H2 molecule involves the linear combination of 1s orbital of one H-atom with 1s-orbital of another H-atom forming two molecular orbitals. The molecular orbital formed by the addition overlap is known as bonding molecular orbital and is represented as σ1s while the other formed due to subtraction overlap is known as antibonding molecular orbital and is represented as σ*1s. Two 1s electrons one from each H-atom, pass into the σ1s orbital (bonding molecular orbital) because it is of lower energy while σ*1s orbital (antibonding molecular orbital) with higher energy remains vacant. Thus, the electronic configuration of the H2 molecule is [σ(1s)]2.
|Bond Order of H2 molecule = nb-na/2 = 2-0/2 = 1|
Thus, the two H-atoms in a molecule of H2 are joined by a single covalent bond. It has no unpaired electron, so it is diamagnetic.
Hydrogen Molecule Ion (H2+):
It is formed by the combination of H-atom containing one electron and hydrogen ion (H+) containing no electron. Thus, this molecule has only one electron which is present in the lowermost molecular orbital i.e. σ1s. So electronic configuration of H2+ is [σ(1s)]1.
|Bond Order of H2+ = nb-na/2 = 1-0/2 = 1/2|
Since bond order is positive, so the molecule is stable, though stability is not very high due to low value. The presence of one unpaired electron shows that it is paramagnetic.
Hydrogen Molecule Ion (H2–):
It is formed by the combination of H-atom containing molecular orbitals of H2+ ions having one electron (in 1s orbital) and hydrogen ion (H–) containing two electrons (in 1s orbital). Thus, the molecule has three electrons, two are present in bonding molecular orbital (σ1s being of lower energy) and one in antibonding molecular orbital (σ*1s). Thus the electronic configuration of H2– ion is [σ(1s)]2 [σ*(1s)]1.
|Bond Order of H2– ion = nb-na/2 = 2-1/2 = 1/2|
Like H2+ ion, its bond order is also +1/2 but this ion, H2–, is less stable than H2+ ion due to the presence of one electron in the antibonding orbital which results in repulsion and decreases the stability. Due to the presence of one unpaired electron, it is paramagnetic.
Helium Molecule (He2):
The electronic configuration of He is 1s2 indicating that in the He2 molecule there will be 4 electrons. Two of these electrons are present in σ1s and two in σ*1s molecular orbital. Therefore, the electronic configuration of the He2 molecule will be [σ(1s)]2 [σ*(1s)]2.
|Bond Order of He2 = nb-na/2 = 2-2/2 = 0|
Since the bond order of the He2 molecule is zero, therefore, this molecule is unstable and does not exist.
Nitrogen Molecule (N2):
The electronic configuration of N is 1s2, 2s2 2px1 2py1 2pz1 which shows that the N2 molecule has 14 electrons which are to be accommodated in orbitals in the increasing order of their energy. The molecular orbital configuration of the N2 molecule is [σ(1s)]2 [σ*(1s)]2 [σ(2s)]2 [σ*(2s)]2 [π(2px)]2 [π(2py)]2 [σ(2pz)]2.
The structure [σ(1s)]2 [σ*(1s)]2 indicated a closed K shell structure which does not enter into bonding and is usually written as KK. The electronic configuration of the N2 molecule may also be written as KK [σ(2s)]2 [σ*(2s)]2 [π(2px)]2 [π(2py)]2 [σ(2pz)]2.
|Bond Order of N2 molecule = nb-na/2 = 10-4/2 = 3|
The two N-atoms in a molecule of N2 are joined by a triple covalent bond (N≡N). It has no unpaired electron and, hence, is a diamagnetic molecule.
Oxygen Molecule (O2):
The electronic configuration of O is 1s2, 2s2 2px2 2py1 2pz1 indicating that it has 8 electrons so that the molecule of O2 has 16 electrons which are to be accommodated in various molecular orbitals in the increasing order of their energy. The molecular orbital configuration of O2 is [σ(1s)]2 [σ*(1s)]2 [σ(2s)]2 [σ*(2s)]2 [σ(2pz)]2 [π(2px)]2 [π(2py)]2 [π*(2px)]1 [π*(2py)]1.
Since 1s electrons are not involved in bonding and so the molecular orbital configuration may also be written as KK [σ(2s)]2 [σ*(2s)]2 [σ(2pz)]2 [π(2px)]2 [π(2py)]2 [π*(2px)]1 [π*(2py)]1.
|Bond Order of O2 molecule = nb-na/2 = 10-6/2 = 2|
Thus, the two O-atom in a molecule of O2 is joined by a double bond (O=O). Since it contains two unpaired electrons (in π*2px and π*2py orbitals) so it is paramagnetic in nature.