: ( s ) ( ∗ s ) ( ∗ s ) ( π2 p x = π 2p y ) or KK ( s ) ( ∗ s ) ( π2 p x = π 2p y ) The bond order of C is ½ ( – ) = and C should be diamagnetic. Diamagnetic C molecules have indeed been detected in vapour phase. It is important to note that double bond in C consists of both pi bonds because of the presence of four electrons in two pi molecular orbitals. In most of the other molecules a double bond is made up of a sigma bond and a pi bond.
In a similar fashion the bonding in N molecule can be discussed. . Oxygen molecule (O ): The electronic configuration of oxygen atom is 1s 2s 2p . Each oxygen atom has electrons, hence, in O molecule there are electrons.
The electronic configuration of O molecule, therefore, is O : ( s ) ( ∗ s ) ( s ) ( ∗ s ) (2p z ) ( π2 p x ≡ π p y ) ( π ∗ p x ≡ π ∗ 2p y ) O : From the electronic configuration of O molecule it is clear that ten electrons are present in bonding molecular orbitals and six electrons are present in antibonding molecular orbitals. Its bond order, therefore, is Bond order = [N b – N a ] = [ – ] = So in oxygen molecule, atoms are held by a double bond. Moreover, it may be noted that it contains two unpaired electrons in π ∗ 2p x and π ∗ 2p y molecular orbitals, therefore, O molecule should be paramagnetic, a prediction that corresponds to experimental observation . In this way, the theory successfully explains the paramagnetic