Question

Determine the bond orders of (a) S2, (b) Cl2, and (c) NO from their molecular orbital

configurations and compare the values with the bond orders determined from Lewis structures. (NO

has orbitals like those of O2)

Answer 1

The S2 will be going to follow the same bond order as the previous group atom O2

Molecular orbital configuration of O2

O2 = 1s2 1s*2 2s2 2s*2 s2p2 p2p4 p*2p2

Number of electrons in bonding orbitals = 4 + 2 = 6

Number of electrons in anti-bonding orbitals = 2

BO = 1/2 * (electrons in bonding orbitals - electrons in antibonding orbitals) = 1/2 * (6-2) = 2

The Cl2 will be going to follow the same bond order as the previous group atom F2

Molecular orbital configuration of F2

O2 = 1s2 1s*2 2s2 2s*2 s2p2 p2p4 p*2p4

Number of electrons in bonding orbitals = 4 + 2 = 6

Number of electrons in anti-bonding orbitals = 4

BO = 1/2 * (electrons in bonding orbitals - electrons in antibonding orbitals) = 1/2 * (6-4) = 1

Hence bond order of Cl2 will be 1

Molecular orbital configuration of NO

NO = 1s2 1s*2 2s2 2s*2 s2p2 p2p4 p*2p1

BO = 1/2 * (electrons in bonding orbitals - electrons in antibonding orbitals) = 1/2 * (6-1) = 2.5

The lewis structure of NO has a coordinate bond in order to satisfy the electron configuration, ut has two fixed bonds and one coordinate bond from O to N