Q17 of 18 Page 226

Explain the formation of the following molecules using valence bond theory

a) N2 molecule


b) O2 molecule

(a) Formation of N2 molecule


Nitrogen (7N14) has configuration 1s22s22p3. Its valence shell contains 5 electrons, i.e. 2 in s and 3 in p orbitals - 2px12py12pz1



if the px orbital of a N atom overlaps with other px orbital of another N atom giving (σpx – px) bond, along the internuclear axis. Similarly py and pz orbitals of one N atom overlap with py and pz orbitals of another N atom laterally and perpendicularly to internuclear axis given by (ϖpy – py) (ϖpz – pz) bonds. So, N2 molecule has a triple bond between 2 nitrogen atoms.



(b) Formation of O2 molecule.


oxygen (8O16) has configuration 1s22s22p4. Its valence shell contains 6 electrons, i.e. 2 in s and 4 in p orbitals - 2px22py12pz1



the py orbital of one oxygen atom overlaps with py orbital of another oxygen atom along the inter nucleus axis, a sigma py – py bond is formed.


Similarly, pz orbital of O atom overlaps with pz orbital of another O atom laterally, perpendicular to the inter nuclear axis giving pi pz-pz bond.


O2 molecule has a double bond between 2 oxygen atom.



More from this chapter

All 18 →