Question

5. Suppose that photoelectron spectroscopy is done on a 2nd-row homonuclear diatomic molecule. Photons with energy 21.21 eV are used, and electrons are ejected with kinetic energies of 11.01 eV, 8.23 eV, and 15.22 eV. (a) Sketch the molecular orbital energy level diagram for the molecule, showing the ionization energies of the three orbitals. Assume that these are the only three filled or partially-filled orbitals and that the molecule is in the ground state. (b) What are the possible identities of the molecule?

Answer 1

1. (a) Molecular Orbital Energy Level diagram for Oxygen Oxygen Molecule with photo election I spectroscopic Characteristics AANLOOA --> I.E = 5.99 ey I Î I V 2p Ibu Internet ---- TT2BC ---- - > I. E = 10.20 ev ---I. E = 12.98 ey ENERGY Iden (,25 as --- T It IS 15

Ionizalian energy of an orbital is the energy required to eject an election from an orbital in the gaseous state . Ionization energy of male and Taly orbitals are same. It is equal to = 21.21 - 15.22 = 5.99 ev V Ionization energy of Tale or Tapy orbital = 21.2. - 11.01 = 10.20 ev :7 Ionizalión energy of Ra orbital > 21.21–6-23 = 12.98 eV [Hint hl= hv + 12 016? (6) Oxygen molecule (Q) contains two Oxygen aloms The subscript a in the formula shows it as a diatomet molecule from two single oxygen aloms. It is a Paramagnetic molecule according to molecular Orbital theory.