Ionization Potential on The Basis of Molecular Orbital Theory(MOT)

The ionization potential of N-atom is 14.53 eV whereas this value for N2 molecule is 15.56 eV

The electronic configuration of N-atom is 1s22s22p3 . The 1st ionization potential of N-atom can be expressed as:

On the other hand, the electronic configuration of N2 molecule according to MOT can be given as:

One electron is eliminated from π2Px bonding orbital during the ionization process.

From the above diagram, we can see that π2Px orbital of N2 molecule is a bonding orbital and this orbital is more stable than that of 2p orbital of N-atom. Therefore, it requires more energy to eliminate an electron from π2Px orbital than that of 2p orbital. This is why the first ionization energy of the N2 molecule is greater than that of N-atom.

Q-2: The first ionization potential of the oxygen atom (O) is 13.61 eV whereas this value for the O2 molecule is 12.21 eV. Explain.

Hints:

  1. Find out the orbital ( bonding or anti-bonding) of the last electron of O2 in the MO diagram.
  2. Compare its energy with atomic orbital of O-atom
  3. Electron presents in anti-bonding orbital has higher energy than that of the corresponding atomic orbital. So, to remove an electron from the anti-bonding orbital of a molecule requires less energy when compared with corresponding atomic orbital.

Source:

  1. www.google.com
  2. ESSENTIALS OF PHYSICAL CHEMISTRY BY B.S. BAHL, G.D. TULI, and ARUN BAHL.
January 2023
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