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 π_2Px bonding orbital during the ionization process.

From the above diagram, we can see that π_2Px 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 π_2Px 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.