Question

What is the correct order of ionisation energy 1) k

Answer 1

The ionization energy of an atom is the amount of energy that is required to remove an electron from a mole of atoms in the gas phase:

M(g)  ®  M+(g)  +  e-

Answer 2

Answer:

1) B < Be < N < O

2) Be < B < N < O

3) B < Be < O < N

4) B < O < Be < N

Logic:

* Ionization energy, in general, increases with decrease in the atomic radius across the period from left to right. However there are exceptions.

* s-orbitals have greater penetration power than p-orbitals. Hence the removal of electrons from s-orbitals require more energy.

* The electrons in half filled orbitals are stable as they experience less repulsion. If there is another electron sharing same orbital, then they will be destabilized due to repulsion from each other.

* Above atoms belong to same period (2nd) of periodic table.

Solution:

* Boron, B is smaller than beryllium, Be atom. Hence we expect increase in ionization energy from Be to B.

However, Be atom has greater ionization energy than B atom. The reason is - in case of Beryllium, the last electron is in the s-orbital and in Boron, the last electron is in the p-orbital. We know that removal of electron from s-orbital requires more energy than the electron from p-orbital.

* 2s22p3 configuration is more stable than 2s22p4 due to half filled p-sublevel. The electrons in the p-orbital in N atom experience less repulsion, thus more stable and more ionization energy. Whereas in case of O atom the 4th p-electron atom can be removed more easily as it experiences more repulsion from the electrons in the p-orbitals already present. Hence nitrogen, Oxygen atom has less ionization energy than Nitrogen atom.

Conclusion:

The correct ionization energy order is: B < Be < O < N

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