Question

Which of the following order or ionization energy is correct?
$
  A)Be > B > C > N > O \\
  B)B < Be < C < O < N \\
  C)B < Be < C < N < O \\
  D)B < Be < N < C < O \\
 $

Answer 1

Hint:We know that ionization energy increases with the decrease in atomic radius across the period from left to right. But there are some exceptions:
Due to higher penetration power of s-orbitals than p-orbitals, electron removal from s-orbital require more energy.
Electrons in half-filled orbitals are stable, they face less repulsion and cannot be easily removed.

Complete step-by-step answer:First let us understand what ionization energy is. Ionization energy is the minimum amount of energy required to remove an electron from an isolated neutral gaseous atom or molecule. In the above options all the atoms belong to the 2nd period in the periodic table.

Elements	$_4Be$	$_5B$	$_6C$	$_7N$	$_8O$
Valence electronic configuration	$2{s^2}$	$2{s^2}2{p^1}$	$2{s^2}2{p^2}$	$2{s^2}2{p^3}$	$2{s^2}2{p^4}$

Ionization energy increases with the decrease in size across the period. All the atoms given in the options belong to the second period. Their size decreases in the order O < N < C < B < Be. Therefore, ionization energy increases in the order Be < B < C < N < O. However, there are some exceptions-
Due to higher penetration power of s-orbitals than p-orbitals, electron removal from s-orbital requires more energy. Therefore, ionization energy of electrons in s-orbital is greater than ionization energy of electrons in p-orbital.
Electrons in half-filled orbitals are stable, they face less repulsion and cannot be easily removed. Therefore, ionization energy of electrons in half-filled configurations is greater.
Comparing Boron and Beryllium, Beryllium is smaller than Boron because moving left to right across the group atomic size decreases. Hence, ionization energy of Boron should be greater than that of Beryllium. However, for Be electrons in s-orbital experience high penetration (electrons are held tightly by the nucleus) and hence require more energy for removal than B electrons.
Therefore, ionization energy of B < ionization energy of Be.
Comparing Nitrogen and Oxygen, p-orbital in Nitrogen is half-filled as it has 3 electrons and p-orbital in Oxygen have 4 electrons. Due to half-filled configuration in Nitrogen, it is more stable than Oxygen. Hence, removing electrons from Nitrogen will require more energy than Oxygen. The order of ionization energy will be – ionization energy of O < ionization energy of N.

Additional information: Penetration is the ability of an electron to get close to the nucleus. It is the ability of the orbital to attract electrons. The penetration effect of s-orbital electrons is maximum due to its closeness to the nucleus than p-orbital.

Therefore, the correct answer is $A)Be > B > C > N > O$.

Note: It is very important to know the exceptions in periodic trends. Although in general ionization energy increases with decrease in atomic size across period as we move from left to right, the exceptions due to penetration effect of electrons in s-orbital and stability of half-filled electronic configuration, ionization energy trend changes.