Question

What is the correct order of second-order ionization energy?

• A. $\mathrm{C}>\mathrm{N}>\mathrm{O}>\mathrm{F}$
• B. $\mathrm{O}>\mathrm{N}>\mathrm{F}>\mathrm{C}$
• C. $\mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C}$
• D. $\mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C}$

Answer 1

The correct option is C

$\mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C}$

The explanation of the correct answer:

• The effective nuclear charge increases as the period progress because valence electrons cannot effectively screen the charge. This results in an increase in ionization potential.
• The second ionization potential of Oxygen is higher than that of Fluorine. As a result of the first ionization, oxygen acquires a stable half-filled configuration.
• Therefore, the Oxygen electron must be removed from $2{\mathrm{p}}^3$ sub-shell for the second ionization. While in the case of Fluorine, it is being expelled from the $2{\mathrm{p}}^4$ subshell.

The explanation of the incorrect answer:

Option (a):

• The ionization energy decreases as the period's order of the periodic table moves from the right to the left direction.

Option (b):

• The 2p sub-shell of Nitrogen is not completely filled so it requires less energy as compared to the Fluorine.

Option (d):

• Fluorine has not a stable electronic configuration so it requires less ionization energy as compared to Oxygen.

Hence, the correct option is (c).