Question

Which is the correct order of second ionization potential of $C$ , $N$ , $O$ and $F$ in the following?
A.$O > N > F > C$
B.$O > F > N > C$
C.$F > O > N > C$
D.$C > N > O > F$

Answer 1

Hint:The energy required removing an electron from an isolated atom or a molecule is known as the ionization potential or ionization energy. Ionization potential is the measurement of ionization energy in electron volts. Helium has the largest ionization energy.

Complete answer:
We know that ionization potential is the energy measured in electron volts which is required to remove an electron from an isolated or neutral atom or a molecule. Ionization energy is the energy that an electron absorbs to come out of the atom or molecule influence. The ionization energy decreases as we move down a group and increases on moving from left to right across a period. Ionization is caused due the charged particles which are formed by collisions and it is always positive because electrons have negative energy.
Second ionization potential is the energy required to remove the second electron after first ionization potential from the outermost shell of the atom. The second ionization energy is always higher than the first because more energy is required in removing an electron once the first electron has been removed because the positive charge binds the electrons strongly after the first ionization.
In the above given question, the second ionization potential of $C$ , $N$ , $O$ and $F$ is
Carbon: $[He]2{s^2}2{p^1}$ . On removing the first electron, the carbon atom occupies ideal gas configuration and the second electron will be removed easily and thus it will have low ionization potential.
Nitrogen: $\left[ {He} \right]2{s^2}2{p^2}$ . After we remove the first electron, only one electron will be left for removal from the outermost shell of the nitrogen atom.
Oxygen: $\left[ {He} \right]2{s^2}2{p^3}$ . After we remove the first electron, two electrons will be left in the outermost p subshell and this will make it very unstable and oxygen atom will have high second ionization potential.
Fluorine: $\left[ {He} \right]2{s^2}2{p^4}$ . On removal of the first electron, the fluorine atom will be half filled which is a stable configuration.
 Thus, the order of second ionization potential is $O > F > N > C$

So, the correct option is (B).

Note:
Helium has the largest ionization energy which means that its electrons require the most amount of energy to move out of the influence of the helium atom and francium has one of the lowest ionization energies.