Question

The ascending order of the first ionization potential of C, N, O and F is _ _ _ _ _.
(A)- C < F < O < N
(B)- O < C < N < F
(C)- C < O < N < F
(D)- C < N < O < F

Answer 1

Hint: The ionisation potential is dependent on the nuclear charge experienced by the atoms, as more the attraction from the nucleus, more the energy required to remove an electron.

Complete answer:
The given elements are present in Period 2, that is, the valence shell is the same, but the number of electrons increases with the increase in the atomic number as we move from left to right.
But, as we move from along the period, with the increase in the electrons, the effective nuclear charge also increases. So, the electrons experience more attraction from the nucleus, causing the decreases in the atomic radius of the atom.
Then, the removal of an electron from an atom with smaller atomic radius experiencing more nuclear charge will require more energy. Thus, increasing its ionisation potential.
The first ionisation potential is known as the amount of energy required to remove the most loosely bound valence electron from an isolated gaseous atom to form a cation.
So, along the period, with the increase in the nuclear charge, the first ionisation energy of the atoms will also increase in the order, as follows: C < N < O < F
But the stability of the electronic configuration of the atom also contributes to its ionisation potential. So, in nitrogen atom having half- filled 2p orbital, that is, $\left[ He \right]2{{s}^{2}}2{{p}^{3}}$ , making it more stable. Hence, the ionisation potential of nitrogen atoms is greater than the oxygen atom.

Therefore, the order of the first ionization potential of the given atom of Period 2 is option (C)- C < O < N < F.

Note:
The oxygen atom can easily lose its first electron, obtaining a stable half-filled configuration. Thus, having lower first ionisation potential.
But, the second ionisation potential is higher, as now the electron is removed from a stable ion. So, the order becomes C < N < F < O.