Which has maximum ionization potential?
(A) N
(B) O
(C) ${{O}^{+}}$
(D) Na

Hint: Ionization energy, also called ionization potential is the amount of energy required to remove an electron from an isolated atom or molecule. First ionization energies of the elements First ionization energies of the elements.

Complete step by step solution:
We have been provided with four elements: N, O, Na and ${{O}^{+}}$,
Nitrogen is the chemical element with the symbol N and atomic number 7,
Oxygen is the chemical element with the symbol O and atomic number 8,
Sodium is a chemical element with the symbol Na and atomic number 11.
We need to tell which of the above elements have maximum ionization potential,
So, for that:
We know that ionization energy is the minimum amount of energy required to remove the most loosely bound electron of an isolated neutral gaseous atom or molecule.
Now, we have nitrogen (N) whose atomic number (Z) is 7
Atomic number (Z) of oxygen is:8
Atomic number (Z) of sodium is 11,
But the valency of Na is one, so it would have the lowest ionization potential,
Ionization potential depends on the value of Z,
Therefore, Z of N (7) < Z of ${{O}^{+}}$(8),
The electronic configuration of ${{O}^{+}}$ is $1{{s}^{2}}2{{s}^{2}}2{{p}_{x}}^{2}2{{p}_{y}}^{1}2{{p}_{z}}^{1}$. Clearly, we can see that there are three p orbitals and all of them are half-filled. We already know that half-filled and full filled p orbitals show great stability, which makes the substance resistive to changes.
So, we can say that, ${{O}^{+}}$ has maximum potential,

Therefore, option (C) is correct.

Note: Determine which element from a list has the highest ionization energy, find the elements' placements on the periodic table. Remember that elements near the top of the periodic table and further to the right of the periodic table have higher ionization energies.