Question

Which has maximum ionization potential?

• A. N
• B. O
• C.
• D. Na

Answer 1

The correct option is C

Lets write electronic configurations first.

N = 1$s^2$ 2$s^2$ 2$p^3$ O = 1$s^2$ 2$s^2$ 2$p^4$

$O^{+}$ = 1$s^2$ 2$s^2$ 2$p^3$ Na = 1$s^2$ 2$s^2$ 2$p^6$ 3$s^1$

As, we can see from the electronic configuration, Atomic radius of Na will be highest, so ionization potential will be the least.

Between O and N, We know, Half-filled p-orbitals are more stable. Though atomic radius of O < atomic radius of N, but because of half-filled orbitals, more energy is needed to remove the electron in N.

So, IE, of $N>IE$, of O.But $O^{+}$ also has half filled orbitals and atomic radius < atomic radius (N)

Therefore, $I{E}_1$ of $O^{+}$ >$I{E}_1$ of N
Here, $O^{+}$ has maximum ionization potential.