Question

Which one of the following elements has the highest ionization energy?
A. $\left[ {Ne} \right]3{s^2}3{p^1}$
B. $\left[ {Ne} \right]3{s^2}3{p^2}$
C. $\left[ {Ne} \right]3{s^2}3{p^3}$
D. $\left[ {Ar} \right]3{d^{10}}4{s^2}4{p^3}$

Answer 1

We need to know that the ionization energy is the energy that is needed to scatter an electron from an iota. The farther away the electron is from the core, the simpler it is to ionize. This is the reason, as you go down the occasional table, the ionization energy diminishes; the electron needs less energy to be released on the grounds that the nuclear radii increments and it is farther away from the core.

Complete step by step answer:
As we know that the ionization energy is the energy that is needed to scatter an electron from a molecule. The farther away the electron is from the core, the simpler it is to ionize. This is the reason, as you go down the occasional table, the ionization energy diminishes; the electron needs less energy to be released on the grounds that the nuclear radii increments and it is farther away from the core. Notwithstanding, as you go from left to directly on the periodic table, the ionization energy increments since you are expanding the quantity of electrons in the external shell. These electrons help to "shield" different electrons and make it harder to scatter an electron.
Among the given options, the most stable configuration is for $\left[ {Ne} \right]3{s^2}3{p^3}$ as its outer subshell is half-filled. Generally, stable configurations are seen in half-filled configurations. The stability is obtained when one orbital is completely filled and the other orbital is half-filled. So, it would be harder to remove an electron and the amount of energy required to remove the electron is also more.
So among the elements, the highest ionization energy is $\left[ {Ne} \right]3{s^2}3{p^3}$.

So, the correct answer is Option C.

Note: We must need to know that the ionization energy of an electron increments with the nuclear number of the particle and diminishes for higher energy orbitals. In the event that we take a gander at the occasional table and move from left to directly across the components, the ionization energy increments because of diminishing nuclear span.