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

Hint: Ionization enthalpy is higher for elements having stable half-filled and full-filled configuration. As the distance of the electron from the nucleus increases, the force of attraction decreases and thus its ionization enthalpy also decreases.

Complete step by step answer:
-First we will see what ionization enthalpy is.
The energy required to remove an electron from the outermost shell of an isolated gaseous atom is known as ionization enthalpy. Its unit is $KJ.mo{l^{ - 1}}$
                                             $X(g) \to {X^ + }(g) + {e^ - }$

-Ionization enthalpy depends on the force of attraction between the electrons and the nucleus and the force of repulsion between the electrons.
-The elements having more stable configurations like the half filled and full filled configurations face more difficulty in losing the electron and so their ionization enthalpy is higher.
-Now we will see the electronic configurations given in the options and check which has the highest ionization enthalpy.
For (A): $\left[ {Ne} \right]3{s^2}3{p^1}$ is neither half filled or fulfilled configuration.
For (B): $\left[ {Ne} \right]3{s^2}3{p^2}$ is also neither half filled or fulfilled configuration.
For (C): $\left[ {Ne} \right]3{s^2}3{p^3}$ is half filled stable configuration and thus has higher ionization enthalpy.
For (D): $\left[ {Ar} \right]3{d^{10}}4{s^2}4{p^3}$ is also stable half filled configuration and thus has higher ionization enthalpy.

From (C) and (D) we know that (C) has its outermost electron in the 3rd shell while (D) has its outermost electron in the 4th shell. More nearer is the outermost shell, more strongly it is attracted towards the nucleus and more difficult it is to remove that electron. Hence the shell which is nearer to the nucleus will have higher ionization enthalpy.
So, the correct answer is “Option C”.

Note: Ionization enthalpy is basically a property of metals because they have the ability to lose electrons. In a periodic table when we move from left to right ionization enthalpy increases because they move towards a more stable fulfilled configuration. From top to bottom the ionization enthalpy decreases because the shell number increases and distance from the nucleus increases.