Question

The electronic configuration with the highest ionization enthalpy is:
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

Find the elements with the above corresponding electronic configurations and arrange them in an order according to their atomic numbers. Ionization enthalpy decreases on moving from top to bottom in the periodic table or in a group and Ionization enthalpy increases on moving from left to right in a row of the periodic table. Use this info to find the configuration with the highest ionization enthalpy.

Complete step by step answer:
We are given four electronic configurations and we have to determine the configuration with the highest ionization enthalpy.
First one is $\left[ {Ne} \right]3{s^2}3{p^1}$. The no. of electrons present in Neon is 10. The no. of electrons present in $\left[ {Ne} \right]3{s^2}3{p^1}$ is $10 + 2 + 1 = 13$. Therefore, the atomic number is 13. The element with atomic number 13 is Aluminum.
Second one is $\left[ {Ne} \right]3{s^2}3{p^2}$. The no. of electrons present in $\left[ {Ne} \right]3{s^2}3{p^2}$ is $10 + 2 + 2 = 14$. Therefore, the atomic number is 14. The element with atomic number 14 is Silicon.
Third one is $\left[ {Ne} \right]3{s^2}3{p^3}$. The no. of electrons present in $\left[ {Ne} \right]3{s^2}3{p^3}$ is $10 + 2 + 3 = 15$. Therefore, the atomic number is 15. The element with atomic number 15 is phosphorus.
Fourth one is $\left[ {Ar} \right]3{d^{10}}4{s^2}4{p^3}$. The no. of electrons present in Argon is 18. The no. of electrons present in $\left[ {Ar} \right]3{d^{10}}4{s^2}4{p^3}$ is $18 + 10 + 2 + 3 = 33$. Therefore, the atomic number is 33. The element with atomic number 33 is Arsenic.
So the elements are Aluminum, Silicon, Phosphorus, and Arsenic.
Aluminum, silicon, and phosphorus are arranged consecutively from left to right in the 3rd period. So ionization enthalpy of Phosphorus is the highest.
Phosphorus is present above Arsenic. So Arsenic will have less ionization enthalpy compared to Phosphorus.
Therefore, Phosphorus with configuration $\left[ {Ne} \right]3{s^2}3{p^3}$ has the highest ionization enthalpy.
The correct option is Option C, $\left[ {Ne} \right]3{s^2}3{p^3}$

Hence we can conclude that option C is correct.

Note:
Neon and Argon are zero group elements (inert gases). Their symbols are used in writing the electronic configurations of other elements because all the zero group elements have octet and duplet (stable) configurations. Do not confuse ionization enthalpy with electron gain enthalpy.