Question

The atomic number of the element with highest ionization energy among the following is:
A. $z = 16$
B. $z = 14$
C. $z = 13$
D. $z = 15$

Answer 1

Hint: The value of ionization energy indicates the ease or difficulty with which the outermost electron can be removed from an atom and thus the order can be deduced from the electronic configuration and other periodic trends such as atomic size.

Step by step answer: We have a modern periodic table in which elements have been positioned in the increasing order of atomic number. As the term “periodic” suggests, we have found that there is a periodicity in properties of elements when they are arranged in such a way. This periodicity can help us in deducing various trends across the rows and columns which are called periods and groups respectively.
For example, atomic radius decreases across a period as there is more effective nuclear charge than there is electronic repulsion. Now we can understand the order of ionization enthalpy by relating that to atomic radius but first let’s understand what ionization energy is. We know that elements combine and form bonds by gaining or losing electrons. In both the manners, ions are formed. Ionization energy refers to the minimum required energy to remove the outermost electron from a gaseous atom to give a gaseous cation. Now, as the electrons are held by nuclear charge so usually, the trend is that smaller the size, higher is the ionization energy. So, the element with $z = 16$ shall have the highest ionization energy based on the discussion so far. However, there are some other factors to look at such as higher stability of half-filled and completely filled orbitals. Let’s understand these with the electronic configurations for the given atomic numbers that we can write as follows:
$
z\left( {13} \right) = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}\\
z\left( {14} \right) = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^2}\\
z\left( {15} \right) = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^3}\\
z\left( {16} \right) = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^4}
$
As we can see, electrons are being added up in the p-orbitals with an increase in atomic number. Now, for $z = 15$, we have half-filled p-orbitals which will give it extra stability and cause more difficulty to remove any electron from such a stable configuration. So, it will have the highest ionization energy among all given elements.

Hence, the correct option is D.

Note: We can deduce many trends from the periodic table as general but there are many exceptions to these general trends as we have seen in this case so we have to take into account all the contributing factors.