Question

Which one of the following has the lowest ionisation energy?
a) \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}\]
b) \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}\]
c) \[1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}\]
d) \[1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}\]

Answer 1

Hint: To answer this question, we should first know about ionisation energy. In simple terms, we can say that it is the energy required in removing an electron from the outer orbital.

Complete step by step solution:
First of all, we should know about ionisation energy. So, ionisation energy in simple terms can be described as a measure of the difficulty in removing an electron from an atom. Or we can say that it is the tendency of an atom or ion to surrender an electron. The loss of electrons usually happens in the ground state of the chemical species. By ionisation energy, we get to know about the measure of strength or attractive forces by which an electron is held in a place.
Let us discuss some factors that govern ionisation energy:
If the configuration is of the noble gas then it will definitely have high ionisation energy.
We should note that when an electron is near to the nucleus then the attraction will be greater than the one when the electron is far away.
We should note that when there are more electrons between the outer level and the nucleus the attraction forces are less.
So, now we will discuss our options:
>In first option, the configuration is of noble gas neon (\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}\]). It will be very difficult to remove an electron from an outer shell. Neon first ionisation energy is 2080.66 kilojoule per mole.
>The second option has electronic configuration of sodium (\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}\]). It will have low ionisation energy. The first ionisation energy of sodium is 496 kilojoule per mole.
>The third electronic configuration is of fluorine (\[1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}\]). We should know that fluorine needs the only electron to have a stable electronic configuration of neon. So, it has a very high first ionisation energy. Fluorine first ionisation energy is 1681.05 kilojoule per mole.
>The fourth option has electronic configuration of nitrogen (\[1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}\]). It has a half-filled orbital in p. So, it will have high ionisation energy. Nitrogen will have 1400 kilojoule per mole.
So, from the above data, we can say that sodium will have the lowest first ionisation energy among the given electronic configuration.

So, option B is the correct option.

Note: We should know about the general trend. It should be noted that, when we move from left to right in a periodic table, atomic radius decreases, and so electrons are more attracted to the (closer) nucleus. This makes an element have higher ionisation energy.
When, we move from top to bottom down a periodic table. Moving down a group, a valence shell is added. The outermost electrons are further from the positive-charged nucleus, so they are easier to remove. And they will have lower ionisation energy.