Question

In the ions ${{P}^{3-}}$, ${{S}^{2-}}$ and $C{{l}^{-}}$ the increasing order of size is:
A. $C{{l}^{-}}<\text{ }{{S}^{2-}}<\text{ }{{P}^{3-}}$
B. ${{P}^{3-}}<\text{ }{{S}^{2-}}<\text{ }C{{l}^{-}}$
C. ${{S}^{2-}}<\text{ }C{{l}^{-}}<\text{ }{{P}^{3-}}$
D. ${{S}^{2-}}<\text{ }{{P}^{3-}}<\text{ }C{{l}^{-}}$

Answer 1

Hint: As we are moving from left to right in the periodic table the size of the elements decreases as the number of protons added to the nucleus increases. As the proton number increases in the nucleus the electrons are more attracted towards the nucleus and the size of the element decreases.

Complete Solution :
- In the question it is given that there are three ions ${{P}^{3-}}$, ${{S}^{2-}}$ , $C{{l}^{-}}$and asking to arrange them as per increasing order of their atomic size.
- We know that phosphorus belongs to V A group, sulphur belongs to VI A group and chlorine belongs to VII A group.
- The atomic size is going to decrease from left to right across the periods in the periodic table.
- So phosphorus atoms have more atomic size than the sulphur and chlorine atoms.
- But in the question it is given that to arrange ${{P}^{3-}}$, ${{S}^{2-}}$ and, $C{{l}^{-}}$ as per increasing order of their atomic size.
- Phosphorus accepts three electrons, three more electrons are added to the outer shell means the added electrons have less effect by the protons and neutrons present in the nucleus. So, the atomic size of the ${{P}^{3-}}$ is very high compared to phosphorus atoms.
- Coming to ${{S}^{2-}}$ , two more electrons are added to the outer shell means the added electrons have less effect by the protons and neutrons present in the nucleus. So, the atomic size of the ${{S}^{2-}}$ is very high compared to a sulphur atom.
- Coming to $C{{l}^{-}}$, one electron is added to the outer shell means the added electron has less effect by the protons and neutrons present in the nucleus. So, the atomic size of the $C{{l}^{-}}$ is very high compared to chlorine atoms.
- Therefore the increasing order of the atomic sizes of the given ions is as follows.
$C{{l}^{-}}<{{S}^{2-}}<{{P}^{3-}}$
So, the correct answer is “Option A”.

Note: If we remove the electrons from the outermost orbit from any atom then the atomic size of the atom decreases due to the enhancement of the attraction of the remaining electrons by the protons which are present in the nucleus.