Question

Why $C{{a}^{2+}}$ has a smaller ionic radius than ${{K}^{+}}$?

Answer 1

Hint: The atomic number of Ca is 20 and that for K is 19.
- The species of atoms, molecules or ions that possess the same valence electrons are called isoelectronic configuration.

Complete Solution :
So in the question, it is asked why calcium ions possess a smaller ionic radius than the potassium ion.
For solving this question we should know the atomic numbers of the elements of the ions given.
And we should know some idea about the electronic configuration of the ions given here.
- We know that the periodic table is the table in which the elements are placed according to the increasing order of atomic numbers. There are 18 groups and 7 periods in a periodic table.
- As in the question the discussion is about the atomic radii, we will discuss how atomic radii change along periods and groups of the periodic table.
- If we move top to bottom along a group, then we can see that the atomic radii increases from top to bottom of the group since a new shell is introduced in the elements as we move down the group.
- If we move from left to right along a periodic table i.e. if we move along a period, the atomic radii decreases along a period. Since in a period no shell is introduced but instead electrons are filled in the same shell which increases the effective nuclear charge and the electron shielding will remain the same.
So here if we take the case of $C{{a}^{2+}}$and ${{K}^{+}}$ions, we know that the atomic number of elemental calcium is 20 hence it possess 20 electrons and for forming $C{{a}^{2+}}$. Ca loses 2 electrons from the electronic configuration hence there will be only 18 electrons in them.
$Ca\to C{{a}^{2+}}+2{{e}^{-}}$
- For K the atomic number is 19 and hence total electrons are also 18 in number. But to form ${{K}^{+}}$ ion, K loses one electron and hence there will be only 18 electrons left in ${{K}^{+}}$.
$K\to {{K}^{+}}+{{e}^{-}}$
Hence in ${{K}^{+}}$ and $C{{a}^{2+}}$ there are only 18 electrons and hence they are isoelectronic species.
- Now we have to check the positions of these two ions, the ${{K}^{+}}$ is placed in the first group and the and $C{{a}^{2+}}$ in the second group and both are in same period.
- Since $C{{a}^{2+}}$ is placed right to the ${{K}^{+}}$ ion, it is evident that the $C{{a}^{2+}}$ will be having the smaller as it possess more electron to pull inward and decreases the size.

Note: The ionic radii is different from that of ionic radii, it’s often seen that we get confused between both.
- The atomic radii is comparatively larger than the ionic i.e. the cationic or anionic radii.