Question

How to arrange the following atoms and ions in order of increasing atomic size? Rb, Ag, ${O^{ - 2}}$, Al, O, Cs, $A{l^{ + 3}}$, Si

Answer 1

Hint: The atomic size is defined as the distance from the center of the nucleus of the atom to the outermost shell. In the periodic table the atomic size of the atom decreases on moving left to right and the atomic size of the atom increases on moving top to bottom.

Complete step by step answer:
The atomic size of the atoms decreases as we move left to right in the periodic table as the attractive force of the protons on the same number of sets of electrons increases from each element as we move left to right.
The atomic size increases on moving top to bottom in the periodic table because of the addition of another set of electrons to each element as we move from top to bottom.
The smallest atoms are present at the top right of the periodic table and the larger atoms are present at the lower left of the periodic table.
So, by this we can predict the order as shown below.
$O < Si < Al < Ag < Rb < Cs$
In the given set two ions are present ${O^{ - 2}}$and $A{l^{ + 3}}$.
The atomic number of aluminium is 13 and the electronic configuration of aluminium is $[Ne]3{s^2}2{p^1}$. By removing three electrons from the valence shell the aluminium forms $A{l^{ + 3}}$. The cation form is smaller than the atom. So it will be placed behind the silicon.
The atomic number of oxygen is 8 and the electronic configuration is $[He]2{s^2}2{p^4}$. Two electrons are added to form ${O^{ - 2}}$. The anion size is greater than the atom and due to electron repulsion the ion is placed before aluminium atom.
The resulting order of the increasing atomic size is shown below.
$O < A{l^{3 + }} < Si < Al < {O^{2 - }} < Ag < Rb < Cs$

Note:
Don’t get confused as in the given set atoms and ions both are given. The cations are generally smaller than the atom as they lose electrons and anions are generally larger than the atom as they gain electrons.
The order is like $Anion > atom > cation$.