Question

Which of the following has the smallest size?
A. $Al$
B. $A{l^ + }$
C. $A{l^{2 + }}$
D. $A{l^{3 + }}$

Answer 1

Hint: On moving from left to right in the periodic table, the atomic radii decrease. One of the points to remember is the size of the cation is smaller than the anion. Anions are larger in size than the corresponding neutral atom, similarly, metal atoms on losing their valence electrons form positive ions called a cation and they are smaller in size compared to the corresponding neutral atom.

Complete answer:
Any of the atoms losing or gaining electrons, will not have the same size of the ion as the original atom.
The average distance of the outermost shell from the nucleus is the measure of atomic radius. In cation, the number of electrons is less than the proton’s number which results in the increase of nuclear charge due to which the electrons of the element are strongly attracted by the nucleus, and hence the atomic radius decreases.
In other words, due to the increased nuclear charge, the same number of protons pull in a relatively smaller number of electrons to the nucleus which decreases the atomic radius.
Therefore greater is the charge on the cation, the smaller will be the size.
So from the options $A{l^{3 + }}$ has the greatest charge on it i.e. +3 (the electrons are strongly attracted by the nucleus which increases the nuclear charge) and $A{l^{3 + }}$ has the smallest size.
So the correct answer is option D.

Note:
In a covalent molecule where two atoms are bonded by a single bond, then the distance between these two atoms are called covalent radius but when two atoms are non-covalently bonded the one-half distance between the two nuclei of the atom is called van der Waals radius.