Question

What is the radius of an iodine atom? (Atomic No.$53$,Mass No.$126$)
A.$2.5 \times {10^{ - 11}}\,m$
B.$2.5 \times {10^{ - 9}}\,m$
C.$7 \times {10^{ - 9}}\,m$
D.$7 \times {10^{ - 6}}\,m$

Answer 1

A chemical element's atomic radius is a measurement of the size of its atoms, usually the mean or typical distance between the nucleus's centre and the boundary of the surrounding electron shells. There are several non-equivalent definitions of atomic radius since the border is not a well-defined physical entity.
Formula used:
$Radius\,({r_n}) = \,0.529 \times \dfrac{{{n^2}}}{z}\,{A^ \circ }$
$Radius\,({r_n}) = \,0.529 \times {10^{ - 10}} \times \dfrac{{{n^2}}}{z}\,\,\,m$
$n =$Principle Quantum Number
$z =$Atomic number

Complete answer:
Van der Waals radius, ionic radius, metallic radius, and covalent radius are four commonly used definitions of atomic radius. Typically, atomic radius is measured in a bound form because to the difficulties of isolating atoms to measure their radii independently; nevertheless, theoretical calculations are much easier when considering atoms in isolation. The interdependence of the environment, probe, and condition results in a plethora of definitions.
Now let's come to the problem:
Atomic Number $z = 53$
Electronic configuration $= 2,8,18,18,7$
So we can see from the above electronic configuration that the value of $n = 5$
Now,
$Radius\,({r_n}) = \,0.529 \times {10^{ - 10}} \times \dfrac{{{n^2}}}{z}\,\,\,m$
$r = \,0.529 \times {10^{ - 10}} \times \dfrac{{{5^2}}}{{53}}\,\,\,m$
$r = 2.5 \times {10^{ - 11}}\,\,m$
So option (A) is correct.

Note:
Atoms can be modelled as spheres for a variety of purposes. This is simply a rough approximation, but it can provide quantitative explanations and predictions for a variety of phenomena, including liquid and solid density, fluid passage through molecular sieves, atom and ion arrangement in crystals, and molecule size and structure.