Question

How do you find atomic radius given the density?

Answer 1

Atomic and ionic radii are used to denote the distance of an atom/ion from the nucleus. Ionic radii are the distance between the nucleus and valence shell electrons of ions whereas atomic radii are the distance between the nucleus and valence shell electrons of an atom.

Complete step by step answer:
The density of the atom can be calculated by using the formula,
$\rho = \dfrac{{nM}}{{{N_A}V}}$
Where,
$\rho$-density of the atom
n-number of molecules present
M-molar mass
${N_A}$-Avogadro’s number
V-volume
It is known that the volume of one atom can be written as,
${V_{(atom)}} = \dfrac{4}{3}\pi {r^3}......(1)$
Where,
V-volume of one atom
r-radius of one tom
A cube contains more atoms based on its arrangement. Thus, its effective volume can be written as,
${V_{(efficiency)}} = {a^3}$ Where, a-side of the cube
$\Rightarrow {V_{(efficiency)}} = {(2r)^3}$
$\Rightarrow {V_{(efficiency)}} = 8{r^3}......(2)$
Thus, from the density of the atom, the volume of the atom will be obtained and from the volume of the atom, the radius of the atom can be calculated.

Note: It is known that the size of atomic radii is less than ionic radii of negative ions and also, the size of atomic radii is greater than ionic radii of a positive ion. A well-known example is a comparison between the sodium atom and sodium ion. The size of sodium atom (Na) is greater than the size of sodium ion ( $N{a^ + }$ ) because the $N{a^ + }$ ion has lost its one electron and it leads to a decrease in size. Similarly, the size of the chlorine atom (Cl) is smaller than the chlorine ion ( $C{l^ - }$ ) because one electron gets added in $C{l^ - }$ and it leads to an increase in size.