Question

CsCl has cubic structure of ions in which ${\text{C}}{{\text{s}}^{\text{ + }}}$ present in the body-centre of the cube. If density is ${\text{3}}{\text{.99}}\,{\text{gc}}{{\text{m}}^{ - 3}}$.
(a). Calculate the length of the edge of a unit cell.
(b). What is the distance between ${\text{C}}{{\text{s}}^{\text{ + }}}$ and ${\text{C}}{{\text{l}}^ - }$ ions?
(c). What is the radius of ${\text{C}}{{\text{s}}^{\text{ + }}}$ ion if the radius of ${\text{C}}{{\text{l}}^ - }$ion is $180$pm?

Answer 1

Hint: To answer this question we should know the density formula of solids. Density of a solid depends upon the number of atoms, mass and length of a unit cell. First we will determine the edge length by using the density formula. Then by using edge length we will determine the distance between ions and atomic radius of caesium ions.

Formula used: \[{{\text{a}}_{{\text{bcc}}}}\,{\text{ = }}\dfrac{{{\text{4r}}}}{{\sqrt {\text{3}} }}\], ${\text{d}}\,{\text{ = }}\dfrac{{{\text{z}}\,{\text{m}}}}{{{{\text{N}}_{\text{a}}}{{\text{a}}^{\text{3}}}}}$

Complete step-by-step answer:
(a)The formula to calculate the density of cubic lattice is as follows:
${\text{d}}\,{\text{ = }}\dfrac{{{\text{z}}\,{\text{m}}}}{{{{\text{N}}_{\text{a}}}{{\text{a}}^{\text{3}}}}}$
Where,
${\text{d}}$is the density.
${\text{z}}$is the number of atoms in a unit cell.
${\text{m}}$is the molar mass of the metal.
${{\text{N}}_{\text{a}}}$is the Avogadro number.
${\text{a}}$ is the length of the unit cell.
The number of molecules in a bcc unit cell is $1$. The molar mass of CsCl is $168.5\,{\text{u}}$.
On substituting $2$ for number of atoms, $168.5\,{\text{u}}$for molar mass of the metal, \[6.02 \times {10^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}\]for Avogadro number, ${\text{3}}{\text{.99}}\,{\text{gc}}{{\text{m}}^{ - 3}}$ for density.
${\text{3}}{\text{.99}}\,{\text{gc}}{{\text{m}}^{ - 3}} = \dfrac{{1 \times 168.5\,{\text{g/mol}}}}{{6.02 \times {{10}^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}\, \times {{\text{a}}^3}}}$
${{\text{a}}^3} = \dfrac{{1 \times 168.5\,{\text{g/mol}}}}{{6.02 \times {{10}^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}\, \times {\text{3}}{\text{.99}}\,{\text{gc}}{{\text{m}}^{ - 3}}}}$
\[{{\text{a}}^3} = \dfrac{{165.8\,}}{{24.02 \times {{10}^{23}}{\text{c}}{{\text{m}}^{ - 3}}}}\]
\[{{\text{a}}^3} = 7.0 \times {10^{ - 23}}{\text{c}}{{\text{m}}^3}\]
\[{\text{a}} = 4.12 \times {10^{ - 8}}{\text{cm}}\]
So, the length of the edge of a unit cell is\[4.12 \times {10^{ - 8}}{\text{cm}}\].

(b)The formula to calculate the atomic radius ions of body-centered cubic unit cell is as follows:
\[{\text{a}}\,{\text{ = }}\dfrac{{{\text{4r}}}}{{\sqrt {\text{3}} }}\]
Where,
${\text{r}}\,$is the atomic radius ions.
${\text{a}}$ is the edge length of the unit cell.
On substituting \[4.12 \times {10^{ - 8}}{\text{cm}}\] for the edge length of the unit cell.
\[4.12 \times {10^{ - 8}}{\text{cm}} = \dfrac{{4 \times {\text{r}}}}{{\sqrt 3 }}\]
\[{\text{r}}\,{\text{ = }}\,\,\dfrac{{4.12 \times {{10}^{ - 8}}{\text{cm}}\, \times \sqrt 3 }}{4}\]
\[{\text{r}}\,{\text{ = }}\,\,\dfrac{{7.13 \times {{10}^{ - 8}}{\text{cm}}}}{4}\]
\[{\text{r}}\,{\text{ = }}\,\,1.78 \times {10^{ - 8}}{\text{cm}}\]
The distance between two bonded atoms is ${\text{2r}}$.
So, on substituting \[1.78 \times {10^{ - 8}}{\text{cm}}\] for r,
\[ = \,\,2\, \times 1.78 \times {10^{ - 8}}{\text{cm}}\]
\[ = \,\,3.568 \times {10^{ - 8}}{\text{cm}}\]
So, the distance between ${\text{C}}{{\text{s}}^{\text{ + }}}$ and ${\text{C}}{{\text{l}}^ - }$ ions is \[3.568 \times {10^{ - 8}}{\text{cm}}\].
Now, we will convert the above distance from cm to pm as follows:
${\text{1}}\,{\text{cm}}\,{\text{ = }}\,{\text{1}}{{\text{0}}^{{\text{10}}}}\,{\text{pm}}$
$3.568 \times {10^{ - 8}}\,{\text{cm}}\,{\text{ = }}\,356.8\,{\text{pm}}$

(c)The distance between ${\text{C}}{{\text{s}}^{\text{ + }}}$ and ${\text{C}}{{\text{l}}^ - }$ion is the sum of atomic radius of ${\text{C}}{{\text{s}}^{\text{ + }}}$ ion and ${\text{C}}{{\text{l}}^ - }$ion. So,
$356.8\,{\text{pm}}\,\, = \,\,{{\text{r}}_{{\text{C}}{{\text{s}}^{\text{ + }}}}}\, + \,{{\text{r}}_{{\text{C}}{{\text{l}}^ - }}}$
On substituting $180$pm for the radius of ${\text{C}}{{\text{l}}^ - }$ion,
$356.8\,{\text{pm}}\,\, = \,\,{{\text{r}}_{{\text{C}}{{\text{s}}^{\text{ + }}}}}\, + \,180\,{\text{pm}}$
$\,{{\text{r}}_{{\text{C}}{{\text{s}}^{\text{ + }}}}}\, = \,\,356.8\,{\text{pm}}\, - 180\,{\text{pm}}$
$\,{{\text{r}}_{{\text{C}}{{\text{s}}^{\text{ + }}}}}\, = \,\,176.8\,{\text{pm}}$
So, the radius of ${\text{C}}{{\text{s}}^{\text{ + }}}$ ion is $176.8\,{\text{pm}}$.

Note: The value of the number of atoms depends upon the type of lattice. For face-centred cubic lattice, the number of atoms is four whereas two for body-centred and one for simple cubic lattice. Here, o${\text{C}}{{\text{s}}^{\text{ + }}}$ present in the body-centre of the cube only so, the contribution of ${\text{C}}{{\text{s}}^{\text{ + }}}$ is one in the unit cell so, we took the number of atoms one. The sum of atomic radius of two ions is known as distance of two bonded atoms.