# Question d2a5c

Mar 5, 2017

WARNING! Long answer! Here's how I would do it.

#### Explanation:

The unit cell of $\text{CsCl}$ (BCC) looks like this:

The $\text{Cs}$ and $\text{Cl}$ ions are touching along the red line connecting diagonally opposite corners of the cube.

Our job is to calculate ${r}_{\textrm{C s}} + {r}_{\textrm{C l}}$, which is half the length of this line.

Step 1. Calculate the mass of the unit cell.

The unit cell contains

$\text{1 Cs atom" + 8 × 1/8 "Cl atom" = "1 Cs atom + 1 Cl atom" = "1 CsCl formula unit}$

"Mass of unit cell" = 1 color(red)(cancel(color(black)("FU"))) × (1 color(red)(cancel(color(black)("mol CsCl"))))/ (6.023 × 10^23 color(red)(cancel(color(black)("FU")))) ×"168.36 g"/(1 color(red)(cancel(color(black)("mol CsCl"))))

= 2.795 × 10^"-22"color(white)(l) "g"

Step 2. Calculate the volume of the unit cell.

Density is an intensive property, so the density of the unit cell is the same as that of bulk $\text{CsCl}$.

V = 2.795 × 10^"-22" color(red)(cancel(color(black)("g"))) × ("1 cm"^3)/(3.97 color(red)(cancel(color(black)("g")))) = color(white)(l)7.042 × 10^"-23"color(white)(l) "cm"^3

Step 3. Calculate the edge length of the unit cell.

The volume of a cubic unit cell with edge length $l$ is given by

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} V = {l}^{3} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

l = root(3)V = root (3)(7.042 × 10^"-23" "cm"^3) = 4.130 × 10^"-8" color(red)(cancel(color(black)("cm"))) × (1 color(red)(cancel(color(black)("m"))))/(100 color(red)(cancel(color(black)("cm")))) × (10^12color(white)(l) "pm")/(1 color(red)(cancel(color(black)("m")))) = "413.0 pm"

Step 4. Calculate the internuclear distance in $\text{CsCl}$.

The length $d$ of the diagonal on a face of the cell is given by

${d}^{2} = {l}^{2} + {l}^{2} = 2 {l}^{2}$

The length $D$ of the diagonal to opposite cornets of the cube is given by

${D}^{2} = {l}^{2} + {d}^{2} = {l}^{2} + 2 {l}^{2} = 3 {l}^{2}$

$D = l \sqrt{3}$

$D = 2 {r}_{\textrm{C s}} + 2 {r}_{\textrm{C l}} = l \sqrt{3}$

r_text(Cs) + r_text(Cl) = (lsqrt3)/2 = ("413.0 pm" ×sqrt3)/2 = "358 pm"#